What is the work done by friction in a sliding and rolling ring?

[vijayram]

Homework Statement

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,

[/B]

Homework Equations

w = fd[/B]

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.

 

[Baluncore]

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

 

[vijayram]

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

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

 

[Baluncore]

Where will that energy go?

 

[BvU]

so work done by friction is -amgd,but given answer is -0.5mv²
Please tell me where have I gone wrong.

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 ...

 

[haruspex]

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 ...

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.

 

[BvU]

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

 

[haruspex]

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

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

 

[Baluncore]

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

 

[haruspex]

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

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

 

[BvU]

Too much liberty for me. It says projected, not dragged.

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

 

[vijayram]

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.

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.

 

[zwierz]

-

 

[haruspex]

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.

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

 

[vijayram]

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

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

 

[haruspex]

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

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

 

[vijayram]

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

I got -0.25mv²

 

[haruspex]

I got -0.25mv²

Right.