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View Full Version : Physics- Torque on a Merry-go-round


ihatephysics111
Dec 2, 2008, 04:41 PM
A teenager pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 15 rpm in 9.6 s. Assume the merry-go-round is a uniform disk of radius 2.3 m and has a mass of 730 kg, and two children (each with a mass of 27 kg) sit opposite each other on the edge.Calculate the torque required to produce the acceleration, neglecting frictional torque. Then find the force required at the edge.

What I have established / found (I think)

Vi= 0
Vf= .25 rps

Moment of Inertia of a uniform disk is I=1/2 M R^2
Moment of Inertia of the merry-go-round without children = 1930.85
Moment of Inertia with children = 2073.68

I know net torque is Tnet = I * alpha, but I don't know what alpha is...

I also know that to find the force it is Fnet = (I * alpha)/r

or so I think...

so what is alpha and how do I find it? Help please! :confused:

ebaines
Dec 3, 2008, 02:30 PM
Alpha is the rotational acceleration, typically in units of radians per second per second. You have found that the final rotational velocity is 0.25 rps, which is equivalent to pi/2 radians/sec. It took 9.6 s to accelerate up to this speed, so assuming constant acceleration:

alpha = (pi/2 radians/sec) /9.6 sec

Remember that for rotational veocity and acceleration:

\omega = \alpha t


This is analogous to the linear acceleration, where for the case of constant acceleration velocity and acceleration are related by: v = at.