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Rotational motion with a constant nonzero acceleration is not uncommon in the world around us

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Rotational motion with a constant nonzero acceleration is not uncommon in the world around us. For instance, many machines have spinning parts. When the machine is turned on or off, the spinning parts tend to change the rate of their rotation with virtually constant angular acceleration. Many introductory problems in rotational kinematics involve motion of a particle with constant, nonzero angular acceleration. The kinematic equations for such motion can be written as

$\theta (t) = \theta_0 +\omega_0t + \frac{1}{2}\alpha t^2$

and

$\omega (t) = \omega_0 + \alpha t$ .

Here, the symbols are defined as follows:

is the angular position of the particle at time .
is the initial angular position of the particle.
is the angular velocity of the particle at time .
is the initial angular velocity of the particle.
is the angular acceleration of the particle.
is the time that has elapsed since the particle was located at its initial position.

In answering the following questions, assume that the angular acceleration is constant and nonzero: $\alpha \neq 0$ .

True or false: The quantity represented by theta is a function of time (i.e., is not constant).

	true
	false

True or false: The quantity represented by theta_0 is a function of time (i.e., is not constant).

	true
	false

True or false: The quantity represented by omega_0 is a function of time (i.e., is not constant).

	true
	false

True or false: The quantity represented by omega is a function of time (i.e., is not constant).

	true
	false

Which of the following equations is not an explicit function of time ? Keep in mind that an equation that is an explicit function of time involves as a variable.

	$\theta=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2$
	$\omega=\omega_0+\alpha t$
	$\omega^2=\omega_0^2+2\alpha(\theta-\theta_0)$

In the equation $\omega=\omega_0+\alpha t$ , what does the time variable represent?

Choose the answer that is always true. Several of the statements may be true in a particular problem, but only one is always true.

	the moment in time at which the angular velocity equals $\omega_0$
	the moment in time at which the angular velocity equals $\omega$
	the time elapsed from when the angular velocity equals $\omega_0$ until the angular velocity equals $\omega$

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to $\theta_{\rm A}(t)=\theta_0+\omega_0t+\frac{_1}{^2}\alpha t^2$ . At time $t=t_1$ , particle B, which also undergoes constant angular acceleration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time $t=0$ .

Which of the following equations describes the angular position of particle B?

	$\theta_{\rm B}(t)=\theta_0+2\omega_0t+\frac{1}{4}\alpha t^2$
	$\theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0t+\alpha t^2$
	$\theta_{\rm B}(t)=\theta_0+2\omega_0(t-t_1) +\frac{1}{4}\alpha (t-t_1)^2$
	$\theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0(t-t_1)+\alpha (t-t_1)^2$
	$\theta_{\rm B}(t)=\theta_0+2\omega_0(t+t_1) +\frac{1}{4}\alpha (t+t_1)^2$
	$\theta_{\rm B}(t)=\theta_0+\frac{1}{2}\omega_0(t+t_1)+\alpha (t+t_1)^2$

How long after the time t_1 does the angular velocity of particle B equal that of particle A?

	$\frac{\omega_0}{4\alpha}$
	$\frac{\omega_0+4\alpha t_1}{2\alpha}$
	$\frac{\omega_0+2\alpha t_1}{2\alpha}$
	The two particles never have the same angular velocity.