Is this angular or linear momentum?

[allok]

hello


1.

a) Does object circling around the Earth have angular momentum or linear momentum?
It seems to me that we could use either of the two momentums to describe the system?

b)How about when you rotate a ball on piece of string? Again we could describe the object as having either angular or linear momentum, depending on preference?

c)Could we say that torque is causing the ball to circle with certain acceleration ?



2)

We have rotational kinetic energy and linear one. But why is potential energy only defined as energy that has ability to transform itself into linear kinetic energy? Why doesn't potential energy also include the ability to transform itself into rotational kinetic energy?

Thus we measure the change of potential energy with regards to by how much has the center of mass change its vertical position. But even if object's COM doesn't change its position, it can still rotate a bit due to gravity so work was done ... it would seem reasonable to think of potential energy also as ability for object to rotate due to force of gravity?!

thank you

 

[nrqed]

hello


1.

a) Does object circling around the Earth have angular momentum or linear momentum?
It seems to me that we could use either of the two momentums to describe the system?

b)How about when you rotate a ball on piece of string? Again we could describe the object as having either angular or linear momentum, depending on preference?

In both cases the object has both angular and linear momentum. But the linear momentum is not a constant vector. If the speed is constant, the angular momentum is a constant vector. Also, if you look at a rigid body in rotation (for example a full wheel), the points located at different radii all have different linear momenta at any given time whereas they all have the same angular momentum.

If a rotating object keeps moving at a fixed radius, the concept of angular momentum is more useful to study the motion. For example, if the net torque is zero, the angular momentum will remain constant. If is is not zero, one can easily calculate the angular acceleration. It would be very complicated and messy to try to use momentum and forces to study such a situation.

Angular momentum is simply a much more useful concept to study rotation.

c)Could we say that torque is causing the ball to circle with certain acceleration ?

If the net torque is non zero, it will indeed cause a nonzero *angular* acceleration. Notice that even if the net torque is zero, any point on a rotating object has a nonzero *linear* acceleration.

2)

We have rotational kinetic energy and linear one. But why is potential energy only defined as energy that has ability to transform itself into linear kinetic energy? Why doesn't potential energy also include the ability to transform itself into rotational kinetic energy?

Thus we measure the change of potential energy with regards to by how much has the center of mass change its vertical position. But even if object's COM doesn't change its position, it can still rotate a bit due to gravity so work was done ... it would seem reasonable to think of potential energy also as ability for object to rotate due to force of gravity?!

thank you

Conservation of energy must take into account both the translational kinetic energy of the COM *and* the kinetic rotational energy. So you are correct, that *must* be taken into account.

An obvious example is when you let a ball roll down an inclined plane. If you calculate the speed at the bottom of the inclined plane just using 1/2 mv^2 (where v is the speed of the COM), you will NOT get the correct answer. You *must* included the energy that went into making the object rotate.

This is why if you let a full cylinder and a hollow cylinder of the same mass and radius roll down an inclined plane, they will not reach the bottom at the same time. The hollow disk has all its mass concentrated on its rim, so it is more "difficult" to rotate (it has a larger moment of inertia). It will "use" more kinetic energy for rotation, so it will have less translational kinetic energy at a given height than its full counterpart, and will therefore take more time to reach the bottom of the plane.

Hope this helps

Patrick