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!
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