Question

A hollow, thin-walled sphere of mass 11.0 kg and diameter 48.0 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by θ(t)=At2+Bt4, where A has numerical value 1.20 and Bhas numerical value 1.60.

At the time 3.00 s , find the angular momentum of the sphere.

At the time 3.00 s , find the net torque on the sphere.

Answer 1

Mass of the hollow sphere = M = 11 kg

Diameter of the hollow sphere = D = 48 cm = 0.48 m

Radius of the hollow sphere = R = D/2 = 0.24 m

Moment of inertia of the hollow sphere = I

I = 2MR²/3

I = (2)(11)(0.24)²/3

I = 0.4224 kg.m²

The angle is given by,

(t) = 1.2t² + 1.6t⁴

Angular velocity = = d/dt

Differentiating the angle equation with respect to time we get,

(t) = 2.4t + 6.4t³

Angular acceleration = = d/dt

Differentiating the angular velocity equation with respect to time we get,

(t) = 2.4 + 19.2t²

At time t = 3 sec

(t) = 2.4t + 6.4t³

(3) = 2.4(3) + 6.4(3)³

(3) = 180 rad/s

(t) = 2.4 + 19.2t²

(3) = 2.4 + 19.2(3)²

(3) = 175.2 rad/s²

Angular momentum of the sphere at time t=3 sec = L

L = I(3)

L = (0.4224)(180)

L = 76.032 kg.m²/s

Net torque on the sphere at time t=3 sec = T

I(3) = T

(0.4224)(175.2) = T

T = 74.005 Nm

a) Angular momentum of the sphere at time t=3 sec = 76.032 kg.m²/s

b) Net torque on the sphere at time t=3 sec = 74.005 Nm