Question

A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 18 rpm, in 13.0 s. Assume the merry-go-round is a uniform disk of radius 2.7 m; and has a mass of 800 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque. What force is required at the edge?

Answer 1

Part A)

First we need to find the angular velocity is rad/s

Given 18 rev/min with 2π radians /rev and 60 sec/min, the angular velcoity, ω, = 1.88 rad/s

Then, using ω_f = ω_o + αt, we can find the angular acceleration

1.88 = 0 + α(13)

α = .145 rad/s²

The formula for torque is τ = Iα

The moment on inertia, I, is that of the disc, plus the kids,

For a disc the moment of inertia is .5mr²

For each child it will just be their individual mr²

So I = (.5)(800)(2.7)² + (25)(2.7)² + (25)(2.7)² = 3280.5 kgm²

Then..

τ = (3280.5)(.145) = 475.7 Nm

Part B)

The force is fouce from another way to write the torque equation

τ = FL

475.7 - (F)(2.7)

F = 176.2 N