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!

Kinetic Energy Ratio/Rotational Equilibrium

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Two equal masses, each m, are resting at the ends of a uniform rod of length 2a and negligible mass. The system is in equilibrium about the center C of the rod. A piece of clay of mass m is dropped down on the mass at the right end, hits it with velocity v as shown below and sticks to it.

    The ratio of the kinetic energy Ef just after the collision to the kinetic energy Ei just before the collision, Ef/Ei, would be:

    (A) 1
    (B) 3/4
    (C) 2/3
    (D) 1/2
    (E) 1/3

    2. Relevant equations

    Not really sure.

    3. The attempt at a solution

    If the system is initially in equilibrium, then it's not moving, and I don't understand how it could have kinetic energy if it's not moving. I want to write: Ef/0, but that's obviously wrong.
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2012 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Victorzaroni! :smile:
    "The system" includes the clay, and that is moving! :wink:
     
  4. Feb 21, 2012 #3
    ooohhh. wow i feel dumb. that makes sense. so its the initial kinetic energy of the system is the kinetic energy of the clay.
     
  5. Feb 21, 2012 #4
    So Ei=(1/2)mv2. Now what? I know torque is involved somehow.
     
  6. Feb 21, 2012 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    torque has nothing to do with it :confused:

    now find Ef (as a function of vf)
     
  7. Feb 21, 2012 #6
    Energy is a function of velocities (kinetic) and positions (potential), energy isn't a function of accelerations and torques
     
  8. Feb 22, 2012 #7
    Alright... So the new kinetic energy after the clay has stuck to the mass would be KE=(1/2)(M+M)r2? How do I account for the mass on the other end of the board?
     
  9. Feb 22, 2012 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    add it on! :smile:

    (it'll have the same speed, won't it? :wink:)​
     
  10. Feb 22, 2012 #9
    Alright so I have: KEi=(1/2)Mv^2, and KEf=(1/2)(M+M+M)v^2. Dividing KEf by KEi yields (3/2)/(1/2), which is 3, which is not a choice? Dividing them the other way yields 1/3, choice E...
     
  11. Feb 22, 2012 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    i'm lost :redface:

    what conservation equation are you using? :confused:
     
  12. Feb 22, 2012 #11
    I'm not using conservation, KEf/KEi is what the question is asking for and the other stuff is what you've been saying. lol
     
  13. Feb 23, 2012 #12

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    how are you going to find the relation between before and after without a conservation law? :confused:
     
  14. Feb 23, 2012 #13
    So how do i use a conservation law?
     
  15. Feb 23, 2012 #14

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    what do you think will be conserved?

    (and i'll repeat my previous question, since i'm still wondering …

    how were you intending to find the relation between before and after without a conservation law?)​
     
  16. Feb 23, 2012 #15
    The kinetic energy of the clay equals the rotational kinetic energy of the system?
     
  17. Feb 23, 2012 #16

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    no, this is a perfectly inelastic collision
    … so energy is not conserved
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Kinetic Energy Ratio/Rotational Equilibrium
  1. Kinetic Energy Ratios (Replies: 2)

Loading...