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.
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
F = 260N/2.2
F = 118N
Hope that helps!
Answer this Question
physics - A dad pushes tangentially on a small hand-driven merry-go-round and is...
physics - You wish to accelerate a small merry-go-round from rest to a ...
Physics - A horizontal 873 N merry-go-round of radius 1.11 m is started from ...
physics - A horizontal 790-N merry-go-round of radius 1.30 m is started from ...
physics - A horizontal 810-N merry-go-round of radius 1.70 m is started from ...
physics - A merry-go-round in a playground has a mass of 200 kg and radius of 1....
Physics - A merry-go-round with r = 4m and a perfect frictionless bearing is ...
Physics - A merry-go-round in the park has a radius of 1.8 m and a rotational ...
College Physics - A 40 kg child is standing on the edge of a merry-go-round in ...
physics - A merry-go-round in the park has a radius of 1.8 m and a rotational ...