# Homework Help: A question of rolling?

1. Feb 25, 2015

### Rishavutkarsh

1. The problem statement, all variables and given/known data
A force(Fi) is acting on the top point of a disc of radius r and mass m. The disc is rolling without slipping. Angular velocity of disc after center has been displaced distance x is?

2. Relevant equations
Energy conservation; Moment of inertia of disc (MR^2)/2

3. The attempt at a solution

By energy conservation
Fx=1/2*Mr^2/2*w^2+1/2mv^2
also v=rw (Rolling)

What am I missing? Any help appreciated.

2. Feb 25, 2015

### Staff: Mentor

It look like you are using work done = mechanical energy gained
That should work. Now go ahead and solve for angular velocity.

3. Feb 25, 2015

### PeroK

Beaten to it by NO!

4. Feb 25, 2015

### haruspex

For Fx to be the work done, what must the relationship between the force F and the distance x be? Is that their relationship in this question?

5. Feb 25, 2015

### Rishavutkarsh

Well I thought so too but that didn't really work out, I got the wrong answer. The answer is root 2 times my answer.
They should be in the same direction and yes they are.

6. Feb 25, 2015

### PeroK

Is F the only force acting on the disc?

7. Feb 25, 2015

### Rishavutkarsh

No, friction must be acting too since it's rolling but it wont do any work during rolling.

8. Feb 25, 2015

### PeroK

Even so, you could take a look at the equations of motion involving both F and the frictional force.

9. Feb 25, 2015

### Rishavutkarsh

F-fr=ma
FR+frR=I*(alpha)
Using this I got the answer... Hurray!!! But I am still curious to know why did energy conservation give me wrong answer.

10. Feb 25, 2015

### PeroK

First, work out over what distance F would have to act to generate the correct KE. Then, see whether you can explain why F does indeed act over that distance.

11. Feb 25, 2015

### Rishavutkarsh

2x? Well yeah that gives me the answer but how does F 'indeed act over that distance'? A hint would help, I can't think of anything.

12. Feb 25, 2015

### PeroK

Perhaps a couple of ways to look at it:

If you're applying a force to something that's moving, then the work done by the force in time $dt$ is $Fvdt$. The top of the disk is moving twice as fast as the centre of mass of the disk. So, the work done by a force applied at the top of the disc is $F2vdt$ (where v is the velocity of the centre of mass).

Alternatively, if you think of the force as a continuous series of impulses $Fdt$ (applied at the top of the disk), you see the same result. Each impulse acts over the same time, but twice the distance, that a similar impulse at the centre of the disk would do.

If you redo your problem with the force at the centre of the disc, you can use work-energy directly over the distance x. But, at the top of the disc, the force acts through a distance of 2x.

It's similar to why the friction force does no work, as the bottom of the disc is stationary when in contact with the ground.

13. Feb 25, 2015

### haruspex

Yes, but there's a bit more to it than that, as PeroK notes, x must be the distance the force acts over:
Suppose you push on a lever with force F, and you make the lever (at the point where you push it) move distance x. The work you have done is Fx. Had you applied the force twice as far from the fulcrum point the force needed would only have been F/2, but you would have needed to advance the point of application 2x in order to achieve the same movement of the lever.

14. Feb 26, 2015

### Rishavutkarsh

Thank you both of you for helping me out I was able to think it as the point above moving twice as fast as center but the analogy of lever and the ground being stationary so that friction does no work helped me understand much more clearly.