Question

A turntable has a moment of inertia of 3.0 × 10⁻² kg·m² and spins freely on a frictionless bearing at 25 rev/min. A 0.60-kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10 m from the center. By what factor does the kinetic energy of the system change after the putty is dropped onto the turntable?

a. 1.0

b. 0.83

c. 1.5

d. 0.91

Answer 1

first we need to find angular speed of turntable after  drop of putty

apply  angular momentum conservation

W = 25 rev/min = 25*2pi/60 =2.61799388 rad/s

I*W = If*Wf

{ If = I-table +I-putty = 3.0 × 10−2 + m*r^2 }

= 3*10^-2*2.61799388 = ( 3*10^-2 +0.60*0.10^2) *Wf

Wf = 3*10^-2*2.61799388 /  ( 3*10^-2 +0.60*0.10^2) =2.18166157 rad/s

now  

KE initial = 1/2*I*Wi^2 = 0.5*3*10^-2*2.61799388^2 =0.102808379 J

KEfinal = 1/2*If*Wf^2 = 0.5* ( 3*10^-2 +0.60*0.10^2)*2.18166157^2 =0.0856736497 J

change factor = 0.0856736497/0.102808379 = 0.833

b. 0.83​

let me know in a comment if there is any problem or doubts