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

where is the photo and the answer choices?

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.