# Physics problem involving rotational motion.

1. Oct 8, 2011

### shivamparke

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

The question is given in the following pictures.
http://imageshack.us/photo/my-images/851/iitjee2011.png/

http://imageshack.us/photo/my-images/844/iitjee20112.png/

This was asked in IIT-JEE 2011 (which is regarded by some as the toughest engineering entrance exam in the world)
The problem is that,after much debate,it was deemed as WRONG (I don't know why).Although the institute's website displayed either A or BOTH A and C as the correct answer(s),grace marks were given to all the students who attempted it.
All I want to know is the MISTAKE in this problem.

2. Relevant equations

I am using - "M,r,v and ω1 and ω2" as the mass,radius,velocity and angular velocity of the ring before and after the collision respectively.
"m,Vx,Vy,x and h" as the mass of the ball,velocity of the ball before collision(x-direction),velocity of the ball after collision (y-direction),perpendicular distance between center of ring and the direction of ball's velocity after the collision and the height above the ground at which the ball hits the ring,respectively.
Given : M = 2 kg , r = 0.5 m , v = 1 m/sec ; m = 0.1 kg , Vx = 20 m/sec , Vy = 10 m/sec , h = 0.75 m , x = sq.root of (1/4 - 1/16) = 0.433 m .
Also I is the moment of inertia of the ring (Mr^2) = 0.5 kg m^2

3. The attempt at a solution
Taking the left direction as negative,upward as positive,anticlockwise rotation as positive and applying the law of conservation of angular momentum,

→ Iω1 + m(Vx)(h-r) - Mvr = m(Vy) + Iω2

→ ω1=v/r ( pure rolling ) = 2 rad/sec
→ 0.5 = 0.433 + Iω2
→ ω2 = 0.134 rad/sec (is in the anticlockwise direction because it is positive) .....1
Applying law of conservation of linear momentum,
→ mV - Mv = 2-2 = 0
So linear velocity of ring after collision will be zero ..... 2
From results ....1 and ....2 we can say that the ring will not move but will only rotate about it center of mass.So option A is correct(according to me)
I think option C is also correct becasue friction will be towards left as it has the ring will have a tendency to slip towards right.
What is confusing me is that during pure rolling(as mentioned in option A) friction does not act so after the collision it shall first slip and then purely rotate about its com .
I want to if I am wrong anywhere and also what is WRONG with this problem ?
Thanks.

2. Oct 8, 2011

### PeterO

If the ring is in pure rotation (no translation) then the bottom of the ring must be slipping on the surface. While it was rolling, it was not slipping, so had no friction force, but while it is rotating there should be a friction fore to the Left.

Clearly the answer will be A & C or B & D.
I see the problem as one of establishing that translational momentum is reduced to zero, but rotational momentum is not - as you did.

3. Oct 9, 2011

### shivamparke

Due to friction (towards left) there will be a clockwise torque and since the ring is rotating initially in the anticlockwise direction it will first come to a stop and then rotate (without slipping) in the clockwise direction.Am I right ?

4. Oct 9, 2011

### PeterO

I don't think so.

The friction force to the Left will gradually accelerate the hoop to the left, so that it eventually resumes rolling to the Left. Not enough information to determine whether it will regain its original speed - it might even end up travelling faster.

I would expect the hoop to behave in a similar way to a cue ball in billiards that has rapid spin before striking another ball full on.
The cue ball first stops [elastic collision with an equal mass object] then the effect of the spin takes over and the cue ball either comes back (if you used back spin) or follows through (if you used top spin)

The back spin example is easy to replicate with a hula hoop. You can toss it forward with rapid back-spin. It first stops, then runs back to you.
The hoop in this problem stops because of the collision with the other mass, not by the cumulative affect of the friction.

5. Oct 9, 2011