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: Rotational Kinetic Energy of a rod

  1. Mar 23, 2008 #1
    1.A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.


    What fraction of the original kinetic energy is converted into internal energy in the collision?

    Here is my attempt:

    Ki = (1/2)mv^2

    kf = (m+M) L^2

    Now I'm a bit confused what they mean by fraction, so I initial thought I had to divide Kf/Ki , which is:

    2(m + M)L^2 / (mv^2)

    but WebAssign marked it wrong. I have one more try and I want to try the other way, Ki/Kf but I don't want to get it wrong.

    Can anyone help me and tell me if I'm doing this right or not. Thanks

  2. jcsd
  3. Mar 23, 2008 #2
    Your final kinetic energy equation is wrong. First off, it doesn't have the right units. It looks like you just used I. The final energy should be 1/2 I w^2, where the equation v = Lw holds.
  4. Mar 23, 2008 #3
    Oh okay I see how I got the final kinetic energy wrong. I did only use I. I fixed it, and now its:

    Kf = (1/2)(m + M)v^2

    Can anyone confirm this?

    And if thats correct, how do I find the fraction?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook