Purely Rotating Bodies

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1. Aug 4, 2016

Milo Martian

1. The problem statement:
Consider a solid sphere and a hollow sphere, of equal mass M and equal radius R ,at rest on top of an incline . If there is no slipping which will reach the bottom faster.

2. Relevant equations:
acm = Fext/M (cm= centre of mass)
angular acceleration= torqueext/ I ( I = moment of inertia)
Ihollow sphere= 2MR2/3
Isolid sphere=2MR2/5

3. The attempt at a solution:
acm should be same for both of them as Fext and M are same, so they should reach at same time. But angular acceleration of solid sphere will be more since Isolid sphere< Ihollow sphereand torque is same for both, so by this info the solid sphere should reach faster. Thus, the confusion.Also, how can acm remain same for both when angular acceleration of solid sphere is higher , the acm of the solid cylinder should be higher since the radius is same for both. But according to the first equation, acm should be same.

2. Aug 4, 2016

drvrm

how the angular acceleration is related with the acceleration of the center of mass of the two spheres ?
i doubt that a(c.m.) is same for both as the rotation and translational motion seems to be related(a case of pure rolling).

3. Aug 4, 2016

Milo Martian

well , if there is no slipping the distance covered in one rotation is 2piR for both the spheres. So, with higher angular acceleration, the number of rotations per second increases at higher rate, leading to higher increase in rate of ground coverage, so shouldn't the acm be higher as well? But, is the first equation not applicable for pure rolling?

4. Aug 4, 2016

Vibhor

No . The net force is not same . Please write equation of motion for both translation and rotational motion .Find expression for acm in terms of given variables .

5. Aug 4, 2016

Milo Martian

aren't the equations given the right one's for translation and rotation? isn't the equation , acm = Net external force/ Mass ,correct for translation + rotation?

6. Aug 4, 2016

haruspex

Yes, but as Vibhor wrote the net force is not the same. What forces act on the spheres?

7. Aug 5, 2016

Milo Martian

Well... there's a force mgsin(theta) acting parallel to the incline along the centre of mass and friction acting at the bottom of the sphere in the opposite direction. Since there's no slipping, the forces should be equal.So, the net external force for both is 0? so, acm is 0? that can't be right. i'm confused.

8. Aug 5, 2016

haruspex

Well, that's the component of the gravitational force in that direction, yes.
No such rule.

9. Aug 5, 2016

Milo Martian

But as the body does not slip , shouldn't the magnitude of static friction be equal to the magnitude of the external force trying to make it move?

10. Aug 5, 2016

haruspex

The fact of non-slipping only tells you that the point of the wheel in contact with the plane has no acceleration or velocity parallel to the plane. The net zero acceleration will be achieved by the linear downslope acceleration of the wheel as a whole being equal and opposite to the tangential acceleration resulting from the wheel's rotational acceleration.

11. Aug 5, 2016

Milo Martian

Yes, i thought i said that only. The linear downslope acceleration comes from the component of gravitational force in that direction ( doesn't it) and the tangential acceleration comes from friction (doesn't it?) Since force= mass X acceleration and the mass and acceleration being equal , the forces will also be equal ( isn't it?).
I am getting really confused, the answer to my doubts seems miles away. plz help

12. Aug 5, 2016

haruspex

The wheel is a rigid body. You cannot figure out the motion of one small piece of it just from looking at external forces applied to the wheel as a whole. There are also forces between the parts of the wheel.
The two accelerations only cancel out at the point of contact.

The normal force and the component of the gravitational force perpendicular to the plane exactly cancel and are in the same line of action, so they produce no net force nor torque. So we can ignore these.

There is the downslope component of gravity, call this force Fgd, and the upslope force of friction, Ff. The difference between them gives the net force which provides the linear downslope acceleration of the wheel.

At the same time, these two forces produce a net torque. That produces the rotational acceleration.
The value of the torque depends where you put your reference axis. If we put it at the centre of the wheel then it will be R.Ff. If we put it at the point of contact it will be R.Fgd. The corresponding moments of inertia for those two axes are also different, and it turns out that the resulting angular acceleration is the same either way.

13. Aug 5, 2016

Milo Martian

Ok.. so to find out acm , need to find out angular acceleration and multiply it by R?
and regarding the equation acm= Fext/M , can't we predict the motion of centre of mass from the knowledge of external forces and the mass? but the forces due to different parts of the sphere are internal forces and are not part of the above equation. Sorry, if my questions are getting repetitive .

14. Aug 5, 2016

Nidum

It is always easier to understand these problems using a diagram . Take a copy of the diagram above and mark in whatever forces that you think are acting

15. Aug 5, 2016

Milo Martian

OK so,m being mass of the body, here normal force, N = mgcosA and since there is no slipping ( so there is only static friction) the static friction, f should be equal to mgsinA i think. There will be a torque due to the combination of friction, f and component of gravitational force mgsinA. Am i wrong so far? If yes, can you tell me where and why?
If i am not wrong then are the equations i gave in the question applicable here? if yes, then since friction, f and mgsinA are equal and opposite so the net force on the body is 0 right? So, acm= 0 ( since Fext=0) which must not be right. Or is Fext= mgsinA as friction is internal force to the system and, thus, the acm= gsinA? Here is my confusion. Can you tell me how to get the right acm? and whats wrong with my method?

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16. Aug 5, 2016

haruspex

You keep repeating that, but it is not true. Your intuition is blocking you. Just work with the free body diagram and the standard laws, Fnet=ma, τnet=Iα, and the rolling condition a=rα.
Fot the torque equation, make sure to use the same axis for both torque and moment of inertia.
When you have worked that through, you can see for yourself how the static frictional force differs from mg sin θ.

17. Aug 6, 2016

Milo Martian

Ok so f not equal to mgsinA. But Fext= mgsinA, right? so acm = gsinA for both the spheres?

18. Aug 6, 2016

haruspex

What is Fext? If you mean Fnet, no.
No.

Quit guessing and follow proper procedure:
• There are two forces parallel to the plane, mg sin A and Ffriction
• What is the net force?
• Write the equation for linear acceleration
• What is the torque about the mass centre?
• Write the equation for angular acceleration
• Because it is rolling contact, the angular acceleration and linear acceleration are related. Write the equation for that.

19. Aug 6, 2016

Milo Martian

Fext= net force external to the system of incline+ sphere. isn't friction internal to the system?if yes then the only external force is mgsinA, isn't it?

20. Aug 6, 2016