Solving for Linear Speed of Spherical Shell After Rolling

[anigeo]

Homework Statement

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.

Homework Equations

1/2mv²=1/2mv'²+1/2Iω²

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

 

[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.

 

[Doc Al]

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.

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
(Please do not create multiple threads on the same problem!)

If you don't care to use conservation of angular momentum, you can also just use Newton's 2nd law for rotation and translation and a bit of kinematics.