# Homework Help: Rotational Mechanics Problem again

1. Dec 6, 2012

Hi friends the problem states like this:

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/s480x480/68815_2639239357533_424911713_n.jpg

When the attached block is horizontal it is applying torque on the ring. Here I am using the formula T = Iα

So,

mg. r = (mr2(ring) + mr2(block))α

so α = g/2r

But it is not the correct answer. The correct answer is g/3r.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 6, 2012

### Saitama

You need to take the torque due to friction force too.

And check your moment of inertia for ring.

3. Dec 6, 2012

### BruceW

hey, man! What values have you used for the mass of the block and the mass of the ring?

4. Dec 7, 2012

Sorry Bruce It happened by mistake.

Taking mass of the ring as 3m the ans is g/4R.

5. Dec 7, 2012

Hi Pranav.

But here μ is also absent. Now about which axis I should take torque ? Because If I am taking it about the central axis μ is also there.

6. Dec 7, 2012

### Saitama

The body rolls without slipping/sliding implies that there is force due to friction too.

You have to take the torque about the CM of ring.

7. Dec 7, 2012

### ehild

It is static friction, which needs to be enough to prevent the bottom point of the ring from sliding. No need to know μ.

ehild

8. Dec 7, 2012

SO what should I do after that. What should I apply to solve it Not getting exactly. Please Help.

9. Dec 7, 2012

mg.r + μ. mg r = (mr2 + 3m r2). α

now sticking with μ

10. Dec 7, 2012

### Saitama

Now make an equation for the translational motion.

11. Dec 7, 2012

Well I did some wrong in the previous equation,

Right one -

mgr + μ. 4mgr = (mr2 + 3mr2) α

i.e.
g + 4g. μ = 4r.α

For translational motion,

4μ mg = 3ma = 3 mrα

Solving both equations,

α = g/r

Please tell me what i am doing wrong now.

12. Dec 7, 2012

### tiny-tim

don't forget that the total centre of mass is moving downward, and the total moment of inertia is changing!

you'll probably need to consider the ring and the mass separately

13. Dec 7, 2012

### Saitama

I should have been a bit more clear in my previous post. You don't know the frictional force here, so instead of μ. 4mgr, write fr where f is the frictional force.

For translational motion, f=ma=mrα (since its given that it rolls without slipping).

Solve the two equations.

14. Dec 8, 2012

### ehild

This time it is easier to use the instantaneous axis.

Apply the parallel axis theorem to get the moment of inertia of the ring with respect to that axis, and add the moment of inertia of the small block.

ehild

15. Dec 9, 2012

So as per you guys instructions I am doing it with full consciousness.

Taking τ = Iα about the center of mass of the ring,

mg.r + f.r = (3mr2 + mr2

takinking F = ma for the com of the ring,

f = 3m.a = 3mr.α

solving both the equations I get the result,
α= g/r.

16. Dec 9, 2012

### ehild

What can be the direction of the force of friction? Think: when the ring rolls clockwise, the centre accelerates to the right. It is only friction that accelerates the ring and block to the right.

The force of friction accelerates the whole system, ring and block. That is 4m.

And I think the given answer (g/3r) is not correct.

ehild

Last edited: Dec 9, 2012
17. Dec 9, 2012

### TSny

I second echild's recommendation of using the instantaneous axis of rotation!

18. Dec 9, 2012

### ehild

Anyway, I do not get the given result. Do you?

ehild

19. Dec 9, 2012

### TSny

Me either. But I do get one of the other choices.

20. Dec 9, 2012

### tiny-tim

thirded!

(and i too don't get the given answer)