I Will a sphere rotate on a frictionless inclined surface?

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1. Dec 25, 2017

Well, my physics teacher taught us about rotation the other day and I came across a scenario where a sphere and a ring roll down a friction-less inclined plane from a point of absolute rest. I found it counter-intuitive as I started to think about why would they start rolling in the first place. There is no off-set force that can provide torque. So, I raised my hand and argued that the normal force and gravity pass through the center of mass of the body, so the sphere ought to just slide down the plane. Then he drew a diagram and a gravity vector going straight down the center of the sphere and said gravity provides torque if we consider the point of contact as the pivot. At that point I didn't know what else to say and pretended to be satisfied by that. But, I'm not and that's why I'm here. Please help!

2. Dec 25, 2017

Staff: Mentor

You are correct and your teacher is wrong. If there is no friction, there will be no torque about the center of mass to produce a rotational acceleration.

His mistake is not realizing that the point of contact is accelerating and thus different rules apply.

3. Dec 25, 2017

Can you please elaborate as to what mathematical statement can be used as an evidence that the ball will definitely slide and not spin about it COM. His thinking doesn't really matter, as long as I understand what is happening.

4. Dec 25, 2017

Staff: Mentor

He should realize and accept that you can always use the COM as your reference point. Since there is no torque about the COM, there will be no rotational acceleration. Simple as that.

If he chooses to use the point of contact as the reference point, that's fine, but a more advanced analysis is required: The bottom line is that a proper analysis using that accelerating reference point will also show that there will be no rotational acceleration about the COM. (This is covered in every Classical Mechanics textbook. It's not really that advanced, just one step above Intro physics.)

5. Dec 25, 2017

Staff: Mentor

That was my first thought, but then I found I myself thinking about the normal force exerted by the surface of the inclined plane.... Does it pass through the center of mass of the sliding object? If I were to try standing on a frictionless incline looking downhill, my feet would go out from under me and the back of my head would hit the surface.

6. Dec 25, 2017

Staff: Mentor

For spheres and rings, yes. (That's what we're talking about here. I think!)

7. Dec 25, 2017

Thank you for your time, my query feels answered. What textbooks would you recommend for one level above the introductory level, as I would like to step into physics at home.

8. Dec 25, 2017

Staff: Mentor

Ah - yes, of course. I posted, went off to do start doing the dishes, and then realized that we're talking about circular spheres and rings so a vector perpendicular to the surface will pass through the center. I suspect that OP's teacher made the same mistake, failing to recognize the the importance of the symmetry in this case.

9. Dec 26, 2017

Mister T

You can easily simulate the situation you proposed in your previous post by standing on roller skates on a ramp or hill. If you stay on your feet a line perpendicular to the ramp will pass through your feet and your center of mass, something that most roller skaters can manage, and then you'll move down the ramp without rotating. On the other hand, if you fell over instead, you would still slide down the frictionless ramp without rotating.

10. Dec 26, 2017

A.T.

He is wrong as Doc Al wrote. Here is another way to think about it: A uniform gravitational field alone never makes stuff spin, only the non-uniformly applied contact forces do, if they create a torque around the COM.

11. Dec 26, 2017

CWatters

Teacher is wrong as others have said.

Draw a free body diagram of the ball.

There is no friction so the reaction force between the ball and slope must be normal to the slope1. This means that the reaction force also passes through the COM of the ball. So both gravity and the reaction force pass through the COM.

1) It's useful to remember this for other situations such as ladders leaning against frictionless walls.

12. Dec 26, 2017

PeroK

Just another take on this. Simply decompose the gravitational force into normal and tangential components. The normal component cancels the normal force from the slope, leaving simple tangential acceleration.

When there is friction, you have an additional force up the slope. Note that this not only provides the torque for rotation, but also reduces the tangential acceleration. In your teacher's analysis, therefore, the energy would not add up. The rotational energy would come with no reduction to linear kinetic energy.

13. Dec 26, 2017

But, shouldn't that be the case anyway? That friction is necessary to BEGIN rolling but once you start rolling, the instantaneous velocity of contact point w.r.t. ground is zero. Hence friction should not cost any energy as it does ZERO work. Am I right?

14. Dec 26, 2017

PeroK

In the case of friction it does zero work. The total kinetic energy of the sphere, rotational plus linear, equals the loss of gravitational potential energy. Friction has the effect of causing some of the energy to be rotational - at the expense of linear kinetic energy.

This scenario is worth studying.

But, if a sphere starts to roll without friction, then there is nothing to reduce the linear acceleration. So, all the potential energy must go to linear kinetic energy and the rotational energy must come for free.

15. Dec 26, 2017

A.T.

Static friction dissipates no energy. Kinetic friction and rolling resistance do.

16. Dec 26, 2017

jbriggs444

Late to the party here. Another way of seeing the error of the teacher's claim...

Yes, if one uses the point of contact as the [instantaneous] axis of rotation then the force of gravity amounts to an unbalanced torque. However, one must realize that not all angular momentum consists of rotation. In this case, the resulting angular momentum increase is accounted for by the linear acceleration of the sphere or ring.

When the center of mass of an object is moving on a line that does not pass through the chosen axis then that motion has non-zero angular momentum given by $\vec{L}=\vec{r}\times\vec{p}$ where L is the angular momentum, r is the offset of the center of mass from the axis of rotation and p is the linear momentum of the object. If the object accelerates, its linear momentum is increasing along with the associated angular momentum.

17. Dec 26, 2017

A.T.

@adesh123 I think the above is the best general explanation. You could just as well drop the ball vertically and calculate a non-zero gravity torque around some random point off the vertical. That torque will not make the ball spin during the fall, even tough it will change the angular momentum around that point.

18. Dec 27, 2017

CWatters

I still think drawing a free body diagram for the ball is the way to go. It's obvious from that there is no torque on the ball.

19. Dec 27, 2017

jbriggs444

Except, of course, for the fact that there is a torque on the ball about the point of contact.

20. Dec 27, 2017