Question

A puck rests on a horizontal frictionless plane. A string is wound around the puck and pulled on with constant force. What fraction of the disk's total kinetic energy is due to the rotation? KErot/KEtot

Answer 1

Answer:

Explanation:

Let v be the linear velocity ,  ω be the angular velocity  and I be the moment of inertia of the the puck.

Kinetic energy ( linear ) = 1/2 mv²

Rotational kinetic energy = 1/2 I ω²

I = 1/2 m r² ( m and r be the mass and radius of the puck )

Rotational kinetic energy = 1/2 x1/2 m r² ω²

= 1/4 m v² ( v = r ω )

Total energy

= Kinetic energy ( linear ) + Rotational kinetic energy

= 1/2 mv² +  1/4 m v²

= 3/4 mv²

rotational K E / Total K E = 1/4 m v² / 3/4 mv²

= 1 /3

So  1 /3  rd of total energy is rotational K E.