1. Not finding help here? Sign up for a free 30min 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!

Rotational Work

  1. Nov 12, 2009 #1
    The pulley in the illustration is a uniform disk of mass 2.40 kg and radius 0.220 m. The block applies a contant torque to the pulley, which is free to rotate without friction, resulting in an angular acceleration of magnitude 0.180 rad/s2 for the pulley. As the block falls 0.500 m, how much work does it do on the pulley?

    The illustration is of a pulley with a rope hanging a block down vertically.

    Here is my attempt:

    work = torque x angular diplacement

    If the block falls then the pulley turns by the same amount, right?

    so arc length = .500m

    theta = arc length / radius = .500 / .220 = 2.27 radians = angular displacement?? (is this right?)

    To find torque, I need F x r

    r = .220

    How do I find the Force?

    Do I use the mass = 2.4 kg and multiply by tangential acceleration?
     
  2. jcsd
  3. Nov 12, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    Yes. It is tangential acceleration.
    How to find out the tangential acceleration?
     
  4. Nov 12, 2009 #3
    tangential acceleration is angular acceleration times radius

    but what about angular displacement? Did I do that right?
     
  5. Nov 12, 2009 #4

    rl.bhat

    User Avatar
    Homework Helper

    Your angular displacement is correct.
    Now the torque = moment of inertia*angular acceleration.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rotational Work
  1. Rotational Work (Replies: 1)

Loading...