# Tough angular motion problem

1. Jul 19, 2006

### nocloud

okay, i've thought about this for a while and still can't figure it out and its absolutely bugging me to death

here's the picture:

Assume that the static friction coefficient between the cylinder and the ground is "u", the gravitational acceleration is "g", the mass of the cylinder is "M", and the radius is "r". What is the maximum force that can be applied without slippage occurring

Last edited: Jul 19, 2006
2. Jul 19, 2006

### Office_Shredder

Staff Emeritus
Well, what forces are preventing the cylinder from rolling?

3. Jul 19, 2006

### Andrew Mason

I don't get this question. It would make more sense if it asked what maximum force can be applied without slippage occurring.

Any force F will cause the cylinder to roll.

AM

4. Jul 19, 2006

### nocloud

AH, oops, my bad, yea, the question is supposed to be what is the maximum force that can be applied without slippage occurring

5. Jul 19, 2006

### quasar987

Under what conditions (force-wise) does an object slides on a surface?

6. Jul 19, 2006

### nocloud

when the torque caused by the force exceeds the torque generated by friction?

7. Jul 19, 2006

### nocloud

ahh, i'm so confused, can somebody help? i'm totally lost now

8. Jul 19, 2006

### Office_Shredder

Staff Emeritus
Draw a free body diagram. The forces are:

F on the top

Force of static friction on the bottom

However, the static friction can only apply so much force, right? So what happens after F is greater than the static friction on the bottom?

9. Jul 19, 2006

### nocloud

it begins to roll?

10. Jul 19, 2006

### Office_Shredder

Staff Emeritus
nocloud, which direction is the static friction pointing in? (Hint: It's the opposite direction of motion).

You should be able to see that the static friction is applied in the correct direction for the cylinder to roll. However, what happens after the static friction is overcome? Then it's kinetic friction... now let's compare the two:

Static friction: Object doesn't slide (hence, it rolls instead)
Kinetic friction: Object slides

11. Jul 20, 2006

### Andrew Mason

Try analysing the forces and torque about the point of instantaneous point of contact. You can see that the force and torque does not change as the cylinder rolls - it is just the point of contact (the fulcrum) moves.

$$\tau_{top} = I_{rim}\alpha = (\frac{1}{2}MR^2 + MR^2)\alpha$$

where $\alpha$ is the angular acceleration of the cylinder about the point of contact, which is just a/R where a is the acceleration of the centre of mass. So:

(1) $$\tau_{top} = \frac{3}{2}MRa$$

Since the net torque produces the angular acceleration of the cylinder about the cylinder's center of mass (which if it rolls is just a/R where a is the acceleration of the centre of mass), we have:

(2) $$\tau_{top} - \tau_{bottom} = I_{cm}\alpha = \frac{1}{2}MR^2\alpha = \frac{1}{2}MRa$$

Subtracting (2) from (1) we have:

(3) $$\tau_{bottom} = MRa \implies F_{bottom} = Ma$$

This makes sense, since the road provides the force which accelerates the centre of mass.

AM

Last edited: Jul 20, 2006
12. Nov 19, 2006

### phalanx123

I don't understand why I(rim)=1/2MR^2+MR^2. shouldn't that be just 1/2MR^2?