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Homework answers / question archive / These problems are on curvilinear one dimensional systems and are giving me a lot of difficulty, if you could provide help along with visuals to help explain that would be very helpful
These problems are on curvilinear one dimensional systems and are giving me a lot of difficulty, if you could provide help along with visuals to help explain that would be very helpful.
1) Which of the following forces is conservative? (a) F = k (x, 2y, 3z) where k is constant. (b) F = k (y, x,
0). (c) F = k (-y, x, 0). For those which are conservative, find the corresponding potential energy U, and
verify by direct differentiation that F = - ∫U.
2) The figure shows a child's toy, which has the shape of a cylinder mounted on top of a hemisphere. The
radius of the hemisphere is R and the CM of the whole toy is at height h above the floor. (a) Write down
the gravitational potential energy when the toy is tipped to an angle from the vertical.(b) For what
values of R and h is the equilibrium = 0 stable?
Please see the attached file.
If F is conservative, then we can write where U is a scalar function.
Therefore the curl of a conservative force must obey:
For a general vector field we have:
Hence:
Therefore is conservative.
Therefore is conservative.
is not a conservative vector field.
Since:
We see that:
For the first vector field, this translates to:
We need now to find f(y,z):
And by the same token:
So finally, the generating potential of the vector field is:
Simple Check:
For the second vector field, this translates to:
We need now to find f(y,z):
Where C is an arbitrary constant.
And by the same token:
So finally, the generating potential of the vector field is:
Simple check:
The concept of center of mass allows us to treat this solid, for all practical reasons, as if all its mass is located at the center of mass.
Hence the potential energy is:
To find the equilibrium point we state that the force there must be zero:
This condition leads to:
The equilibrium angle is when the solid not tipped.
A stable equilibrium occurs when this extremum point is a minimum.
This means that we require the second derivative of the potential energy to be positive at the equilibrium point:
Hence:
As long as R>h this equilibrium is stable.
