Problem Concerning Rotational Kinetic Energy

[Curtiss Oakley]

Homework Statement

A solid sphere of mass m and radius r rolls without slipping along the track shown below. It starts from rest with the lowest point of the sphere at height h=3R above the bottom of the loop of radius R, much larger than r. Point P is on the track and it is R above the bottom of the loop. The moment of inertia of the ball about an axis through its center is I =2/5 mr2. The ball should not be treated as a point mass. For the following parts, you can express your answers intermsofm,g,r,andR. Find
(a) the center of mass speed of the ball when it is at point P;
(b) the angular speed of the ball when it is at point P;
(c) the angular acceleration of the ball when it is at point P;
(d) the tangential acceleration of the ball when it is at point P;
(e) the static frictional force acting on the ball when it is at point P.

Relevant Equations

Ke=1/2mv^2
Rotational Energy=1/2Iw^2
Change in height=mgh
Torque=rFsin(•)
Fnet=ma
a=alpha(r)

For parts A and B I used energy to find the vcom and omega, but that won’t work for C. I have an answer by combining the three formulas that use acceleration above. My answer for alpha=-5g/3r. The next two are easily solvable if you find C, but I still feel like I’m missing something. Any help would be appreciated, but don’t give me the answer. Let me try and work through it with whatever tips and pointers I can get. There is a picture attached of the problem

 

[RPinPA]

Angular acceleration is produced by torque. What forces are acting on the ball in a direction to cause a torque?

 

[Curtiss Oakley]

Angular acceleration is produced by torque. What forces are acting on the ball in a direction to cause a torque?

The only force causing torque is friction. With torque net = r(fs)sin(-90) = I(-alpha)

 

[vela]

I would suggest trying to calculate $\alpha = d\omega/dt$. (I haven't solved the problem, but it's the tack I'd try.)

 

[Curtiss Oakley]

I would suggest trying to calculate $\alpha = d\omega/dt$. (I haven't solved the problem, but it's the tack I'd try.)

That sounds overly complicated, especially since the problem didn’t give me anything for t to go off of. If you wanted to do that, how would you start. Maybe I’m missing something.

 

[vela]

Considering you don't know the force of static friction so you can't calculate the torque, how else can you do it?

You could, equivalently, find $a_t = dv/dt$ and then divide by $R$. To figure out the time derivative, you can use the chain rule, e.g. $\frac{dv}{dt} = \frac{dv}{dy} \frac{dy}{dt}$.

 

[haruspex]

The only force causing torque is friction.

That's the only force causing torque about the mass centre of the ball, but you are free to choose other axes that are fixed in space (so long as you use the appropriate MoI).

 

[Curtiss Oakley]

Considering you don't know the force of static friction so you can't calculate the torque, how else can you do it?

You could, equivalently, find $a_t = dv/dt$ and then divide by $R$. To figure out the time derivative, you can use the chain rule, e.g. $\frac{dv}{dt} = \frac{dv}{dy} \frac{dy}{dt}$.

That still seems a lot longer than considering both torque and force to solving static friction out. That’s how I worked it out (I attached my work to this reply). I still just feel like I’m missing something.

 

[Curtiss Oakley]

That's the only force causing torque about the mass centre of the ball, but you are free to choose other axes that are fixed in space (so long as you use the appropriate MoI).

Why change the axis though? To avoid considering a force that you don’t know? Even if you changed the axis, gravity still wouldn’t impact the rotational motion, correct?

 

[vela]

I still just feel like I’m missing something.

You're not. I didn't actually look into the problem all that much and write the equations of motion out.

I do think there is a problem with your work, however. You used $a$ to represent the magnitude of the acceleration so it should come out positive, yet your answer for $\alpha = a/R$ is negative. Are you sure about the direction of the frictional force?

 

[Curtiss Oakley]

You're not. I didn't actually look into the problem all that much and write the equations of motion out.

I do think there is a problem with your work, however. You used $a$ to represent the magnitude of the acceleration so it should come out positive, yet your answer for $\alpha = a/R$ is negative. Are you sure about the direction of the frictional force?

The frictional force would oppose my rotational and translational motion, which is my reason for making it negative.

 

[vela]

At point P, isn't the ball rotating clockwise as it rolls up the track?

 

[haruspex]

Why change the axis though? To avoid considering a force that you don’t know?

Yes.

Even if you changed the axis, gravity still wouldn’t impact the rotational motion, correct?

No. If the axis is not on a vertical line through the mass centre then the gravitational force will have a torque about the axis.

 

[Curtiss Oakley]

At point P, isn't the ball rotating clockwise as it rolls up the track?

Yes, but I flipped which is postive. Didn’t have to but I did when I first solved it because I mentally saw it as slowing down-making it negative.

 

[Curtiss Oakley]

Yes.

No. If the axis is not on a vertical line through the mass centre then the gravitational force will have a torque about the axis.

But isn’t torque completely separate from the axis and is instead dependent on the how tangential it is to the circle?

 

[vela]

Yes, but I flipped which is postive. Didn’t have to but I did when I first solved it because I mentally saw it as slowing down-making it negative.

You have the frictional force pointing down on your free-body diagram. Wouldn't that cause its rotation to speed up?

 

[Curtiss Oakley]

You have the frictional force pointing down on your free-body diagram. Wouldn't that cause its rotation to speed up?

That does make sense. So friction would be pulling in the opposite direction I have indicated? The ball would lose tangential velocity in the loop which means it loses rotational velocity in the hoop. That means that the alpha would have to be negative by some force and my only force is friction. Is there any way the ball becomes quicker through point P on the loop?

 

[vela]

That does make sense. So friction would be pulling in the opposite direction I have indicated? The ball would lose tangential velocity in the loop which means it loses rotational velocity in the hoop. That means that the alpha would have to be negative by some force and my only force is friction. Is there any way the ball becomes quicker through point P on the loop?

Wouldn't that violate conservation of energy?

 

[Curtiss Oakley]

Wouldn't that violate conservation of energy?

It for sure would. In my understanding of this question, the frictional force goes with the rotational motion but gravity and friction slows down both the rotational motion and translational motion. Is that fair to say?

 

[haruspex]

But isn’t torque completely separate from the axis and is instead dependent on the how tangential it is to the circle?

Torque is (almost) always only meaningful relative to a stated axis.
The exception is when there are two equal and opposite forces not in the same line, i.e. antiparallel. In that case the torque is independent of the axis.