A 1.0 kg ball and a 2.0 kg ball are connected by a 1.0-m-long rigid, massless rod. The rod is rotating cw about its center of mass at 20 rpm. What torque will bring the balls to a halt in 5.0 s?

torque is 0.2324 N-m

Explanation:

given data

mass m1 = 1 kg

mass m2 = 2 kg

length r2 = 1.0 m

angular velocity = 20 rpm = 20 * 2π/60sec = 2.092 rad/s

to find out

What torque will bring the balls to a halt in 5.0 s

solution

we know center of mass of rod is = m1r1 + m2r2 / (m1+m2)

= 1 (0) + 2 (1.0) / (1+2)

= 2/3 = 0.666 m

so distance from center of rod (d) = 1 - 0.66 = 0.333

so that we say

distance of axis rotation to m1 is r1 i. e = 0.666

distance of axis rotation to m2 is r2 i. e = 0.334

so we now find moment of inertia of rod about center of mass

inertia = m1r1² + m2r2²

inertia = 1 (0.666) ² + 2 (0.334) ²

inertia = 0.4435 + 0.1115 = 0.555 kg m²

so

angular acceleration = angular velocity / time

angular acceleration = 2.0932 / 5 = 0.4187 rad/s²

we apply torque formula

torque = inertia * angular acceleration

torque = 0.555 * 0.4187

torque is 0.2324 N-m