Question

A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4.7 kg⋅m2 . A second student tosses a 1.7 kg mass with a speed of 3.0 m/s to the student on the stool, who catches it at a distance of 0.50 m from the axis of rotation.

A) Calculate the initial kinetic energy of the system.

B) Calculate the final kinetic energy of the system.

Answer 1

moment of inertia of student stool system I = 4.7 kg.m²

m = 1.7 kg

v = 3m/s

distance , r = 0.50 m

law of conservation of momentum

L1 = l2

m*v*r = ( I+m*r² )*w

w = m*v*r /(I +m*r²)

w = 1.7*3*0.50 / ( 4.7 + 1.7*0.5*0.5)

w = 0.497 rad/s

(1)

initial kinetic energy

Ki = 1/2*m*v² = 0.5*1.7*9

Ki = 7.65 J

(b)

final kinetic energy

Kf = 1/2*I*w² = 0.5*4.7*(0.497)²

Kf = 0.580 J