Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Math & Sciences (https://www.askmehelpdesk.com/forumdisplay.php?f=402)
-   -   Physics- Torque on a Merry-go-round (https://www.askmehelpdesk.com/showthread.php?t=287595)

  • Dec 2, 2008, 04:41 PM
    ihatephysics111
    Physics- Torque on a Merry-go-round
    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:
  • Dec 3, 2008, 02:30 PM
    ebaines

    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:


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

  • All times are GMT -7. The time now is 06:34 AM.