Question

1.An automobile engine slows down from 3500 rpm to 1200 rpm in 2.1 s .Calculate the total number of revolutions the engine makes in this time. Express your answer as an integer.

2.Determine the moment of inertia of a 9.50 kg sphere of radius 0.387 m when the axis of rotation is through its center.

Answer 1

1) initial angular speed(U or _i ) = 3500 rpm

final angular speed (V or _f) = 1200 rpm

time = 2.1 sec = (2.1/60) minute

apply equation of motion in circular form  

S = (1/2) *( V+U)*t

S = (1/2 ) * (3500+1200)*(2.1/60)

S= 82.25 revolution

2) this sphere can be hollow or solid

in case of hollow sphere

moment of inertia = (2/3) *M*R² = 2/3 * 9.50 *0.387² = 0.9485 ​ kg.m²

in case of solid sphere

moment of inertia = (2/5) *M*R² = 2/5 * 9.50 *0.387² = 0.569 kg.m²