# Homework Help: Rotational Mechanics Question.

Tags:
1. Feb 6, 2017

### Yodaa

1. The problem statement, all variables and given/known data
A tangential force F acts at the top of a thin spherical shell of mass m and radius R. Find the acceleration of the shell if it rolls without slipping.

2. Relevant equations
Since the shell is rolling, friction does not act at the bottom.
So equating the torque,
F*R= I*α

3. The attempt at a solution
So putting in value of Moment of inertia for a hollow sphere, I=2/3mR^2
and α=a(COM)/R
And solving for a, i got a=3F/2m
But the correct answer is : 6F/5m
Where am i going wrong?

2. Feb 6, 2017

### malemdk

Since the tangential force F acts on Top of the shell, torque is equal to F x 2R, and the correct answer is 3F/m

3. Feb 6, 2017

### Yodaa

Ive taken the centre of mass as the axis of rotation. So i think it will still be F x R?
Also the correct answer given is 6F/5m

4. Feb 6, 2017

### malemdk

No, you said the sphere is rolling without slipping, since sphere has a contact at the bottom torque arm is 2R,if it's freely rotating about an axis which passes through the center of spherical shell then the torque arm will be R,please recheck

5. Feb 6, 2017

### ehild

This is not true if the shell is accelerating.

The static friction counts both in the acceleration of the CoM, and in the rotation. When you consider rotation about the centre of mass, you have to take the torque of the friction force into account.

Last edited: Feb 6, 2017
6. Feb 6, 2017

### haruspex

You are taking the moment about the point of contact - a good idea since that means you don't have to worry about the frictional force - but then you must use the moment of inertia about that point, not the MoI about the sphere's centre.

7. Feb 6, 2017

### malemdk

Since we have to find the acceleration of the axis of the sphere it is enough to calculate the MoI about the axis, center of mass not rotating about the point of contact which is at bottom

8. Feb 6, 2017

### TomHart

I would have thought if there was no friction the shell would be slipping.

9. Feb 6, 2017

### ehild

The center of mass rotates about the point of contact when the shell rolls.

10. Feb 6, 2017

### TomHart

"@ehild, is your fs pointing in the correct direction?" Tom sheepishly asks.

11. Feb 6, 2017

### ehild

The static friction prevents relative motion between the surfaces in contact. Without friction, the bottom part of the shell would slip forward with acceleration equal to that of the CoM. So what is the direction of the force acting here, to prevent slipping?
Edit: not true as the force also rotates the shell with angular acceleration FR/I. That would make the rim accelerate with FR2/I with respect to the axis of the shell. The center of the shell would accelerate forward with at=F/m, the bottom part of the rim would accelerate with ar=FR2/I backward with respect to the center, that is with F/m(1-R2m/I ) with respect to the ground. As I < mR2, the contact point would slip backwards, so the force of friction should point forward.

Last edited: Feb 7, 2017
12. Feb 6, 2017

### malemdk

Since shell is rolling on horizontal plane no rotation of axis takes place with respect to the bottom

13. Feb 6, 2017

### malemdk

Rolling resistance, that prevents the wheel from slipping

14. Feb 6, 2017

### ehild

No, rolling resistance decelerates both rotation and translation, and it acts when the objects rolls already.

15. Feb 7, 2017

### TomHart

Please humor me on this. I'm still learning. Consider the force acting at a distance r/100 instead of r. In that case I can easily see that the friction force would act in a direction opposite of force F. But with the force F at a distance r, I think my calculations show that, without friction (Edit added "without friction"), the torque will produce an angular acceleration such that the acceleration of the perimeter of the sphere exceeds the linear acceleration of the sphere. The result of that would be that the friction force would act in the same direction as force F. Please let me know what you find. My humble disclaimer: I am still learning. :)

16. Feb 7, 2017

### malemdk

The applied force balances rolling resistance force which is acting vertically and offset from the center of axis, ITS NOT HORIZONTAL

17. Feb 7, 2017

### haruspex

You mean, acceleration about the axis, not acceleration of the axis.
The angular acceleration is the same no matter which axis you use. It's a rigid body. But you must use the same axis for the torque that you use for the moment of inertia, and that axis must either be the mass centre or one that is (momentarily at least) fixed in space.
And you must include all torques. If you select the centre of the sphere as axis then there will be a torque from the friction.
By definition, rolling contact means that the piece of the sphere in contact with the ground is instantaneously stationary relative to the ground. It can have a relative acceleration but not a relative velocity. Consequently it is the instantaneous centre of rotation. If you are not familiar with the concept of instantaneous centre of rotation Google it.
No, friction with the ground prevents slipping. Since it is static friction, while rolling, it does no work.
Rolling resistance is something else. It does negative work on the body and can apply whether purely rolling or rolling while skidding. It acts like, and can include, axle friction.

Last edited: Feb 7, 2017
18. Feb 7, 2017

### TomHart

Do I understand you correctly that you are saying rolling resistance is a vertical force? Maybe I misunderstood, but rolling resistance is a horizontal force.

19. Feb 7, 2017

### ehild

You are right (I am still learning at this old age :) ) You can take the direction of fs either forward or backward, you get the same acceleration of the CoM, but the direction of the static friction proves to be forward in this case.

20. Feb 7, 2017

### haruspex

You are right that the net effect of rolling resistance is antiparallel to the motion (not necessarily horizontal, of course), but its nature is more complicated than you might guess. See section 4 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/.

21. Feb 7, 2017

### haruspex

The applied force acts horizontally. A horizontal force cannot, of itself, balance a vertical force. There would have to be some kind of pivot mediating, turning them into opposing torques.
Anyway, there is no suggestion in the setting of the question that rolling resistance is to be taken into account.

22. Feb 7, 2017

### TomHart

@haruspex, I am trying to get on board with this rolling resistance concept.

The link you posted said: "Because of the imperfect elasticity, the normal force is stronger in the front half than in the rear. This leads to a torque opposing rotation."

So is that basically the bottom line reason for rolling resistance - that a counter torque is produced because more of the normal force is on the front part of the tire than the rear part of the tire? Or I suppose if it was a very hot day where the roads were kind of sticky, that could also produce an opposing torque from the rear part of the tire.

23. Feb 7, 2017

### Yodaa

So from what i have understood, as @ehild mentioned, i had taken centre of mass as the axis which means i should have taken the static friction acting at the bottom into count(and should not have said that the friction does not act at the bottom what was i even thinking)
And approaching as @haruspex said, by taking the bottom point as the axis, we dont need to worry about the friction acting
Solving, Tx2R = I*(a/R) [ where I= 2/3MR^2 + MR^2 by parallel axis theorem]
Thus we get a=6F/5m, the correct answer
Thanks for putting things into perspective!

24. Feb 7, 2017

### malemdk

It's not vertical force balances the horizontal force rather F x R = Fr x s
Where Fr is rolling resistance and s the distance Fr offset from the axis

25. Feb 7, 2017

Yes,