# Rolling and sliding ring

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1. Mar 27, 2017

### vijayram

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

A uniform ring of mass m and radius r is projected horizontally with velocity v on a rough horizontal floor,so that it starts off with a pure sliding motion and it acquires a pure rolling motion after moving a distance d.If the coeffcient of friction between ground and ring is a,the work done by the friction in this process is,

2. Relevant equations
w = fd

3. The attempt at a solution

Since the ring starts with a pure sliding motion,the point of contact should move,since it is moving f =aN,N
=mg
so, f=amg
and it begins to roll,after rolling motion starts,friction won't do any work so w=fd.
so work done by friction is -amgd,but given answer is -0.5mv2

Please tell me where have I gone wrong.

2. Mar 27, 2017

### Baluncore

What is the initial kinetic energy of the ring as it is projected?

3. Mar 27, 2017

### vijayram

Sir,since initially no rotation motion would be present so it must be 1/2mv2

4. Mar 27, 2017

### Baluncore

Where will that energy go?

5. Mar 27, 2017

### BvU

How did you convince yourself your answer is really wrong ? It may well be that it is in fact the same answer, only expressed in different variables ...

6. Mar 28, 2017

### haruspex

Yes, if they wanted the answer in terms of m and v they should have said so, or not provided the variable names d and a.
But surely there is something wrong with the given answer. The initial energy is mv2/2. If friction does work -mv2/2 there should be none left, yet it is rolling.

7. Mar 28, 2017

### BvU

From the problem statement you could read that v is kept constant. But it takes some liberty...

8. Mar 28, 2017

### haruspex

Too much liberty for me. It says projected, not dragged.
I think the answer is supposed to be -mv2/4.

9. Mar 28, 2017

### Baluncore

Yes, I agree, the correct answer should be – 0.25 m v2
Half the initial energy is converted to rotational energy.

10. Mar 28, 2017

### haruspex

No, half is lost to friction and a quarter converted to rotational.

11. Mar 28, 2017

### BvU

Agree. How about interpreting $v$ as the final velocity

12. Mar 28, 2017

### vijayram

Thank you very much sir for replying but it actually happened that both options were there in the question when my answer matched with one of those,I didn't check the others.

13. Mar 28, 2017

### zwierz

-

14. Mar 28, 2017

### haruspex

Are you saying it was multiple choice, with -mv2/2 being only one of the possible answers?

15. Mar 29, 2017

### vijayram

yes sir,the others being none of the above and insufficient information

16. Mar 29, 2017

### haruspex

Strange.
Leaving that aside, what answer can you get expressed only in terms of m and v?

17. Mar 29, 2017

### vijayram

I got -0.25mv2

18. Mar 29, 2017

Right.