Question

A 120-kg hollow spherical ball 1 m in diameter accelerates at a constant rate from rest to 5 rpm in 20 s and then continues to rotate at constant speed. The ball can be treated as a spherical shell.

a. What is the moment of inertia of the disco ball? (3)
b. What is the angular acceleration of the ball in rad/s?? (5)
c. What is the torque delivered by the motor? (2)
d. Through how many turns does the ball rotate during the 20-s acceleration period? (7)
e. What is the tangential speed at t=20 s of a point on the surface of the ball at a distance 0.5 m (measured perpendicularly) from the axle? (3)

Answer 1

Step-by-step explanation:

a )

moment of inertia of hollow ball

= 2 / 3 mR² , m is mass and R is radius of the ball

= 2 / 3 x 120 x .5²

= 20 kg m²

b )

5 rpm = 5 / 60 rps

n = .0833

angular velocity ω = 2πn= 2 x 3.14 x .0833= .523 rad /s

angular acceleration = increase in angular velocity / time

= .523 - 0 / 20

α = .02615 rad /s²

c )

Torque = moment of inertia x angular acceleration

= 20 x .02615

= .523 Nm

d )

θ = 1/2 α t²

= .5 x .02615 x 20²

= 5.23

2π n = 5.23 where n is required number

n = .83