Question

Two 0.2-kg masses are located at either end of a 1-m long very light and rigid rod, as shown.

a) What is the rotational inertia of this system about an axis through the center of the rod?

b) What is the rotational acceleration of this system if a net torque of 10 N∙m is applied to it?

Answer 1

Mass located on each end of the rod = m = 0.2 kg

Length of the rod = L = 1 m

The axis of rotation passes through the center of the rod therefore the masses are at a distance equal to L/2 from the axis.

Moment of inertia of the system = I

I = 0.1 kg.m²

Torque applied to the system = T = 10 N.m

Angular acceleration of the system =

I = T

(0.1) = 10

= 100 rad/s²

a) Rotational inertia of the system = 0.1 kg.m²

b) Rotational acceleration of the system =100 rad/s²