1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Angular speed of merry-go-round

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A playground merry-go-round has a radius 2.40 m and a moment of inertia 2100 kgm^2 about a vertical axle through its center, and it turns with negligible friction.

    A child applies an 18.0 N force tangential to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s interval?

    2. Relevant equations
    [tex]\vec \tau = \vec r \times \vec F[/tex]

    3. The attempt at a solution

    I have absolutely no idea what to do. I know the answer is .309 rad/s. From playing around with the numbers I know that (2.40 m * 18.0 N * 15.0 s)/2100 kg*m^2 = 0.309 rad/s, but I don't know why. I can't find any sort of relationship between what I have and what I need.
  2. jcsd
  3. Mar 24, 2010 #2
    Immediately after clicking post, it clicked in my mind. Wow. For anyone else who is having trouble with this problem:

    \vec \tau = \vec r \times \vec F\\
    \sum \vec \tau = I \vec \alpha\\
    \omega_f = \omega_i + \alpha t\\\\
    \vec \tau = (2.40 m)(18.0 N)(sin 90) = 43.2 N \cdot m\\
    \sum \vec \tau = 43.2 N \cdot m = (2100 kg \cdot m^2)(\alpha)\\
    \Rightarrow \alpha = 0.021 rad/s^2\\
    \omega_f = 0 + (0.021 rad/s^2)(15.0 s)\\
    \Rightarrow \omega_f = 0.309 rad/s
    Last edited: Mar 24, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook