A1.0 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

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