**
Fill This Form To Receive Instant Help**

Homework answers / question archive / Unit 21 Rotating bodies and angular momentum Introduction Introduction When an ice skater is performing a spin, if she brings her arms in and folds them across her chest, her rate of rotation will increase

Unit 21 Rotating bodies and angular momentum Introduction Introduction When an ice skater is performing a spin, if she brings her arms in and folds them across her chest, her rate of rotation will increase. Why is this? Take another example. Suppose that two cylindrical objects have equal size and mass, but one is hollow and the other is solid. If they are released together from the top of a slope, will they roll down at the same rate? If not, which cylinder will reach the bottom of the slope ?rst? To answer such questions, we need to bring together a number of ideas that have been introduced earlier in the module. In earlier mechanics units, such as Units 3 and 9, we modelled moving objects as particles. However, this approach is inadequate for dealing with the questions above, because we are now concerned with extended bodies, that is, objects that have size, and we are interested in aspects of their motion where that size is important. In Unit 19, you saw that the motion of an extended body can often be modelled by the motion of a representative particle located at the centre of mass of the body. But in the case of the spinning skater, the centre of mass may well be more or less stationary – it is the skater’s rotation about the centre of mass that is of interest. For the rolling cylinders, it is perhaps less obvious that the particle model is inappropriate, but again, rotation about the centre of mass is a crucial part of the motion. In this unit we deal with the motion of extended bodies, and in particular with their rotational motion. In Unit 2, we considered such bodies when stationary: you learned that for a rigid body in equilibrium, the sum of all the external forces must be zero, and that the sum of the external torques must be zero. In Unit 20, you saw that if a non-zero torque is applied to a particle, then this changes the rotational motion of the particle. Now we combine these ideas and consider the motion of an extended body subject to a non-zero torque. An extended body has one or more of length, breadth and depth. A rigid body is an extended body whose shape does not change. In Section 1 we give an overview of the unit and freely apply principles that are not formally established until Section 4. Section 2 begins by reviewing concepts from earlier units, and then goes on to develop a theoretical basis for modelling the motion of extended bodies. Section 3 looks at the rotation of extended bodies about an axis that is ?xed, such as the spinning ice skater. In Section 4 we explore situations where rotational motion is combined with other types of motion. For example, consider a diver in ?ight after leaving a high diving board (see Figure 1). From Unit 19 we would expect the diver’s centre of mass to follow a parabolic trajectory (if we ignore air resistance), but the diver’s rotation about her centre of mass is also a major factor in the success of the dive (as is the change of shape of the body during the dive). Figure 1 Diver with centre of mass following a parabolic path. Note that the diver is an extended body but not a rigid body. 183 Unit 21 Rotating bodies and angular momentum 1 Rotating bodies Rotation is an important aspect of motion in many situations. Understanding such motion involves various mechanical concepts that you have met earlier in the module: these include torque (Unit 2), moment of inertia (Unit 17) and angular momentum (Unit 20). To get an idea of some of the factors involved in analysing rotational motion, consider someone pushing a roundabout in a playground so as to make it move (see Figure 2). Initially the roundabout is stationary, but when it is pushed, it rotates with increasing rotational speed. Even after the person stops pushing, the roundabout will continue to rotate. Figure 2 Playground roundabout being pushed While the roundabout is being pushed, the total force on it is zero: the force supplied by the pusher is balanced by a force exerted by the support at the centre of the roundabout. The roundabout ‘as a whole’ is not going anywhere, that is, its centre of mass is not moving. However, although the two horizontal forces shown in Figure 2 are equal in magnitude and opposite in direction, they have di?erent lines of action. As a result, there is a torque on the roundabout. This torque initiates the rotation of the roundabout and gives it angular momentum. In Unit 20, you saw that for a particle, if there is no torque being applied, then the angular momentum is constant. As you will see later, this result can be generalised to extended bodies. This means that even when the pushing stops, the roundabout will maintain its angular momentum and will continue to rotate; indeed, in the absence of resistive forces, it would go on rotating forever without the need for further pushing (but in practice resistive forces are always present). Later in this unit we will develop quantitative models of rotational motion. But at this stage you should note the following general aspects of rotational motion. You will see later, in Subsection 3.1, that this scalar equation can be obtained from a vector equation by resolving in the direction of the (?xed) axis of rotation. 184 • The motion of an extended body can be treated in two parts: the motion of an equivalent particle located at the centre of mass, and rotation about the centre of mass. This result will be established in Section 4. • A torque applied to an extended body that is initially stationary will initiate rotation and supply angular momentum. • Once an extended body has angular momentum, that momentum will remain constant provided that no further torque is applied to the body. • The component L of the angular momentum along the axis of rotation of an extended body that is rotating about a ?xed axis is the product of the angular speed ω and the moment of inertia I about that axis, that is, L = Iω. This scalar equation is su?cient for the needs of this section. (1) 1 • Rotating bodies An ice skater can vary her speed of rotation during a spin by changing body shape. This is in accord with equation (1) since although the angular momentum L is constant, the moment of inertia is changing. When modelling rigid bodies, the moment of inertia is a constant, but the human body is ?exible and its moment of inertia can change. In the following examples and exercises we will analyse some sporting situations, thus exploring further the relationships between angular momentum, moment of inertia and rotational motion. Example 1 (a) A diver is executing a simple dive in which her body shape remains constant, as illustrated in Figure 3(a). She starts in a handstand position. The subsequent motion can be divided into two phases: ?rst, rotation about the point O while the diver remains in contact with the diving board; second, motion in ?ight after the diver lets go of the board, but before she enters the water. (i) In the ?rst phase, how is the diver’s angular momentum about O changing? (ii) In the second phase, what would you expect to happen to the angular momentum about the diver’s centre of mass? Assume that resistive forces are negligible. (b) Suppose that the diver goes into a tuck position (see Figure 3(b)) in the second phase. What aspect of the motion will be di?erent from that in part (a)? O (a) (b) Figure 3 Diver performing a dive: (a) with body shape constant, (b) taking a tuck position 185 Unit 21 Rotating bodies and angular momentum Solution (a) (i) In the ?rst phase, the diver’s weight provides a torque about O that will act to increase the angular momentum about O. During this phase of the motion, the angular momentum (and angular speed) about O are increasing. (ii) In the second phase, the only force on the diver is her weight, which acts through her centre of mass. This means that there is no torque about the centre of mass, and consequently the angular momentum about the centre of mass will be constant. (b) In the situation in part (a), the diver’s body shape, and hence her moment of inertia, remain constant throughout the dive, so the angular speed of rotation about the diver’s centre of mass does not change after she has let go of the board. However, if the diver adopts a tuck position after letting go of the board, this will reduce her moment of inertia, and will increase her angular speed of rotation about her centre of mass. In Example 2 and Exercise 1, we will make use of the idea that is implicit in Example 1(a)(ii): that angular momentum about the centre of mass is constant for an (e?ectively rigid) extended body in ?ight (assuming that the body is subject only to gravity, that is, that resistive forces are negligible). This result will be established in Section 4. Example 2 O R 1.2 cos π3 π 3 π 6 4 m s−1 2.4 m 3m Figure 4 shows a gymnast rotating anticlockwise around a bar that is 3 m above the ground, at the point of letting go of the bar and dismounting. Assume that the gymnast does not change body shape, so he can be modelled as a rigid rod of length 2.4 m (with his arms extended). Also assume that his centre of mass is at his midpoint, that is, a distance R of 1.2 m from O, while he is contact with the bar. Just before he lets go of the bar, his centre of mass is moving in a circle at a speed of 4 m s−1 . At the moment of release, his body makes an angle of π3 with the vertical. (a) What is the gymnast’s angular speed about his centre of mass just after he releases the bar? Figure 4 Gymnast rotating anticlockwise around a bar 186 (b) If the gymnast is to land successfully, that is, on his feet with his body in a vertical position, through what angle must he rotate while in the air? How long will it take him to rotate through this angle? (c) Consider the vertical movement of an equivalent particle located at the centre of mass of the gymnast. What is the vertical component of the velocity of his centre of mass just after the gymnast releases the bar? Through what distance will his centre of mass fall during the time calculated in part (b)? 1 Rotating bodies (d) Will the gymnast complete a dismount successfully using this approach? If not, what can he do to achieve a successful dismount? Solution (a) Just after releasing the bar, the gymnast’s hands are stationary, and his centre of mass has the same velocity (say u) as immediately before leaving the bar. So, relative to his centre of mass, his hands have velocity −u. (To determine motion relative to the centre of mass, add −u to the velocities throughout the body, as shown in Figure 5, so that the point of view is taken in which the centre of mass is ‘?xed’.) At that moment, the gymnast is rotating about his centre of mass with angular speed |u|/R in the direction in which he was circling the bar before letting go. (Recall from Unit 20 the relationship between speed and angular speed.) Since |u| is 4 m s−1 and R is 1.2 m, the −1 angular speed of rotation is 4/1.2 = 10 3 (in rad s ). (b) At the time of release, the gymnast’s body makes an angle of π3 with the vertical, so the gymnast must rotate through 2π − π3 = 5π 3 in order to achieve a vertical landing. At the angular speed calculated in 10 π part (a), this will take 5π 3 / 3 = 2 (in seconds). R u 2u −u −u −u Figure 5 Velocities of parts of the gymnast (c) At the moment of release, the gymnast’s centre of mass an upward √ has π −1 vertical component of velocity of 4 cos 6 , which is 2 3 m s . Now, taking into account the constant downward √ acceleration of magnitude g, and substituting x0 = 0, v0 = 2 3, a0 = −g and t = π2 into the constant acceleration equation x = x0 + v0 t + 12 a0 t2 , we ?nd that the vertical component of position will increase in π2 seconds by √ $ "2 2 3 π2 − 12 g π2 ) −6.66. The constant acceleration equation was derived in Unit 3. Therefore in the time that it takes for the gymnast to attain a vertical body position for landing, his centre of mass will have fallen through a distance of about 6.66 m. (d) The gymnast cannot complete a dismount successfully in this way. His centre of mass starts at 3 − 1.2 cos π3 , which is 2.4 m above the ground. But as he would have to fall through 6.66 m before his body was vertical, he would hit the ground before he had completed the necessary rotation. If the gymnast were to adjust his body position while in the air, so as to reduce his moment of inertia, he could increase his angular speed of rotation about his centre of mass. This might allow him to complete the necessary rotation before reaching the ground. 187 Unit 21 Rotating bodies and angular momentum Exercise 1 Suppose that the gymnast in Example 2 adjusts his body position while in the air so as to achieve a successful dismount, landing on his feet in a vertical position. When he lands, his body is extended, with his centre of mass at a height of 1.2 m above the ground. (a) How long is the gymnast in ?ight between letting go of the bar and landing? Figure 6 Gymnast in the tuck position (b) Assume that the gymnast is able to adjust his body position instantaneously, so that he is in a tuck position (see Figure 6) for all the time that he is in ?ight. What is the ratio of the moments of inertia of his body in the tuck and extended positions? We end this section by discussing qualitatively, and in some detail, the Highland sport of tossing the caber. Tossing the caber can be divided into ?ve phases (see Figure 7). Note that for the toss to be successful, the caber must land with the end originally held by the competitor pointing away from him. The ?ve phases are as follows. (i) The competitor runs forward with the caber held in a vertical position, and then stops. (ii) The caber rotates forwards about the competitor’s hands, which are stationary. (iii) The competitor pushes upwards on the bottom of the caber, and releases it. (iv) After the competitor releases the caber, it is in ?ight prior to hitting the ground. (v) The end of the caber strikes the ground and remains stationary, but the motion of the rest of the caber continues until the caber falls down ?at. (i) (ii) (iii) Figure 7 Tossing the caber in ?ve phases 188 (iv) (v) 1 Rotating bodies In the ?rst phase, the whole caber gains a uniform horizontal speed. There is no rotation. In the second phase, the bottom end of the caber is stationary. The upper part of the caber retains forward momentum. The result is to turn the forward motion into rotation about the bottom end of the caber. As the caber topples forwards, gravity supplies a torque about the hands, increasing the rate of rotation. In the third phase, as the competitor pushes upwards, the resulting upward force applies a torque about the centre of mass, further increasing the rotation of the caber until the moment of release. (By waiting for the caber to rotate forward in the second phase before applying this force, the competitor increases the distance of the centre of mass from the line of action of the upward force that he applies, so increasing the torque.) In the fourth phase, the centre of mass moves like a projectile. If resistive forces are ignored, the centre of mass follows a parabolic path and the angular speed about it is constant. In the ?nal phase, the end of the caber strikes the ground and becomes stationary (assuming that it does not skid or bounce). The caber will, however, retain some forward rotation about the end that hits the ground. If it lands as illustrated in Figure 7(v), with the upper end yet to reach the vertical, then the weight due to gravity will provide a torque about the end in contact with the ground that slows this rotation. Whether or not the caber will rotate past the vertical, ensuring that the toss is successful, will depend on the angle at which the caber lands, and on how much angular momentum it has from the preceding phase. Typically the caber is tapered, with the thinner end held by the competitor. There are several good reasons for this. As the centre of mass of a tapered caber is nearer to the thicker end of the caber than to the thinner end being held by the competitor, in the third phase the centre of mass is further from the line of action of the force (applied by the competitor) than if the caber were not tapered, so increasing the applied torque about the centre of mass. Moreover, for a tapered caber, the centre of mass has further to fall during the fourth phase. This increases the time of ?ight, and hence increases the angle through which the caber rotates while in ?ight. Finally, suppose that the caber strikes the ground, thicker end ?rst, before it has rotated past the vertical. Then, in the ?fth phase, the centre of mass is closer to the ground than if the caber were symmetric, thereby reducing the torque that is slowing down the caber’s rotation. Consequently the chances of a tapered caber reaching the vertical are greater. A more quantitative mathematical treatment of various aspects of tossing the caber is given in Section 4. 189 Unit 21 Rotating bodies and angular momentum 2 Angular momentum In Unit 20 we obtained the torque law for a particle, L? = Γ, where L is the particle’s angular momentum, and Γ is the torque applied to the particle (each taken about the same point). In this section, you will see that this result can be extended to a system of particles; in that case, L is the total angular momentum of all the particles in the system, and Γ is the total torque exerted by all the external forces on the particles in the system. These results can be applied to extended bodies by modelling them as systems of particles. This section starts by reviewing a number of ideas that you have met earlier: in Subsection 2.1 we review Newton’s third law of motion, and in Subsection 2.2 we review angular momentum and the torque law for a particle. Then in Subsection 2.3, we extend the torque law to a system of particles. 2.1 Newton’s third law of motion revisited In order to extend the results of Unit 20 from particles to extended bodies, we need to look again at Newton’s third law, which deals with the interaction between particles. In the Introduction to Unit 2, Newton’s third law was stated as follows: Law III To every action (i.e. force) by one body on another there is always opposed an equal reaction (i.e. force) – that is, the actions of two bodies on each other are always equal in magnitude and opposite in direction. In symbols this can be expressed by considering two particles, particle 1 and particle 2. Let I12 be the force exerted on particle 1 by particle 2, and let I21 be the force exerted on particle 2 by particle 1. Then the statement of Newton’s third law above can be written as I12 = −I21 . Alternatively, I12 + I21 = 0. Thus the sum of the forces is zero. Recall that these forces between the particles in a system were called internal forces in Unit 19. Exercise 2 considers how pairs of inter-particle forces that satisfy equation (2) might be represented diagrammatically. Exercise 2 Figure 8 shows examples of two forces of equal magnitude acting on particles A and B. For each pair of forces, state whether equation (2) is satis?ed. 190 (2) 2 A B (a) A (b) B A B B (c) Angular momentum A (d) Figure 8 Pairs of forces acting on particles A and B What we did not mention in Unit 2 was that Newton’s third law also states that the pair of equal and opposite forces should have the same line of action (this was because the concept of line of action had yet to be introduced in Unit 2). Try the following exercise to see what this means. Exercise 3 Look back at Figure 8 and ?nd an example of a pair of forces that satisfy equation (2) but where the forces do not have the same line of action. I12 So there are pairs of forces that satisfy equation (2) but where the forces do not act in the same straight line. We therefore need a mathematical condition to test whether forces are acting in the same straight line, and hence whether Newton’s third law is applicable. We will now develop such a condition. Figure 9 illustrates a situation where Newton’s third law applies: the force I12 exerted on particle 1 by particle 2 is equal and opposite to the force I21 exerted on particle 2 by particle 1, with I12 and I21 having the same line of action. Let r1 and r2 be the position vectors of particles 1 and 2, respectively, with respect to some origin O. The vectors I12 and r1 − r2 have the same direction, so their cross product must be 0, that is, (r1 − r2 ) × I12 = 0. particle 1 r1 − r 2 r1 O particle 2 r2 I21 Figure 9 Application of Newton’s third law Recall from Unit 2 that the cross product of parallel vectors is zero. By equation (2) we have I12 = −I21 , so this can be rewritten as (r1 − r2 ) × I12 = r1 × I12 − r2 × I12 = r1 × I12 + r2 × I21 = 0. (3) Let Γ12 = r1 × I12 be the torque about O exerted on particle 1 by the force from particle 2, and similarly let Γ21 = r2 × I21 be the torque about O exerted on particle 2 by the force from particle 1. Then from equation (3) we have Γ12 + Γ21 = 0, that is, the sum of the inter-particle torques is zero when equal and opposite inter-particle forces have the same line of action. In other words, when Newton’s third law applies, the sum of the inter-particle torques is zero. Recall from Unit 2 that a force F whose point of action has position vector r exerts a torque Γ = r × F about the origin O. 191 Unit 21 Rotating bodies and angular momentum Conversely, if the sum of the inter-particle torques is not zero, then the line of action of the forces is not the same (by reversing the above argument). Newton’s third law can therefore be expressed in terms of the properties of the forces and torques between two interacting particles, as follows. Newton’s third law re-stated The force I12 exerted on particle 1 by particle 2 is equal in magnitude but opposite in direction to the force I21 exerted on particle 2 by particle 1, with both forces acting along the line joining the two particles. An equivalent condition in symbols is These equations state that the sum of the inter-particle forces is zero, and the sum of the torques exerted (about the origin) by those forces is also zero. In electromagnetism there is a type of inter-particle force that does not obey Newton’s third law, but this exception need not concern you here. I12 + I21 = 0 and Γ12 + Γ21 = 0, where Γ12 = r1 × I12 and Γ21 = r2 × I21 , with r1 and r2 being the position vectors of particles 1 and 2, respectively, relative to the origin. Almost all inter-particle forces conform to Newton’s third law. The gravitational and electrostatic forces between particles obey this law, as do the forces in model springs or model rods joining two particles. All the systems that you will meet in this module conform to Newton’s third law. 2.2 Torque law for a particle Recall, from Unit 20, the de?nition of angular momentum for a particle, which is as follows: for a particle of mass m that has linear momentum mr? and position vector r relative to an origin O, its angular momentum L about O is L = r × mr?, (4) that is, the angular momentum is the cross product of the particle’s position vector and its linear momentum. Since the position vector is part of the de?nition, the angular momentum is dependent on the choice of origin (which is not the case for linear momentum). So the angular momentum will usually not be the same relative to di?erent choices of origin. Exercise 4 A particle of mass 20 has position vector r = 3 cos(2t)i + 4 sin(2t)j + 5k with origin O (working in SI units), where i, j and k are Cartesian unit vectors. What is the angular momentum of the particle about O at time t = 0? 192 2 Angular momentum Exercise 5 A particle of mass m moves anticlockwise in a circle of radius R at constant speed v. The circle has its centre at the origin O and lies in the (x, y)-plane. Let i, j and k be the Cartesian unit vectors. (a) What is the angular velocity ω of the particle? (b) What is the angular momentum L of the particle about O? This result is also derived in Unit 20. (Hint: Work from the de?nition of angular momentum, and assume that the particle has position R(cos(ωt)i + sin(ωt)j), where ω is the angular speed.) (c) What is the moment of inertia I of the particle about an axis through O in the k-direction? (d) Show that L = Iω. In Unit 20 we showed that if a particle is subject to a torque, then the rate of change of the particle’s angular momentum L about a ?xed point is equal to the applied torque Γ about that point, that is, L? = Γ. The derivation of this important result from Unit 20, known as the torque law for a particle, is repeated below. Using equation (4), we ?nd d d L = (r × mr?) dt dt = (r? × mr?) + (r × mr?) = r × mr?. (5) Since r? and mr? are parallel vectors, r? × mr? = 0. Now, by Newton’s second law, F = mr?, where F is the total force on the particle. Substituting for mr? in equation (5), we have L? = r × F. But the total torque on the particle (relative to the chosen origin) is r × F = Γ. So we obtain the torque law for a particle: L? = Γ. (6) An important special case of this law is when the total external torque is zero. In this case equation (6) becomes L? = 0, so L is a constant vector. This is called the law of conservation of angular momentum for a particle. As an example of a case where angular momentum is conserved, consider any particle moving at constant velocity. By Newton’s ?rst law, the total force acting on the particle must be zero, so the total torque acting on the particle will also be zero. So by the torque law, the angular momentum of the particle about any point will be constant. However, there is an alternative way of understanding this conservation of angular momentum that comes directly from de?nition (4) and proceeds as follows. 193 Unit 21 Rotating bodies and angular momentum path r? r O a Figure 10 Particle moving with constant velocity r? By de?nition, the particle has angular momentum L = r × mr?, where r gives the particle’s position relative to the origin. Since the velocity r? of the particle is constant, the particle will follow a straight-line path, as illustrated in Figure 10. The position vector r is varying, but wherever the particle is on the path, the vector r × r? will have magnitude a|r?|, where a is the perpendicular distance from O to the particle’s path, and direction normal to the plane shown in Figure 10 (and out of the page in the case illustrated). Thus the angular momentum is constant, since both a and |r?| are constant and its direction is constant. Exercise 6 (a) If i, j and k are Cartesian unit vectors and r = xi + yj, show that r × r? = (xy? − y x?)k. (b) Suppose that a particle of mass m is moving in the (x, y)-plane. By expressing x and y in terms of polar coordinates r and θ, show that the angular momentum L of the particle about the origin O is L = mr2 θ?k. This result was also derived in Unit 20. (7) Exercise 6(b) provides an expression for the angular momentum about the origin of any particle moving in the (x, y)-plane. It generalises the result obtained in Exercise 5 for the angular momentum of a particle travelling in a circle at constant speed. In Exercise 5 the particle followed a circle whose centre was at the origin. The next example concerns motion in a circle whose centre is not the origin (indeed, the circle is not in the (x, y)-plane). Example 3 A particle of mass m moves with a constant angular velocity ω = ωk in a circular path of radius R, centred on a ?xed point with Cartesian coordinates (0, 0, h). At t = 0, the particle is at the point (R, 0, h). Let i, j and k be Cartesian unit vectors in the x-, y- and z-directions, respectively, and let O be the origin. (a) Express the particle’s position vector r as a function of t. (b) (i) Calculate the particle’s velocity r? by di?erentiation. (ii) Recall from Unit 20 that a particle executing circular motion with angular velocity ω has velocity r? = ω × r. Verify that this is consistent with your result in part (b)(i). (c) Calculate the angular momentum of the particle about O. (d) Show that the torque about O acting to sustain the motion of the particle is Γ = −mωhr?. 194 2 Angular momentum Solution (a) The angular velocity is in the k-direction, so the particle must be moving parallel to the (x, y)-plane. The plane of motion is z = h. The particle is travelling with constant angular velocity ωk in a circle of radius R, and is at x = R, y = 0 when t = 0. Therefore its position vector is r = R cos(ωt)i + R sin(ωt)j + hk. (b) (i) (8) In Unit 20, you saw that the angular velocity is perpendicular to the plane of motion. See Unit 20. Di?erentiating equation (8) gives the particle’s velocity as r? = −Rω sin(ωt)i + Rω cos(ωt)j. (9) (ii) Using r from equation (8) and ω = ωk, we substitute into r? = ω × r and obtain $ " r? = ωk × R cos(ωt)i + R sin(ωt)j + hk = Rω cos(ωt)j − Rω sin(ωt)i, which is consistent with the result in part (b)(i). (c) By de?nition (equation (4)), the angular momentum of the particle about O is L = r × mr?. On substituting for r and r? from equations (8) and (9), respectively, we obtain $ " L = R cos(ωt)i + R sin(ωt)j + hk $ " × m −Rω sin(ωt)i + Rω cos(ωt)j $ = m R2 ω cos2 (ωt)k + R2 ω sin2 (ωt)k " − hRω sin(ωt)j − hRω cos(ωt)i $ " = mR2 ωk − mhRω cos(ωt)i + sin(ωt)j . (10) (d) From the torque law for a particle, we have Γ = L?. Now, by di?erentiating equation (10) we get $ " L? = −mhRω2 cos(ωt)j − sin(ωt)i . (11) Comparing this with equation (9), we see that L? = −mhωr?, hence Γ = −mhωr?. To sum up: in both Exercise 5 and Example 3 we looked at a particle moving at constant speed in a circle. In Exercise 5 you saw that, relative to the centre of the circle, angular momentum is constant and the total torque acting on the particle is zero. In Example 3, we considered angular momentum and torque relative to a point on the axis of rotation but not at the centre of the circle. Relative to that point, the angular momentum is not constant (although its component in the direction of the angular velocity is constant); consequently, there must be a non-zero total torque on the particle about that point if such motion is to be sustained. Equation (10) shows that the component of L in the k-direction is constant. 195 Unit 21 Rotating bodies and angular momentum Exercise 7 For a particle moving as described in Example 3, calculate the acceleration r?. Use Newton’s second law to ?nd the total force F acting on the particle. Then calculate the total torque Γ on the particle by using Γ = r × F, and show that you obtain the same result as in Example 3(d). 2.3 Torque law for an n-particle system We now move on to consider angular momentum and torque for systems involving more than one particle. First, we look at systems that have two particles, then we go on to examine a general system containing an arbitrary number of particles. Two-particle systems F1 I12 particle 1 r1 O particle 2 r2 I21 F2 Figure 11 Two-particle system Consider a two-particle system, as illustrated in Figure 11, to which Newton’s third law applies. Particle 1, at position r1 relative to a ?xed origin O, may be acted on by any number of external forces, but we will consider only the resultant of all these external forces, F1 . The only other force on particle 1 is the internal force I12 exerted on it by particle 2. The total external force on particle 2, at position r2 , is F2 , and the only other force on particle 2 is the internal force I21 exerted on it by particle 1. Let the angular momentum of particle 1 relative to O be L1 , and that of particle 2 relative to O be L2 . Applying the torque law for a particle (equation (6)) to particle 1 gives L?1 = r1 × (F1 + I12 ) = r1 × F1 + Γ12 , where Γ12 is the torque about O acting on particle 1 by the internal force from particle 2. Similarly, for particle 2 we have L?2 = r2 × (F2 + I21 ) = r2 × F2 + Γ21 , where Γ21 is the torque about O acting on particle 2 by the internal force from particle 1. Adding these two equations, we obtain L?1 + L?2 = r1 × F1 + r2 × F2 + Γ12 + Γ21 . Newton’s third law implies that Γ12 + Γ21 = 0, so this reduces to L?1 + L?2 = r1 × F1 + r2 × F2 . (12) In equation (12), the right-hand side represents the total torque exerted on the two-particle system by the external forces, while the left-hand side gives the rate of change with time of the total angular momentum of the system. Thus equation (12) extends the torque law to a system of two 196 2 Angular momentum particles. The torque law for a two-particle system can then be stated as follows. For a two-particle system, the rate of change of the total angular momentum of the particles in the system is equal to the total torque exerted on the system by external forces (where angular momentum and torque are determined relative to the same ?xed point). Exercise 8 Figure 12 is a schematic representation of a type of children’s playground roundabout. Two children sit on seats at A and B, and the roundabout is set in motion by pushing at C. It rotates horizontally about a ?xed spindle at O. The distances AO, BO and CO are a, b and c, respectively. The mass of the child and seat at A is m1 , and the mass of the child and seat at B is m2 . The positions of the seats are balanced so that m1 a = m2 b. The mass of the rest of the roundabout is negligible. a k m1 A b ω O m2 B c C Figure 12 Two children on a playground roundabout (a) Suppose that the roundabout is rotating anticlockwise at an angular speed ω. Model each ‘child plus seat’ as a particle, and show that the total angular momentum of the roundabout about O is Iω, where I = m1 a2 + m2 b2 and ω = ωk, with k being a unit vector pointing vertically upwards. (b) Suppose that an adult pushes the roundabout so as to apply a force of constant magnitude F at C, in a horizontal direction at right angles to OC. Assuming that resistive forces are negligible, show that I ω? = cF. (Note that the roundabout is rotating anticlockwise, so θ? ≥ 0, thus ω = θ? and ω? = θ?.) (c) If the distance AO is 1 m, BO is 0.75 m and CO is 1.5 m, and m1 is 45 kg, m2 is 60 kg and F is 315 N, then for how long will the adult need to push in order to get the roundabout rotating at 0.5 revolutions per second when the roundabout has started from rest? The next example concerns a two-particle system executing a more complicated motion. 197 Unit 21 Rotating bodies and angular momentum Example 4 x m1 A r1 θ r2 O m2 B Figure 13 Two particles connected by a spring Consider two particles of masses m1 and m2 , connected by a model spring and resting on a smooth horizontal table, as shown in Figure 13. The system is in motion: the two particles are rotating at the same angular velocity about the centre of mass of the system, and simultaneously they are oscillating as the spring extends and contracts. The total external force on the system is zero: the weight of each particle is balanced by the normal reaction from the table, and we assume that no other external forces are acting. The centre of mass is stationary initially. Therefore it remains stationary throughout the motion because the acceleration of the centre of mass is zero (since the total external force is zero). (This was shown in Unit 19.) Take the origin O to be at the centre of mass, and suppose that particle A has polar coordinates (r1 , θ), and particle B has polar coordinates (r2 , θ − π). Let x be the distance between the two particles, and suppose that k is a unit vector pointing vertically upwards. Use the torque law for a two-particle system to show that the quantity x2 θ? is constant. Use this equation to describe the motion qualitatively. Solution The only external forces acting on the system are the weights of the particles and the normal reactions of the table balancing each weight. These normal reactions have the same points of action as the weights; consequently, not only is the total external force on the system zero, but the total external torque on the system is also zero. (The only other force acting on each particle is that from the spring, and this is an internal force.) It follows from the torque law that as the total external torque is zero, the angular momentum L of the system is a constant. From equation (7), the angular momentum of particle A about O is m1 r12 θ?k, and that of particle B is m2 r22 θ?k. So the total angular momentum of the system is L = m1 r12 θ?k + m2 r22 θ?k = (m1 r12 + m2 r22 )θ?k. But L is a constant, as noted above, therefore (m1 r12 + m2 r22 )θ? = c, (13) where c is a constant. Because O is at the centre of mass of the two-particle system, m1 r 1 = m2 r 2 . (14) Now x = r1 + r2 , so we can substitute for r1 in equation (14) to obtain m1 (x − r2 ) = m2 r2 . Then solving for r2 gives m1 x . (15) r2 = m1 + m2 198 2 Substituting this into equation (14) gives r1 in terms of x: m2 x r1 = . m1 + m2 Angular momentum (16) Substituting from equations (15) and (16) into equation (13) yields + ' m21 x2 m22 x2 θ? + m2 c = m1 (m1 + m2 )2 (m1 + m2 )2 m1 m22 + m2 m21 2 = x θ? (m1 + m2 )2 m1 m2 = x2 θ?. m1 + m2 Since the masses m1 and m2 are constant, we can deduce that x2 θ? must be constant. As the system rotates, the spring extends and compresses as the particles oscillate in and out. The rotational motion of the particles is connected to these oscillations since x2 θ? is a constant. When the spring is extended (x is larger) the angular speed |θ?| decreases, and vice versa. Exercise 9 A binary star system consists of two stars of masses m1 and m2 . Model each star as a particle, and assume that all the forces exerted on the system, other than the gravitational attraction between the two stars, can be ignored. Assume also that the distance d between the stars is constant, and that the common centre of mass of the two stars is ?xed. Let the total angular momentum of the binary system have magnitude L. Express the period of rotation of the system in terms of m1 , m2 , d and L. n-particle systems You have seen above that the torque law for a two-particle system has the same form as that for a single particle. This is because the sum of the internal torques between the two particles is zero. The situation is analogous for a system of more than two particles. Therefore we can generalise the torque law to any number of particles. Consider a system of n particles, which we will call particles 1, 2, 3, . . . , n, to which Newton’s third law applies. Particle 1 may be acted on by various external forces, but we will denote the resultant of all the external forces on particle 1 by F1 . In general, for i = 1, 2, . . . , n, Fi denotes the resultant of all the external forces on particle i. As well as being subject to forces external to the system, particle 1 may be acted on by internal forces 199 Unit 21 Rotating bodies and angular momentum exerted by each of the other particles in the system: denote by I12 the force exerted on particle 1 by particle 2, denote by I13 the force exerted on particle 1 by particle 3, and so on. In general, let Iij denote the force exerted on particle i by particle j (for i and j between 1 and n, with i *= j). For i = 1, 2, . . . , n, let the position of particle i be ri (relative to a ?xed origin O), and let the angular momentum of particle i relative to O be Li . The torque law for a particle when applied to particle 1 gives L?1 = r1 × (F1 + I12 + I13 + I14 + · · · + I1n ). (17) If Γij is the torque exerted on particle i by the internal force from particle j, then Γij = ri × Iij , and equation (17) becomes L?1 = r1 × F1 + Γ12 + Γ13 + Γ14 + · · · + Γ1n . (18) Similarly, the torque law for a particle when applied to particle 2 gives L?2 = r2 × (F2 + I21 + I23 + I24 + · · · + I2n ) = r2 × F2 + Γ21 + Γ23 + Γ24 + · · · + Γ2n . (19) For particle i, we get Note that the sum does not contain a term Γii . L?i = ri × (Fi + Ii1 + Ii2 + Ii3 + · · · + Iin ) = ri × Fi + Γi1 + Γi2 + Γi3 + · · · + Γin . (20) Now, to obtain the total angular momentum of the system, we can sum versions of equation (20) for each of i = 1, 2, . . . , n. When we do this, all the internal torques can be ‘paired up’: thus the term Γ12 from equation (18) can be paired with the term Γ21 from equation (19); the term Γ13 from equation (18) can be paired with the term Γ31 from the equivalent equation for particle 3; and generally, the term Γij from the equation for particle i (where i *= j) can be paired with the term Γji from the equation for particle j. Each pair will sum to 0, since from Newton’s third law, Γij + Γji = 0 (for i *= j), so the total of all the internal torques for the n particles comprising the system will be 0. If L is the total angular momentum of the system (where L = L1 + L2 + · · · + Ln ), then from equation (20) (for i = 1, 2, . . . , n) we have L? = r1 × F1 + r2 × F2 + · · · + rn × Fn . (21) Here, r1 × F1 is the torque exerted on particle 1 by the external forces acting on it, r2 × F2 is the torque exerted on particle 2 by the external forces acting on it, and so on. Therefore the total external torque on the system is Γ = r1 × F1 + r2 × F2 + · · · + rn × Fn . So equation (21) can be written as L? = Γ. 200 3 Rigid-body rotation about a ?xed axis Thus we have shown that the rate of change of the total angular momentum of an n-particle system is equal to the total external torque acting on the system. This extends the torque law to an n-particle system, which we now refer to as just the torque law. Torque law Consider a system of n particles. Let ri be the position vector (relative to a ?xed origin O) of particle i, and let Fi be the total external force on particle i, for i = 1, 2, . . . , n. The total external torque on the system about O is Γ= n ) ri × Fi . (22) i=1 The rate of change of the total angular momentum L of the system about O equals the total external torque acting on the system, that is, L? = Γ. (23) In particular, when the total external torque about O is zero, the total angular momentum vector about O is conserved, that is, it is constant. Exercise 10 Suppose that all the external forces on a system of particles are applied at the same ?xed point X. What can you deduce about the angular momentum of the system? 3 Rigid-body rotation about a ?xed axis In this section we model a rigid body as an n-particle system where all the inter-particle distances remain constant. To simplify matters, we will consider rigid bodies that are rotating about a ?xed axis. For such a body, the rotational motion can be expressed in a very convenient way by using the moment of inertia. This is done in Subsection 3.1. Rigid bodies were introduced in Unit 2. Moments of inertia were introduced in Unit 17. 201 Unit 21 Rotating bodies and angular momentum Unit 17 showed how to use multiple integrals to calculate the moments of inertia of complicated shapes. In this unit we model systems using simple geometric shapes, and we will give a table of moments of inertia for these in Subsection 3.1. This table covers only moments of inertia about an axis through the centre of mass of the body. You may want to ?nd moments of inertia about other axes, and this is covered in Subsection 3.2. The kinetic energy of a rotating rigid body can also be expressed in terms of its moment of inertia, and we consider that in Subsection 3.3. 3.1 Angular momentum and moments of inertia In this section we con?ne our attention to motion in which a rigid body is rotating about a ?xed axis. For simplicity, we choose a frame of reference in which the z-axis is the axis of rotation. We choose x-, y-, z-axes that are ?xed in space, and an origin O that is ?xed and on the axis of rotation. Here k is a unit vector in the z-direction. Now, in order to apply the torque law L? = Γ to a rotating rigid body, we need to ?nd an expression for the angular momentum of the body (relative to a point on the axis of rotation). Suppose that the body is composed of particles A, B, C, and so on. The particles that lie on the axis of rotation, like C and D in Figure 14(a), do not move. On the other hand, particles like A and B do move, but their distances from the z-axis remain constant, so they travel in circles centred on a point on the z-axis, and parallel to the (x, y)-plane. The angular velocity of the body (and of each of its constituent particles) is ω = θ?k, where θ is the angle through which the body is rotated about its axis of rotation, as shown in Figure 14(b). z i B z k ω j D d1 A C O θ r1 O y x (a) ω y x (b) Figure 14 Rigid body rotating about the z-axis, showing (a) some individual particles, (b) the relationship with the angular displacement θ 202 3 Rigid-body rotation about a ?xed axis First consider particle A, which has mass m1 and position vector r1 . From Unit 20, particle A has velocity r?1 = ω × r1 . Therefore its angular momentum is L1 = r1 × m1 r?1 = r1 × m1 (ω × r1 ) = r1 × m1 θ?(k × r1 ). (24) If r1 = x1 i + y1 j + z1 k, we have k × r1 = k × (x1 i + y1 j + z1 k) = x1 j − y1 i. Then, on substituting for r1 and k × r1 in equation (24), we obtain L1 = (x1 i + y1 j + z1 k) × m1 θ?(x1 j − y1 i) = m1 θ?(x21 k + y12 k − x1 z1 i − y1 z1 j). (25) Let d1 be the perpendicular distance of particle A from the axis of rotation. Then d21 = x21 + y12 , and equation (25) simpli?es to L1 = m1 θ?(−x1 z1 i − y1 z1 j + d21 k). Suppose that the rigid body consists of n particles, where the ith particle (for i = 1, 2, . . . , n) has mass mi and position vector ri = xi i + yi j + zi k. The total angular momentum L of the body will be the sum of the angular momenta of all these particles, that is, L = θ? n ) mi (−xi zi i − yi zi j + d2i k), (26) i=1 where di is the perpendicular distance of the ith particle from the z-axis. This angular momentum has non-zero components perpendicular to the axis of rotation (in the i- and j-directions). However, in many applications it is only the component in the direction of the axis of rotation that is needed (the situation in Exercise 11 below is an exception). With our choice of axes, this component is in the z-direction, and is Lz = θ? n ) mi d2i . i=1 This is very important. For any particular rigid body, the quantity ,n result 2 = I is a constant, since both m and d are constant for all i. It m d i i i=1 i i depends on the distribution of mass about the axis of rotation, and as you saw in Unit 17, it is called the moment of inertia. Hence for a rigid body spinning about a ?xed axis, the component of the angular momentum in the direction of the axis can be conveniently expressed in terms of its moment of inertia. 203 Unit 21 Rotating bodies and angular momentum Angular momentum of a rigid body rotating about a ?xed axis We have established this result with the z-axis as the axis of rotation. However, for any rigid body rotating about a ?xed axis, we could choose axes such that the z-axis is the axis of rotation, so there is no loss of generality in our argument. Suppose that a rigid body is rotating about a ?xed axis with angular velocity ω. Let L be the angular momentum of the body about a point O on the axis, and let Laxis be the component of L in the direction of the axis. Then Laxis = I θ?, (27) where I is the moment of inertia of the body about the axis of rotation. Note that the magnitude of the component of the angular momentum in the axis of rotation can be written as |Laxis | = Iω, where ω is the angular speed. Exercise 11 Throughout this exercise, torque and angular momentum are measured about the origin O; also, i, j and k are Cartesian unit vectors, with i and j in a horizontal plane and k pointing vertically upwards. z (a) Consider a rigid body modelled as a system of n particles lying on a curve C embedded in a vertical plane as shown in Figure 15. The curve C does not cross the z-axis or extend below the (x, y)-plane, and is moving anticlockwise with constant angular speed ω about the z-axis. ω C O θ x (i) k eθ y er Figure 15 Rigid body modelled as n particles placed on a curve C embedded in a vertical plane rotating about the z-axis Show that the angular momentum of the system can be expressed as L = Iωk − Aωer , (28) where I is the moment of inertia of the body about the z-axis, A > 0 is a constant, and er = cos θ i + sin θ j is the unit vector pointing horizontally from the z-axis within the vertical plane containing the rigid body, as shown in Figure 15. (You met this use of er in Unit 20.) (Hint: Express the x- and y-coordinates of each of the particles in polar coordinates.) (ii) Hence show that L? = −Aω2 eθ , (29) where the unit vector eθ = − sin θ i + cos θ j lies in the (x, y)-plane and is normal to er , pointing in the direction of increasing θ. (You met this use of eθ in Unit 20.) Deduce that the body must be subject to a non-zero torque in the tangential direction but pointing in the opposite direction to eθ . 204 3 Rigid-body rotation about a ?xed axis (b) Consider a skater whose centre of mass is moving with constant angular speed ω in a circle parallel to the (x, y)-plane. (i) By analysing the motion of an equivalent particle located at the skater’s centre of mass, deduce the direction of the total external force F on the skater. (ii) Suppose that the skater tries to follow the circle while remaining perfectly vertical. Use equation (29) to show that this is impossible. What must the skater do if she is to follow such a circle? (Model the skater as a rigid system of particles lying on a curve, as discussed in part (a). Assume that the only forces on the skater are her weight and the force exerted by the ice on her skates.) Suppose that k is a unit vector along the axis of rotation of a rigid body, so ω = θ?k. If we look at just the k-components, then the torque law and equation (27) yield Γ · k = L? · k d = (L · k) dt d = (I θ?) dt = I θ?. So in this situation the torque law leads to the following important and elegant result. Equation of rotational motion Suppose that a rigid body is rotating about a ?xed axis, and that its angular displacement (from some ?xed line normal to the axis) is θ. Then I θ? = Γaxis , (30) where Γaxis is the component of the total external torque on the body in the direction of the axis of rotation. The moments of inertia of a number of common regular geometric shapes are given in Table 1. In each case we assume that the rigid body is continuous and of uniform density, and we consider only objects of this type. The moments of inertia given in the table have been found by integration, as discussed in Unit 17. You may use these moments of inertia when doing the exercises in this unit. The objects are therefore homogeneous rigid bodies, as de?ned in Unit 19. 205 Unit 21 Rotating bodies and angular momentum Table 1 Moments of inertia of homogeneous rigid bodies Object Axis Dimensions Moment of inertia Figure Solid cylinder Axis of cylinder Radius R 1 2 2MR Solid cylinder Normal to Radius R, axis of length h cylinder 1 2 4MR Hollow cylinder Axis of cylinder 1 2 2 M (R Hollow cylinder 16(a), axis AB + 1 2 12 M h 16(a), axis CD + a2 ) 16(b), axis AB Normal to Inner radius a, axis of outer radius R, cylinder length h 1 2 2 4 M (R + a ) 1 2 + 12 M h 16(b), axis CD Solid rectangular cuboid Normal to Faces normal to one pair axis have sides of faces of lengths a, b 1 2 12 M (a 16(c), axis AB Thin straight rod Normal to Length h rod 1 2 12 M h Solid sphere Through centre Radius R 2 2 5MR Hollow sphere Through centre Inner radius a, outer radius R 2 5M Radius R 2 2 3MR Inner radius a, outer radius R Thin spherical Through shell centre + b2 ) R 5 − a5 R 3 − a3 In each case, the mass of the object is M and the axis passes through its centre of mass G (see Figure 16). C A C R G B A b a R G h (a) Figure 16 cuboid D A (b) D B a G h (c) B (a) Solid cylinder, (b) hollow cylinder, (c) solid rectangular The moment of inertia of the ‘thin straight rod’ in Table 1 is calculated using the assumption that all its particles lie on a straight line. While this 206 3 Rigid-body rotation about a ?xed axis will never be absolutely true for a three-dimensional object, we sometimes use such limiting cases as convenient models. Similarly, a ‘thin spherical shell’ is an object whose particles are con?ned to the surface of a sphere. The moment of inertia of a thin spherical shell is, in fact, the limit of that for a hollow sphere as a → R, but that limit is not obvious, so this case is given separately in the table. The next example illustrates the use of Table 1 to ?nd the moment of inertia of a ‘thin disc’, which is another example of a two-dimensional model of a three-dimensional object. Example 5 Consider a uniform solid circular disc of mass M and radius R, and negligible thickness. What is its moment of inertia about an axis normal to the disc and through its centre? Solution The distribution of the mass of the disc relative to the speci?ed axis is the same as for a solid cylinder rotating about its own axis. From Table 1, the moment of inertia in this case is 21 M R2 . Exercise 12 Show that the moment of inertia about the axis of a thin cylindrical shell of mass M and radius R is M R2 . To ?nd the moment of inertia of a compound object formed from several simple shapes joined together, we ?nd the moment of inertia of each component part separately, and then sum these. We use this approach in the next example. Example 6 A playground roundabout can be modelled as a uniform solid circular disc of radius 1.2 m and mass 240 kg. The disc is horizontal and rotates about a vertical axis through its centre, making 0.5 revolutions per second. A man of mass 80 kg is standing stationary with respect to the disc at a position 0.2 m from the centre O of the disc. Suppose that the man moves towards the edge of the disc, stopping when he is 1 m away from the centre. Assume that the external forces on the roundabout are such that the component of the total external torque about O in the direction of the axis of rotation is zero. Assume also that the man can be modelled as a particle. At what rate will the disc be rotating when he is in his new position? 207 Unit 21 Rotating bodies and angular momentum Solution Since Γaxis , the component of the total external torque about O in the direction of the axis of rotation, is zero, from equations (27) and (30) it follows that Laxis , the angular momentum about the axis of rotation through O, is conserved. To proceed, we need to know the moment of inertia of the combined ‘man-plus-roundabout’ system for each of the two positions of the man. Now, the moment of inertia of the roundabout about the axis through O (in kg m2 ) is 2 1 2MR = 1 2 × 240(1.2)2 = 172.8. If we model the man as a particle of mass m located at a distance r from O, then his moment of inertia is mr2 . So when he is 0.2 m from the centre of the disc, the moment of inertia of the combined system is I1 = 172.8 + 80(0.2)2 = 176. With the man 1 m from the centre of the disc, the moment of inertia of the combined system is I2 = 172.8 + 80(1)2 = 252.8. When the man is nearer to the centre of the disc, the angular speed ω1 is 0.5 revolutions per second, that is, ω1 = π (in rad s−1 ). Suppose that the angular speed is ω2 after the man has moved closer to the edge of the disc. Conservation of angular momentum gives I 1 ω1 = I 2 ω2 , hence 176 π ) 2.19. 252.8 Therefore after the man has moved closer to the edge of the roundabout, the roundabout rotates more slowly, at about 2.19 rad s−1 , that is, at about 0.35 revolutions per second. ω2 = X O The following example and exercise consolidate ideas presented in this subsection. Example 7 h T W Figure 17 Bucket attached to a wheel centred at O, drawing water from a well 208 A bucket used to draw water from a well has mass m and is attached to a light inextensible rope that is wound round a heavy wheel (see Figure 17). The wheel is a uniform solid disc, with centre O, radius R and mass M . The distance from O to the surface of the water in the well is h. Suppose that the bucket is released from rest at a point X, level with O, and allowed to fall down the well, causing the wheel to rotate. Model the bucket as a particle and the rope as a model string, and assume that the force supporting the wheel acts at the point O and that there is no friction there. 3 Rigid-body rotation about a ?xed axis (a) If T denotes the tension in the rope, and z denotes the distance that the bucket has travelled down the well in time t, write down the equation of motion for the bucket in terms of m, |T| and z. (b) What is the torque acting on the wheel about its centre? (c) Suppose that the wheel has rotated through an angle θ while the bucket has been falling. What is the relationship between z and θ? (d) Write down the equation of rotational motion for the wheel. (e) Find an expression for the acceleration z? of the bucket by eliminating |T| and θ from the equations of motion obtained in parts (a) and (d), and by using the result from part (c). (f) Hence ?nd the time taken for the bucket to descend the well. Solution (a) Choose Cartesian unit vectors with k pointing vertically downwards, j in the direction of the axis of rotation of the wheel, and i horizontal, as shown in Figure 18. The origin O is at the centre of the wheel. |T|k Resolving in the k-direction gives the equation of motion: (31) (b) The force exerted on the wheel by the rope is |T|k, acting at the point X whose position relative to O is −Ri (see Figure 18). Therefore the torque acting on the wheel about its centre is (−Ri) × |T|k = R|T|j. (d) The wheel is a disc turning about an axis through its centre and normal to its plane. It has moment of inertia I = 12 M R2 (from Example 5) about the axis of rotation. Then from equation (30), the equation of rotational motion of the wheel (in the j-direction) is z j k i Figure 18 Force on the wheel (c) If the bucket has fallen a distance z, then the quantity of rope that has unwound from the wheel as the bucket falls must also be of length z. As the bucket falls, the wheel turns through an angle θ anticlockwise, so the part of the perimeter of the wheel that has moved past the point X in Figure 19 has length Rθ. Hence z = Rθ. z R X z θ O x y z Figure 19 Rotation of the wheel I θ? = Γaxis , and from part (b), Γaxis = R|T|. So we obtain 2 1 2 M R θ? x y mz?k = mgk − |T|k. mz? = mg − |T|. O R X The forces acting on the bucket are the tension force in the string, −|T|k, and the force due to gravity, W = mgk. So by Newton’s second law, = R|T|, or equivalently, M Rθ? = 2|T|. (32) 209 Unit 21 Rotating bodies and angular momentum (e) From part (c) we have z = Rθ, and since R is constant, z? = Rθ?. On substituting in equation (32), we ?nd M z? = 2|T|. (33) Now from equation (31), |T| = mg − mz?. So substituting for |T| in equation (33) yields M z? = 2(mg − mz?). Hence 2mg . (34) M + 2m (f) The expression for z? in equation (34) is a constant, so using the constant acceleration formula z = z0 + v0 t + 21 a0 t2 with z = h, z0 = 0, v0 = 0 and a0 given by the right-hand side of equation (34), we have mg t2 . h= M + 2m Rearranging this % gives the time for the bucket to reach the surface of the water as h(M + 2m)/mg. z? = Exercise 13 z y F 1.2 m O R 1.5 m 0.1 m X x Figure 20 Roundabout rotating about O and pushed by force F with resistive force R A roundabout of mass 250 kg is modelled as a uniform solid disc of radius 1.2 m, which turns in a horizontal plane about a vertical axis through its centre O. It is being pushed with a force F, using a handle at X as in Figure 20. This force, which has constant magnitude 100 N, is applied at a distance of 1.5 m from O, and is horizontal and normal to OX. A force R resists the rotational motion. It has magnitude cω, where c is a constant and ω is the angular speed of the roundabout; its point of action is 0.1 m from O, and its direction is opposite to the velocity of that point on the roundabout. Assume throughout that θ? ≥ 0 (anticlockwise rotation), so ω = θ?. (a) Obtain a di?erential equation in terms of ω for the rotational motion of the roundabout. (b) If the pushing starts at time t = 0 with the roundabout at rest, ?nd ω as a function of t. (c) If c = 10, what is the maximum possible angular speed that the roundabout could reach according to this model? Do you think that this could be achieved in practice? 3.2 Parallel axes theorem The moments of inertia of various objects about their centres of mass were given in Table 1. To ?nd the moments of inertia about some other axis, we can use the parallel axes theorem. 210 3 Rigid-body rotation about a ?xed axis Parallel axes theorem Suppose that IAB is the moment of inertia of a rigid body of mass M about a line AB (see Figure 21(a)). Let EF be a line through the centre of mass of the body that is parallel to AB, let the distance between the lines AB and EF be d, and let IEF be the moment of inertia of the body about EF . Then IAB = IEF + M d2 . (35) To see this, we follow the argument set out below. Figure 21(a) shows a rigid body of mass M , with the centre of mass O taken as the origin. Suppose that we want to ?nd the moment of inertia of the body about the line AB. As shown in Figure 21(b), choose as the z-axis a line EF through O that is parallel to AB. Let the perpendicular distance between these two parallel lines be d. Choose as the x-axis a line through O that intersects the line AB and is normal to AB. E z A E A y Pi Zi O O Qi x d d (a) F B (b) F B Figure 21 Rigid body, showing (a) two axes of rotation, EF and AB, and (b) the coordinate axes and a particular particle Pi Suppose that the rigid body is composed of n particles Pi , where for i = 1, 2, . . . , n the ith particle has mass mi and position (xi , yi , zi ). The distance of particle Pi from the z-axis is di (Pi Zi in Figures 21(b) and 22), where d2i = x2i + yi2 . Also, the particle’s distance from the line AB is si (Pi Qi in Figures 21(b) and 22), where s2i = (d − xi )2 + yi2 = d2 − 2xi d + x2i + yi2 = d2 − 2xi d + d2i . Pi di Zi yi xi si d − xi Qi Figure 22 Part of a section through the rigid body in Figure 21(b), parallel to the (x, y)-plane 211 Unit 21 Rotating bodies and angular momentum By de?nition (see Unit 17), the moment of inertia of the rigid body about the line AB is n ) IAB = mi s2i i=1 = n ) mi (d2 − 2xi d + d2i ) i=1 = d2 ,n n ) mi − 2d n ) mi x i + i=1 i=1 n ) mi d2i . (36) i=1 ,n 2 i=1 mi di Now i=1 mi = M and = IEF , where IEF is the moment of inertia of the object about the z-axis (the line EF ). Also, from Unit 19, $,n " Subsection 2.1, i=1 mi xi /M is the x-coordinate of the centre of mass of the body. Since the origin was chosen at the centre of mass, we have ,n m x = 0. Hence equation (36) becomes i i i=1 IAB = IEF + M d2 , giving the parallel axes theorem. Example 8 E A 0.08 m 0.66 m 0.09 m F B Figure 23 Skater with outstretched arm An ice skater is rotating about a ?xed axis AB at an angular speed of 8 rad s−1 . One of the skater’s arms is modelled as a uniform solid cylinder of mass 5.5 kg, length 0.66 m and diameter 0.08 m. The cylinder is normal to the axis of rotation AB, and the end of the cylinder is 0.09 m from the axis (see Figure 23). Find the magnitude of the component of the angular momentum in the direction of the axis of rotation for this model of the arm. Solution From Table 1, the moment of inertia of the cylinder about an axis EF through its centre of mass and normal to the cylinder is IEF = 14 M R2 + 2 1 12 M h = 1 4 × 5.5(0.04)2 + 1 12 × 5.5(0.66)2 ) 0.202 (in kg m2 ). The axis of rotation AB is 0.33 + 0.09 = 0.42 (in m) from the centre of mass of the cylinder, so from the parallel axes theorem, the moment of inertia of the cylinder about the axis of rotation AB is IAB ) 0.202 + 5.5(0.42)2 ) 1.172 (in kg m2 ). As the skater’s angular speed is ω = 8 (in rad s−1 ), it follows from equation (27) that for this model of the arm, the component of angular momentum in the direction of the axis of rotation has magnitude |Laxis | = IAB ω ) 1.172 × 8 = 9.376 (in kg m2 s−1 ). 212 3 Rigid-body rotation about a ?xed axis Note that we could treat other parts of the body in a similar fashion and, in principle, sum up the moments of inertia of all the individual body parts to obtain the moment of inertia of the whole body. Exercise 14 The mace used by the leader of a troupe of drum majorettes can be modelled as a sphere of radius r attached to the end of a uniform cylindrical rod of length d and radius R. The mass of the sphere is M , and the mass of the rod is m. Find the moment of inertia of the mace about an axis AB through the end of the rod and normal to it (as shown in Figure 24). A R d m M r B Figure 24 Mace rotating about AB 3.3 Kinetic energy of a rotating rigid body You saw in Unit 17 how the kinetic energy of a rotating particle can be expressed in terms of its moment of inertia. Recall that the kinetic energy of a particle with mass m moving at speed v is given by T = 21 mv 2 . If the particle is moving in a circle of radius d (the distance from the axis of rotation) with angular speed ω, then v = dω, so we have T = 21 md2 ω2 = 12 Iω2 , where I = md2 is the moment of inertia of the particle. The kinetic energy of a rotating rigid body can be expressed in a similar way. Consider a rigid body rotating about a ?xed axis with angular speed ω. The kinetic energy of the body is the sum of the kinetic energies of all its constituent particles. If the ith particle is a distance di from the axis of rotation, then its speed is di ω and its kinetic energy is 21 mi (di ω)2 . Hence the total kinetic energy of the body is * n # n ) ) 2 2 1 1 mi di ω2 = 12 Iω2 , 2 mi (di ω) = 2 i=1 i=1 where I is the body’s moment of inertia about the axis of rotation. Kinetic energy of a rigid body rotating about a ?xed axis Suppose that a rigid body is rotating with angular speed ω about a ?xed axis. Let I be the moment of inertia of the body about the axis of rotation. Then the kinetic energy of the body is T = 12 Iω2 . (37) Note that as long as rotation is about a ?xed axis, the expression 21 Iω2 gives the total kinetic energy of the body. 213 Unit 21 Rotating bodies and angular momentum Example 9 A planet is modelled as a uniform solid sphere of radius 6400 km and mass 6.0 × 1024 kg. It is turning on its axis once every 24 hours. If the axis of rotation is ?xed, what is the kinetic energy of the planet? Solution The planet’s kinetic energy is 12 Iω2 , where ω= 2π 24 × 602 (in rad s−1 ), and from Table 1, I = 52 M R2 = 2 5 × 6 × 1024 × (6.4 × 106 )2 (in kg m2 ). Therefore the kinetic energy of the planet (in joules) is ' +2 2π 24 6 2 1 2 ) 2.6 × 1029 . 2 × 5 × 6 × 10 × (6.4 × 10 ) 24 × 602 0.1 m Exercise 15 A drum majorette’s baton is modelled as a uniform cylindrical rod with spheres of equal mass at either end. The rod has mass 0.1 kg, length 0.8 m and diameter 0.04 m. Each sphere has diameter 0.1 m and mass 0.25 kg. The baton is rotated at 1 revolution per second about a ?xed axis that is normal to the axis of the cylinder and through the centre of mass (see Figure 25). 0.8 m 0.04 m 0.1 m Figure 25 Drum majorette’s baton rotating about an axis through its centre of mass Determine the kinetic energy of the baton. Exercise 16 (a) In the situation considered in Example 6, determine the total kinetic energy of the system (man plus roundabout) when the man is stationary at 0.2 m from the centre of the roundabout, and then when he is stationary at 1 m from the centre. (Use the results found in the solution to Example 6, as needed.) (b) In the reverse of the situation in Example 6, the man starts 1 m from the centre of the roundabout and moves inwards until he is 0.2 m from the centre. What happens to the kinetic energy of the system when he does this? In a system such as that dealt with in Example 6 and Exercise 16, the angular momentum is constant because the total external torque is zero (both quantities are measured about the centre of the roundabout and refer to components in the direction of the axis of rotation). However, the internal forces can change the kinetic energy of the system. This means that the kinetic energy is not necessarily constant, as you saw 214 3 Rigid-body rotation about a ?xed axis in Exercise 16(b), where the e?ort put in by the man in moving across the roundabout increased the total kinetic energy of that system. Exercise 17 A rigid body of mass M is rotating about a ?xed axis with angular speed ω. Suppose that the centre of mass is following a circle of radius R at speed v, and that IG is the moment of inertia of the body about an axis parallel to the axis of rotation and through the centre of mass. Show that the kinetic energy of the body is 2 1 2 IG ω + 12 M v 2 . Exercise 18 A diver begins a dive at rest and vertical in the handstand position. She starts to rotate from this position with negligible angular speed. She lets go of the diving board when her body makes an angle α with the vertical (where 0 < α ≤ π2 ), and she then starts the ‘in-?ight’ part of the dive. Let ω be the angular speed of the diver at the moment of letting go of the board. Assume that while she is in contact with the board at O, the diver is a rigid body able to rotate about an axis through O, normal to the plane of Figure 26. Use the principle of conservation of mechanical energy to obtain ω in terms of α and the following parameters: l, the distance from the diver’s hands (in the handstand position) to her centre of mass; m, the mass of the diver; IG , the moment of inertia of the diver about an axis through her centre of mass when her body is straight. start release α l l O Figure 26 Diver beginning a dive in the handstand position Exercise 19 y Consider the diver from Exercise 18 when she is part way through her dive (as shown in Figure 27). Suppose that she has rotated through an angle θ (θ < α) in time t, and in that position the force exerted on her hands by the board has component vectors R1 i and R2 j, as shown in the ?gure. k (a) With the origin and axes shown in Figure 27, let xi + yj be the position of the diver’s centre of mass. Express x and y in terms of θ. By di?erentiating, show that x? = lθ?2 sin θ − lθ? cos θ, y? = −lθ?2 cos θ − lθ? sin θ. θ (38) (b) Let IO be the diver’s moment of inertia about an axis through O. Give the equation of rotational motion for the diver in terms of θ, IO , m and l. (c) Obtain another equation for the rotational motion by applying the principle of conservation of mechanical energy to the diver. Verify that di?erentiation of this equation leads to the equation of rotational motion that you found in part (b). j −mgj l R2 j R1 i O i x Figure 27 Diver part way through her dive 215 Unit 21 Rotating bodies and angular momentum (d) (i) Write down Newton’s second law for the motion of the diver’s centre of mass, and obtain two equations of motion. (ii) Use equations (38) to substitute for x? and y? in these equations of motion. (e) (i) Use the results of parts (b) and (c) to substitute for θ?2 and θ? in your equation for R1 in part (d)(ii). Hence show that R1 = mg ml2 sin θ (2 − 3 cos θ). IO (39) (ii) Deduce that for angles θ between 0 and π/2, the horizontal component R1 of the force exerted by the diving board is zero when θ = 0 or when θ = arccos 23 . (f) Estimate IO by modelling the diver as a uniform thin straight rod of length 2l. Then estimate, as a multiple of the magnitude of her weight, the magnitude of the horizontal force that the diver would need to exert on the board to be able to remain in contact with it until θ = π2 . (g) At what point is the diver likely to lose contact with the board? Assume that there is nothing on the diving board on which the diver can grip, so she cannot pull on the board but can only push on it. 4 Rotation about a moving axis We will not consider the motion of a rigid body when the direction of the axis of rotation varies. The motion of an extended body may be much more complicated than that considered in Section 3, where we con?ned our attention to a rigid body rotating about a ?xed axis. In this section we look at a more general situation where the axis of rotation is moving. The general case is considered in Subsection 4.1, and here we derive the result used in Section 1 that the motion can be decomposed into the motion of the centre of mass together with the rotation about the centre of mass. In Subsection 4.2, we consider the motion of a rigid body when the body is rotating about an axis whose direction is ?xed, though the centre of mass may be moving. In Subsection 4.3, we look at the motion of rolling objects, such as cans rolling down slopes. 4.1 Torque law relative to the centre of mass That is not to suggest that the centre of mass will always travel in a straight line! 216 In Unit 19 we showed how Newton’s second law of motion can be extended to an n-particle system. You saw there that the centre of mass of the system moves as if it is a single particle with the same mass as the whole system, and with all the external forces applied to this particle. The motion of the centre of mass is often referred to as the linear motion of the system, to distinguish it from the rotational motion of the system. It is convenient to analyse complicated motion in two parts: the linear motion, and the motion relative to the centre of mass. Now, to apply the 4 Rotation about a moving axis torque law as stated in Section 2, we need to work relative to a ?xed origin. However, as you will now see, we can still use the torque law when these quantities are calculated relative to a point that is moving, as long as that point is at the centre of mass. To demonstrate this, we need some notation. Throughout this subsection, we will consider an extended body modelled as a system of n particles, with total mass M and centre of mass at a position R relative to some ?xed point O. The ith particle of the system (for i = 1, 2, . . . , n) has mass mi and position relative to O given by ri = R + rrel i , (40) where rrel i denotes the position of the particle relative to the centre of mass. In the following discussion it is also useful to obtain a relationship between the velocities of the particles relative to the di?erent origins by di?erentiating equation (40) to give r?i = R? + r?rel i . (41) There is an additional equation involving rrel i that results from the fact that R is the centre of mass of the system. From Unit 19, we have n n ) ) $ " MR = mi ri = mi R + rrel i i=1 i=1 = n ) mi R + i=1 = MR + n ) n ) This is where we use the fact that R is the centre of mass of the system. mi rrel i i=1 mi rrel i . i=1 Thus, as one would expect, n ) mi rrel i = 0. (42) i=1 This takes us to the following important de?nitions. This equation states that if position vectors are taken relative to the centre of mass, then the centre of mass is at the origin. Torque and angular momentum relative to the centre of mass If rrel i denotes the position of the ith particle relative to the centre of mass, and Fi denotes the total external force on the ith particle, then the total external torque on the system relative to the centre of mass is de?ned by n ) rel Γ = rrel (43) i × Fi , i=1 and the total angular momentum relative to the centre of mass is de?ned as n ) rel rel (44) L = rrel i × mi r?i . i=1 217 Unit 21 Rotating bodies and angular momentum We aim to relate Γrel and Lrel to the corresponding quantities calculated relative to the ?xed origin O, and start by looking at the total angular momentum relative to O, which is de?ned as L= n ) ri × mi r?i . i=1 Substituting for ri and r?i , using equations (40) and (41), gives L= n ) rel (R + rrel i ) × mi (R? + r?i ). i=1 Expanding the brackets gives L= n ) " $ rel rel rel . R × mi R? + R × mi r?rel i + ri × mi R? + ri × mi r?i i=1 The last term can be recognised as the total angular momentum relative to the centre of mass (equation (44)), so we have n ) " $ rel rel L= R × mi R? + R × mi r?rel i + ri × mi R? + L . i=1 As R is independent of i, this expression can be written as # # * n # * n * n ) ) ) mi rrel × R? + Lrel . mi r?rel + mi R? + R × L=R× i i i=1 i=1 i=1 The ?rst bracketed term above is the total mass M of the system. The third bracketed term is zero by equation (42), and the second bracketed term is zero by di?erentiating equation (42). So we arrive at the result L = R × M R? + Lrel . The next exercise asks you to derive a similar relationship that holds between the torque relative to the centre of mass and the torque relative to O. Exercise 20 Starting from the de?nition of Γ, the total external torque about O, and using equations (40) and (43), show that Γ = R × F + Γrel , , where F = ni=1 Fi is the total external force on the system. The two results derived above are worth re-stating formally. 218 4 Rotation about a moving axis Decomposition theorems Let R be the position vector of the centre of mass of a body relative to some ?xed point O, let M be the total mass of the body, and let F be the total external force on the body. If the total angular momentum of the body relative to the centre of mass is Lrel , then the total angular momentum of the body relative to O is given by L = R × M R? + Lrel . (45) If Γrel is the total external torque on the body relative to the centre of mass, then the total external torque on the body about O is given by Γ = R × F + Γrel . (46) This states that both the total angular momentum and the total external torque of an n-particle system can be decomposed into the corresponding quantity for an equivalent particle located at the centre of mass plus the corresponding quantity for the rotational motion relative to the centre of mass. Now we move on to derive the central result of this subsection, and one of the key results of the unit, which is the torque law relative to the centre of mass. To derive this result, we start by di?erentiating equation (45) to obtain " d$ L? = R × M R? + L?rel . dt Using the torque law relative to a ?xed origin O gives L? = Γ, so " d$ R × M R? + L?rel . Γ= dt Using the product rule for di?erentiating the cross product gives See Unit 20. rel Γ = R? × M R? + R × M R? + L? . The ?rst term on the right-hand side is zero, since R? × R? = 0, so Γ − R × M R? = L?rel . By Newton’s second law for n-particle systems, we have F = M R?, hence Γ − R × F = L?rel . The result F = M R? was established in Unit 19, Subsection 2.1. Now we use equation (46) to obtain the desired relationship between the torque and angular momentum relative to the centre of mass: Γrel = L?rel . Torque law relative to the centre of mass The total external torque on an extended body relative to its centre of mass is equal to the rate of change of the total angular momentum relative to the centre of mass. So if Γrel is the total external torque on the body relative to the centre of mass, and Lrel is the total angular momentum of the body relative to the centre of mass, as de?ned in equations (43) and (44), then Γrel = L?rel . Remember that we are modelling the extended body as a system of n particles. (47) 219 Unit 21 Rotating bodies and angular momentum This result shows that we can extend the torque law of Section 2 by considering torques and angular momentum relative to the centre of mass. Therefore to model rotational motion, we can work in a frame of reference where the centre of mass is taken as the origin and thought of as ?xed for the purpose of treating the relative motion. We can then deal with motion of the centre of mass separately; this can be done by applying Newton’s second law to an equivalent particle located at the centre of mass, that is, F = M R?. We can use the torque law relative to the centre of mass to justify an assumption that we made in Section 1, that the angular momentum relative to the centre of mass is conserved for projectiles in ?ight. Conservation of angular momentum: special case Suppose that each particle in a system of n particles is subject to an external force of the form cmi k, where c is a constant, mi is the mass of the ith particle, and k is a ?xed vector, and that there are no other external forces on the system. Then Γrel = 0, so Lrel is constant. The next exercise asks you to establish this result. Exercise 21 (a) Use equation (43) and the torque law relative to the centre of mass to establish the boxed result above. (b) Show that the boxed result applies to a body (such as a diver or gymnast) in ?ight and subject only to gravity. Exercise 22 Suppose that all the external forces acting on a system of particles are directed towards the origin. Show that Γrel = −R × F. Recall that |r?i |2 = r?i · r?i . We end this subsection with another decomposition theorem, this time for kinetic energy. This theorem is valuable if we wish to tackle a problem about rotational motion by using conservation of mechanical energy (rather than by using equations of motion). In terms of the vectors de?ned at the start of this subsection, the square of the speed of the ith particle is r?i · r?i , so the total kinetic energy of the system is T = n ) i=1 1 2 mi r?i · r?i . Using equation (41) gives T = n ) i=1 220 1 2 mi $ " $ " R? + r?rel · R? + r?rel . i i 4 Expanding the brackets gives n ) $ " rel rel rel 1 T = . 2 mi R? · R? + 2R? · r?i + r?i · r?i Rotation about a moving axis Here we have used a · b = b · a in order to collect terms. i=1 Using the fact that R? is independent of i allows us to rearrange to # * n * n # n ) ) ) rel 1 + T = 12 mi r?rel mi R? · R? + R? · mi r?rel i i · r?i . 2 i=1 i=1 i=1 The ?rst bracketed term in this equation is the total mass M of the system. By di?erentiating equation (42) with respect to t, we can show that the second bracketed term is zero. So the equation reduces to n ) rel (48) mi r?rel T = 21 M R? · R? + 12 i · r?i . i=1 This result can be stated in words as follows. Kinetic energy decomposition theorem The kinetic energy of an extended body is equal to the kinetic energy of an equivalent particle that has the velocity of the body’s centre of mass, plus the sum of the kinetic energies due to the motion, relative to the centre of mass, of all the body’s constituent particles. The term ‘equivalent particle’ is used to mean a particle of the same mass as the body located at its centre of mass. 4.2 Rigid body rotating with ?xed orientation Consider a rigid body that is in motion, rotating about an axis that may itself move but remains pointing in the same direction. A cylindrical can rolling down a slope, with its axis pointing in the same horizontal direction throughout, provides an example of such motion. In this situation, the centre of mass of the rigid body may be moving, but the motion relative to the centre of mass is of the type considered in Section 3. The position of each particle in the rigid body, relative to the centre of mass, is constrained in the same way that the position of each particle was in our discussion in Subsection 3.1. This means that arguments similar to those in Section 3 can be used to deduce expressions, in terms of the moment of inertia, for the angular momentum and kinetic energy of the rigid body, relative to its centre of mass. To illustrate these points, take a rigid body of mass M that is rotating about an axis of ?xed orientation through its centre of mass with angular velocity ω = θ?k, where k is a ?xed unit vector and θ is the usual anticlockwise angular displacement relative to the axis. Let I be the body’s moment of inertia about that axis. Then the k-component of the angular momentum of the body relative to the centre of mass is I θ?, while the kinetic energy of the body relative to the centre of mass is 21 Iω2 , where ω = |ω| = |θ?| is the angular speed. Combining these results with the torque law relative to the centre of mass (equation (47)) and the kinetic energy decomposition theorem (equation (48)) leads to the following. 221 Unit 21 Rotating bodies and angular momentum Rigid body rotating with ?xed orientation A rigid body of mass M is rotating about an axis of ?xed orientation through its centre of mass, with angular velocity ω = θ?k, where k is a ?xed unit vector and θ is the anticlockwise angular displacement relative to the axis of rotation. Let I be the moment of inertia of the body about the axis of rotation. rel · k) of the angular momentum of the The k-component Lrel axis (= L body relative to the centre of mass is given by Lrel axis = I θ?. (49) The equation of relative rotational motion of the body is Γrel axis = I θ?, (50) rel where Γrel · k) is the k-component of the total external axis (= Γ torque relative to the centre of mass. The kinetic energy T of the body is the sum of the kinetic energy of an equivalent particle at the centre of mass and the rotational kinetic energy relative to the centre of mass, that is, ( (2 T = 21 M (R?( + 12 Iω2 , (51) where R is the position vector of the centre of mass and ω (= |θ?|) is the angular speed. Exercise 23 You considered an identical baton in Exercise 15. Use any results from the solution to that exercise that you ?nd useful. A drum majorette’s baton is modelled as a uniform cylindrical rod with spheres of equal mass at each end. The rod has mass 0.1 kg, length 0.8 m and diameter 0.04 m. Each sphere has diameter 0.1 m and mass 0.25 kg. The baton has been thrown upwards and is rotating at 1 revolution per second about a horizontal axis through its centre of mass and normal to the axis of the cylinder. Its centre of mass, which was initially at O, is moving vertically upwards at 5 m s −1 . (a) Find the kinetic energy of the baton at the instant described. (b) Find the magnitude of the k-component of the angular momentum of the baton, relative to O, where k is a unit vector in the direction of the axis of rotation. Recall the Highland sport of tossing the caber. The process of tossing the caber can be divided into ?ve phases, as shown in Figure 7. In the following exercise, we consider an early part of the toss, that is, the transition between phases (i) and (ii). 222 4 Rotation about a moving axis y Exercise 24 (a) During a caber-tossing competition, a competitor runs forward holding the caber (carrying the end X shown in Figure 28) and stops suddenly. While the competitor is running, the caber is vertical, and the whole caber has the same forward speed v. When the competitor stops, a large force is exerted at X for a very short time, with the e?ect that the end X of the caber becomes stationary. Model the caber as a uniform thin rod of length 2l and mass m, and assume that the motion of the caber is con?ned to two dimensions, in the (x, y)-plane in Figure 28. (i) What will be the angular speed of the caber just after the competitor stops? (ii) What will be the kinetic energy of the caber just after the competitor stops? v after v 2l before v j X x i k z Figure 28 The instant of change between phases (i) and (ii) of Figure 7 (b) Suppose that after stopping, the competitor holds the end X of the caber stationary while the caber falls forward under gravity. Assuming that resistive forces are negligible, estimate the angular speed of the caber when it makes an angle θ with the vertical. We now consider a much later part of the motion of the caber: the start of phase (v) (see Figure 7) or the moment at which it strikes the ground (see Figure 29). The aim of the toss is to ensure that the caber ?nishes lying on the ground with the end X, which was originally being held by the competitor, now furthest away from him. Even if the caber strikes the ground as shown in the ?gure, before it has rotated su?ciently for X to have moved to the right of Y , it may maintain su?cient rotation after impact for X to swing past Y , and for the caber to fall with X pointing away from the competitor. How might we model the e?ect of the caber hitting the ground on its rotational motion? When an object hits the ground, there is an impact, during which the object is subject to forces of great magnitude over a short period of time. These forces drastically change the motion of the object. To model the impact, we assume that after hitting the ground, the end Y of the caber is stationary, and that during impact, all the forces on the caber are acting at the point Y . Take an origin at the point Y , and consider the torque and angular momentum about that point. Since we are assuming that all the external forces act at Y during the impact, the torque about Y is zero. Then by the torque law, the angular momentum about Y is conserved during the impact. X direction of motion Y Figure 29 Instant where the caber strikes the ground at the start of phase (v) We are assuming that during the impact, the force due to gravity is negligible compared with the forces acting at Y . You are asked to develop this model further in the following exercise. 223 Unit 21 Rotating bodies and angular momentum Exercise 25 y X j ω i k 2l G vx i + vy j θ Y O x Figure 30 Coordinate set-up at the start of phase (v) A caber is in ?ight, and its end Y is about to strike the ground at O at an angle θ from the vertical. Use the axes shown in Figure 30, and assume that the motion of the caber is con?ned to the (x, y)-plane throughout. Just before the end Y comes in contact with the ground, the centre of mass G of the caber has velocity vx i + vy j, and the angular speed of the caber about G is ω, rotating clockwise. Model the caber as a uniform thin straight rod of length 2l and mass m. (a) Use a decomposition theorem to ?nd the k-component of the angular momentum of the caber about O just before it hits the ground. (b) Let ω1 be the angular speed of rotation of the caber about O just after it hits the ground. Show that 3 ω1 = 41 ω + (vx cos θ + vy sin θ). (52) 4l (c) To rotate past the vertical, the caber must have su?cient kinetic energy after impact that it reaches the vertical with non-zero kinetic energy. Use this fact to show that in order for the caber to rotate past the vertical, we must have that ω21 > 3g(1 − cos θ) . 2l (53) The values of θ, vx , vy and ω, as de?ned in Exercise 25, when the caber strikes the ground will be determined by the way that the competitor launches the caber. If we know these values and the length 2l of the caber, then we can calculate ω1 from equation (52). Then condition (53) enables us to determine whether or not the caber will pass the vertical. 4.3 Rolling objects It is interesting to compare how solid and hollow cylindrical cans roll down an inclined plane – a situation mentioned in the Introduction. We are now in a position to model that situation quantitatively. To do so, we will assume that there is no loss of mechanical energy when a cylinder rolls down a slope. This is an important assumption that will be fully justi?ed later on. Consider a cylinder of mass M and radius R rolling down a plane inclined at an angle α to the horizontal (Figure 31(a)). We will ?rst look at the behaviour of a uniform solid cylinder. Its moment of inertia about an axis through its centre of mass is 12 M R2 (from Table 1). The cylinder starts from rest at the origin O, and we want to ?nd how long it will take to reach the point A, where the distance OA is l. 224 4 Rotation about a moving axis M B O R O x sin α l B X α R i α A (a) j x θ A (b) Figure 31 Cylinder rolling down an inclined plane: (a) initial position, (b) after it has rolled a distance x along the plane Take the situation where the cylinder has rolled as far as X (Figure 31(b)), and choose Cartesian unit vectors i and j as shown in the ?gure. Let OX = x, and suppose that in rolling from O to X the cylinder has turned through an angle θ, without any slipping having occurred. Note that since the cylinder is rotating clockwise, θ becomes increasingly negative. The distance OX must be equal to the length of the circumference of the cylinder from B to X, where B is the point of contact between the cylinder and the slope at the outset. Thus taking 0 as the initial value of θ, so that its subsequent values are negative, we have x = −Rθ. Note that, as usual, θ is measured positive in an anticlockwise sense. (54) We refer to this equation as the rolling condition. When the cylinder is at the point shown in Figure 31(b), its centre of mass has position xi + Rj. Hence the velocity of the centre of mass is x?i, since R is constant. The cylinder is rotating clockwise at an angular speed |θ?|. Therefore by equation (51), the cylinder has kinetic energy $ " T = 21 M x?2 + 12 12 M R2 θ?2 . (55) The rolling condition is often referred to as the no slip condition. From equation (54) we have R|θ?| = |x?|, so T = 21 M x?2 + 14 M x?2 = 34 M x?2 . Because the cylinder starts from rest at O, its initial kinetic energy is zero. In moving from O to X, the centre of mass of the cylinder descends a vertical distance x sin α. So the potential energy of the cylinder is reduced by M gx sin α. Then, if mechanical energy is conserved, we have 2 3 4 M x? = M gx sin α. (56) By di?erentiating each side of this equation with respect to t, we obtain 3 2 M x?x? = M g x? sin α. (57) On dividing by M x? and rearranging, we obtain x? = 23 g sin α. (58) The cylinder is not stationary, so x? is not zero. 225 Unit 21 Rotating bodies and angular momentum Now, if the cylinder were simply sliding down the slope (without friction) rather than rolling down, it would have acceleration g sin α. But from equation (58) we see that the rolling cylinder has a lower acceleration than if it were to slide down the slope. This is because in the case of the rolling object, some of its potential energy has been converted into kinetic energy of rotation, while for a sliding object all the kinetic energy is associated with the linear motion of the centre of mass. You saw this in Unit 3. Exercise 26 (a) Adapt the foregoing argument to obtain an expression for the acceleration of a hollow cylinder of mass M and radius R when, starting from rest, it is rolling down the same slope as considered above. (Model the hollow cylinder as a thin cylindrical shell, all of whose mass is at a distance R from its axis.) (b) In a ‘race’ over a distance of 2 m down a slope angled at π6 to the horizontal, how much faster will a solid cylinder travel than a hollow one? Do the masses of the cylinders matter? We assume that air resistance is negligible. If there were no friction between the slope and the cylinder, the cylinder would simply slide down the slope without rolling. N O x F i X W α Figure 32 Forces on a rolling cylinder 226 j A We now consider the forces on a cylinder as it rolls down a slope without slipping. Since the cylinder starts from rest, it is gaining angular momentum about its centre of mass as it rolls, so there must be some force supplying a torque relative to the centre of mass. This cannot be the weight W – you saw in Exercise 21 that the force of gravity on a body gives a zero torque relative to the centre of mass. Another force on the cylinder is that from the slope: the normal reaction N of the slope acts through the centre of mass, so again it gives a zero torque about the centre of mass. That leaves only a force between the slope and the cylinder in the direction of the slope, which is supplied by friction. Therefore we must include friction F in our model if the model is to have any hope of predicting the motion that is observed. Exercise 27 (a) Use Newton’s second law (for the motion of an equivalent particle at the centre of mass) and the equation of relative rotational motion (equation (50)) to ?nd an expression for the acceleration of a solid cylinder of mass M and radius R when it is rolling (without slipping) down a slope of angle α to the horizontal, as in Figure 32. (b) The condition for the cylinder to roll without slipping is |F| ≤ µ|N|, where µ is the coe?cient of static friction. Obtain a condition relating α and µ that must hold if only rolling is to occur. Learning outcomes In a system where there is friction, you might reasonably expect there to be a loss of mechanical energy. However, this is not necessarily the case. A cornering car needs a sideways frictional force between each tyre and the road to avoid skidding, as you saw in Unit 20, but as long as the car does not skid sideways, there is no loss of mechanical energy due to this force. The point of contact between the tyre and the road does not move in the direction of the frictional force, thus no mechanical energy is lost to friction. The situation is similar for the rolling cylinder, though this is perhaps more di?cult to see. The point on the cylinder that is in contact with the slope at any instant does not move relative to the slope (if it did, the cylinder would skid and the rolling condition would not hold). Hence there is no loss of mechanical energy due to the frictional force. The truth of this assertion was demonstrated in Exercise 27(a), where you saw that the equation for the acceleration of the centre of mass obtained using Newton’s second law, and making no assumption about the mechanical energy, is the same as equation (58) obtained earlier in this subsection using the assumption that mechanical energy is conserved. That is, the frictional force F is normal to the velocity r?, so F · r? = 0. Therefore the work done by F is zero (see Unit 16). Learning outcomes After studying this unit, you should be able to: • model the motion of an extended body as the motion of an equivalent particle at the centre of mass combined with the motion of the body relative to its centre of mass • ?nd the moments of inertia of rigid bodies of common geometrical shapes about axes of symmetry by reference to Table 1, and use the parallel axes theorem to ?nd the moments of inertia of such bodies about other axes • determine the angular momentum and kinetic energy of a rigid body rotating about an axis whose direction is ?xed, using the appropriate decomposition theorem if necessa...

Already member? Sign In