Where Did I Go Wrong in Calculating the Acceleration of a Rolling Shell?

[Yodaa]

Homework Statement

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.

Homework Equations

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

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?

 

[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

 

[Yodaa]

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

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

 

[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

 

[ehild]

Homework Statement

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.

Homework Equations

Since the shell is rolling, friction does not act at the bottom.

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.

 

[haruspex]

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

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.

 

[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

 

[TomHart]

Since the shell is rolling, friction does not act at the bottom.

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

 

[ehild]

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

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

 

[TomHart]

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

 

[ehild]

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

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 a_t=F/m, the bottom part of the rim would accelerate with a_r=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.

 

[malemdk]

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

 

[malemdk]

Rolling resistance, that prevents the wheel from slipping

 

[ehild]

Rolling resistance, that prevents the wheel from slipping

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

 

[TomHart]

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?

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. :)

 

[malemdk]

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

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

 

[haruspex]

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

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.

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

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.

Rolling resistance, that prevents the wheel from slipping

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.

 

[TomHart]

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

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.

 

[ehild]

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. :)

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.

 

[haruspex]

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.

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/.

 

[haruspex]

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

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.

 

[TomHart]

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/.

@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.

 

[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 don't 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!

 

[malemdk]

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.

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

 

[malemdk]

@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.

Yes,

 

[haruspex]

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.

Yes, each of those mechanisms would provide an opposing torque. But to slow the ball that torque must somehow be turned into a horizontal force opposing the linear motion.
A third way such a torque can arise is from friction at the axle.
As I explain in the next paragraph at that link, the torque opposes the rotation of the ball, leading to a rotational deceleration. Since rotating more slowly but maintaining the same linear speed would lead to skidding, a horizontal frictional force arises. This is antiparallel to the linear motion, tending to reduce the angular deceleration while causing a linear deceleration. In this way, the rolling relationship between linear and rotational speeds is maintained.
And as the paragraph after that explains, there is another, rather different way that rolling resistance can act. In this case, it is deformation of the substrate. The normal forces are now at a range of angles to the "normal" (or, to put it another way, there is a range of normals). The normal forces at the front of the contact area tend to be stronger than those at the rear, and they point backwards. Thus, the net normal force has a component antiparallel to the linear motion. But all these normal forces point towards the ball's centre (if it it does not deform), so no torque this time.

 

[TomHart]

As I explain in the next paragraph at that link

You wrote that? Great job! But having read it, I realize that one time through is not enough. I am going to have to spend some mulling time.