Forces in a loop on a rolling body

[rbergs215]

Homework Statement

A sphere of mass M and radius r is released from rest at the top of a track, which travels around a loop of R (r <<R). The sphere rolls without slipping during the entire motion. Point A is at height R, one-quarter of the way around the loop, and Point B is at the top of the loop. Rotational inertia of the sphere is 2Mr^2 /5.

What are the forces acting on the sphere at points A and B?

Relevant Equations

Free-Body Diagram is only asked

So looking at the solution, FBD for A makes sense: Normal points inwards, gravity and friction oppose each other, with friction pointing up.

B confuses me: the solution says there is only Normal and gravity pointing downwards, but if "the sphere rolls without slipping during the entire motion" why is there no friction at the top? Otherwise what is causing the rotation? Normal and Gravity supply no torque at point B.

My guess is the solution is wrong and there should be friction acting in the direction of motion.

 

[BvU]

Hello rbergs, $\qquad$ $\qquad$ !

Otherwise what is causing the rotation?

Just like for linear motion: if there is no force, there is no acceleration -- but momentum present remains unchanged (Newton).
If the speed does not change at B(*) , the rate of rotation isn't changed -- so no friction is required.

(*) so if this is plausible is what you have to find out .

 

[TSny]

My guess is the solution is wrong and there should be friction acting in the direction of motion.

Another way to think about it: In what direction is the friction force when the ball is coming back down after passing the top? What is the direction of the friction force just before reaching the top and just after passing the top?

 

[rbergs215]

There's nothing in the problem that states the speed is constant around the loop, so while I understand TSny, and I understand Newton's 2nd Law, still not sure why friction has to be zero. Could there also not be some friction, since the track and loop are not friction-less?

 

[haruspex]

There's nothing in the problem that states the speed is constant around the loop, so while I understand TSny, and I understand Newton's 2nd Law, still not sure why friction has to be zero. Could there also not be some friction, since the track and loop are not friction-less?

At the top of the loop the GPE is at a local maximum, so the KE must be at a local minimum. Hence the acceleration is zero.

 

[BvU]

nothing in the problem that states the speed is constant around the loop

It definitely isn't constant around the loop!
But for the friction to be zero it's enough that the acceleration is zero (in the direction tangent to the loop).
Well, at the top there is no force component that is tangent to the loop, so at that point and at that point only (*) , there is no friction needed.

(*) i.e. in the top half -- what about the opposite point at the bottom point of the loop ?

 

[TSny]

Being overly pedantic, you can argue that there could be friction acting at the top if you allow for "rolling friction". https://en.wikipedia.org/wiki/Rolling_resistance This type of friction is usually not covered in an introductory mechanics course.

However, with rolling friction, mechanical energy is not conserved. Is this an exercise where you are expected to use conservation of mechanical energy? @haruspex gave the argument for why conservation of energy implies zero tangential acceleration at the top (or bottom). So, energy conservation implies zero net tangential force at the top and bottom.

 

[rbergs215]

The later parts of the question involve Point A, not B, but the solution does point to conservation of energy throughout.

I'm being pedantic on purpose I guess. Just want to be able to explain this to my students, because they will definitely have the same question. I guess my only problem was I thought "rolling without slipping" = "friction". Thanks to all the insights and help.

 

[TSny]

The later parts of the question involve Point A, not B, but the solution does point to conservation of energy throughout.

I'm being pedantic on purpose I guess. Just want to be able to explain this to my students, because they will definitely have the same question. I guess my only problem was I thought "rolling without slipping" = "friction". Thanks to all the insights and help.

I did not mean to imply that you were being overly pedantic. I was referring to my post which was a little bizarre in bringing in "rolling friction".

"Rolling without slipping" is generally interpreted as implying no dissipation of mechanical energy to "heat". Thus, in the problem at hand, you can assume conservation of mechanical energy. haruspex gave a nice argument for why this implies that the speed cannot be changing at the top. So, there can't be any friction acting at that point. If you want to put that in more mathematical form:$$E = KE + PE = \frac{7}{10} mv^2 + mgy$$ where the 7/5 factor takes into account both the translational and rotational KE. Assuming E is constant and taking the time derivative of this equation: $$0 = \frac{7}{5}mv \dot{v} + mg \dot y$$ Since $\dot y = 0$ and $v \neq 0$ at the top, you see that $\dot v = 0$ at the top. So, the tangential acceleration must be zero at the top.

 

[jbriggs444]

However, with rolling friction, mechanical energy is not conserved.

If one contemplates rolling resistance, that manifests as something akin to a retarding torque on the ball. The retarding torque tends to make the ball roll too slowly for its forward velocity. The resulting frictional force on the ball is directed rearward -- in a direction so as to reduce its forward velocity while increasing its rolling rate. Thereby preventing any slippage.

By contrast, if one contemplates air resistance to the ball's linear motion, that manifests as a rearward linear force. The result is that the ball is moving too slowly for its roll rate. The resulting frictional force on the ball is directed forward -- in a direction so as to increase its forward velocity while reducing its rolling rate. Again, preventing any slippage.

If one decides to account for dissipation but does not decide what effect is responsible for it, the direction of the frictional force between ball and track at the top of the loop cannot be predicted.

 

[haruspex]

rolling resistance, that manifests as something akin to a retarding torque on the ball.

Quite so. The normal force becomes displaced forwards of the mass centre. Whether it is because of compression of the object or of the surface on which it rolls, the "spring constant " is greater during compression than during relaxation.