# Pure rolling

1. Dec 13, 2011

### anigeo

1. The problem statement, all variables and given/known data
A thin spherical shell lying on a rough horizontal floor is hit by a cue in such a way that the line of action of force passes through the centre of the shell.as a result the shell starts moving with a linear speed v without any initial angular velocity.find the linear speed of the shell after it starts purely rolling.

2. Relevant equations

1/2mv2=1/2mv'2+1/2Iω2

3. The attempt at a solution
i would use the principle of conservation of energy here as angular momentum is not conserved here because there is an external torque acting on it by friction.

1/2mv2=1/2mv'2+1/2Iω2 (v' is the linear velocity of the cue after it starts purely rolling)
v2=(3/2)v'2 (I=1/2mr22)
But the answer happens to be v'=3v/5
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 13, 2011

### gnulinger

Your moment of inertia is wrong. For a spherical shell, the moment of inertia is
$$I = \frac{2}{3}MR^2$$

Plug this into your equation, and you will get the right answer.

3. Dec 13, 2011

### Staff: Mentor

Mechanical energy is not conserved, so that approach won't work. (There's friction!)

You already have a thread open on this very problem where it was explained that angular momentum is conserved if you choose the correct reference point: https://www.physicsforums.com/showthread.php?p=3665682#post3665682