# Rotational Motion Problem - 2

1. Dec 16, 2013

### coldblood

Hi friends,

The problem is as:

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-frc3/q71/s720x720/1471110_1461726330721139_1078073816_n.jpg

Attempt:

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn2/1486585_1461727017387737_725847173_n.jpg

2. Dec 16, 2013

### haruspex

I don't think there's any shortcut here. Write out the usual force/torque/acceleration equations, assuming rolling contact and taking the point of application of F to be some height x above the centre of mass. It doesn't say whether the cylinder is uniform, so maybe put I as the Moment of Inertia for now.

3. Dec 17, 2013

### coldblood

Taking torque about the centre point and let the cylinder to be uniform,

F. (x) + f(R) = (I).α
=> F(x) + f(R) = 1/2 MR2
=>Fx + fR = MR2 α/2 - - - - - - -(1)

Force equation,

F - f = ma (a be the acceleration of c.o.m.) and for no slipping, a = Rα

So, F - f = mRα - - - - - - -- - (2)

4. Dec 17, 2013

### haruspex

So eliminate α to find f as a function of x.

5. Dec 17, 2013

### coldblood

f = F - [2F(R+x)]/ 3MR2

6. Dec 17, 2013

### haruspex

No, that's dimensionally incorrect. Try it again.

7. Dec 17, 2013

### coldblood

Yes, Sorry for this.

The correct one is,

f = F - [F(x + R)]/ 2R ; 0≤x≤R

Here, x can not exceed R.

For x = 0, f = F/2 i.e. translational motion is happening.
For x = R, f = 0 i.e. pure rolling

But for any value of x < R, f will have some positive value. i.e. it will act backwards.

8. Dec 17, 2013

### haruspex

Still not right. You should get a term like 3Rf.

9. Dec 17, 2013

### tiny-tim

hi coldblood!
yes there is

take torque about the bottom of the cylinder (the point of contact and centre of rotation)

[you can do τ = Iα about a moving point provided

i] it is the centre of mass

or

ii] it is the centre of rotation and the centre of rotation is moving in a straight line parallel to the centre of mass ]​

10. Dec 17, 2013

### haruspex

I merely meant that you can't reason it out without getting into equations (as had been attempted in the OP). Yes, some routes to the equations are distinctly faster than others.

11. Dec 17, 2013

### coldblood

are the equations which I wrote in post 3 right?

12. Dec 18, 2013

### coldblood

Yes I got the correct equation,

3Rf = F (R - 2x)

13. Dec 18, 2013

### haruspex

Good. Now, what is the range of possible values of x?

14. Dec 18, 2013

### coldblood

3Rf = F (R - 2x)

x lies between 0 to R
when x = 0, f = F/3
when x = R, f = - f/3
when x = R/2, f = 0

15. Dec 18, 2013

### haruspex

Right. Can you select an answer now?

16. Dec 19, 2013

### coldblood

Well the confusion which is arising is that I got the function

3Rf = F (R - 2x)

x lies between 0 to R
when x = 0, f = F/3
when x = R, f = - f/3
when x = R/2, f = 0

using the condition of pure rolling, I used a = Rα for the bottom point.

17. Dec 19, 2013

### haruspex

Right, and that does match one of A, B, C, D. Which one?

18. Dec 19, 2013

### coldblood

Last one when x = R/2, f = 0, Because during pure rolling contact point does not slip. Hence friction is zero.

19. Dec 19, 2013

### haruspex

No, x can be anything from 0 to 2R. You need an answer that's valid for any x. If x is not R/2 there will be a frictional force.

You said friction = 0 was the last choice. It's the third choice of four.

20. Dec 19, 2013

### coldblood

I got it. That means Any of the choice could be there. Hence Can't be interpreted?