A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 12 in 9.0 . Assume the merry-go-round is a uniform disk of radius 2.2 and has a mass of 650 , and two children (each with a mass of 29 ) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque.

1 answer

Step One: Convert rpm to radians per second

(12 rpm)(2pi rad/1rev)(1min/60s) = 1.26 rad/s = w

Step Two: Determine the Angular Acceleration

w = alpha t + w

1.26 rad/s = alpha (9.0 s) + 0

Alpha = 0.140

Step Three: Determine the moment of inertia of the disk and children

Moment of inertia = Inertia of disk + inertia of child + inertia of child = 1/2mass of disk * r^2 + 2mass of child * r^2.

So:
I = 1/2(650 kg)(2.2)^2 +2(29kg)(2.2)^2

Inertia total = 1573kg + 280.72kg = 1853.72 kg

Step Four: Determine the magnitude of the torque.

torque = Inertia * alpha

torque = (0.140)(1853.72) = 259.5208

So rounded, torque = 260 N.

Some questions also required the following: What force is required at the edge?

If so, force can be found using the following equation:

torque = rFsin(theta)

260N = (2.2)Fsin(90)

Sin(90) = 1

Therefore

F = 260N/2.2

F = 118N

Hope that helps!