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 collision

  1. Aug 11, 2005 #1
    A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at finitial = 152 rpm. A second disk of mass M2 = 247 g and radius R2 = 5 cm, initially not rotating, is dropped on top of the first. Frictional forces act to bring the two disks to a common rotational speed f(final).

    I already found f(final) to 129.2 rpm.

    next the problem ask for the amount of Kinetic energy lost due to friction.

    so i subtract initial KE from final KE to get the change in KE.

    did i approach this wrong? b/c i kept getting the wrong answers....and all the numbers i got were pretty big...
     
  2. jcsd
  3. Aug 11, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks OK. Show how you calculated the KE.
     
  4. Aug 11, 2005 #3

    Fermat

    User Avatar
    Homework Helper

    The two masses rotate, when in final contact, at different angular velocities.

    You didn't use the same (common) angular velocity for both , did you?
     
  5. Aug 12, 2005 #4
    but it says that friction brings the two disks to a common rotational speed final. shouldn't i be using the same though?
     
    Last edited: Aug 12, 2005
  6. Aug 12, 2005 #5

    Doc Al

    User Avatar

    Staff: Mentor

    For the final KE, of course you should use a common rotational speed (which you've already calculated). And for the initial KE, you would use the given initial rotational speed.

    Again: How did you calculate the rotational KE of the disks? (Be sure to use [itex]\omega[/itex] in radians/sec, not rpm.)
     
  7. Aug 12, 2005 #6
    hold on...it says it's rotating about its symmetry axis, doesn't that make I = 3/2 m r^2?

    anyways.....if we assume I to be .5 m r^2, i calculate the KE like this:

    KE initial = .5 I(1) omega(initial)^2 + .5 I(2) omega(initial)^2
    = .5 [.5 m(1) r(1)^2] omega(initial)^2
    = .25 m(1) r(1)^2 omega(initial)^2

    KE final = .5 I(1) omega(final)^2 + .5 I(2) omega(final)^2
    = .25 m(1) r(1)^2 omega(final)^2 + .25 m(1) r(1)^2 omega(final)^2


    KE final - KE initial =...

    convert rpm to rad/s....and plug in numbers....i get a range of different numbers, big to small..
     
  8. Aug 12, 2005 #7

    Doc Al

    User Avatar

    Staff: Mentor

    No. For a disk, [itex]I = 1/2 M R^2[/itex].

    Nothing wrong with this.

    I assume you meant the second term to be for the second disk: m(2), r(2).


    Looks OK to me. How can one method give you a range of answers?
     
  9. Aug 12, 2005 #8

    Fermat

    User Avatar
    Homework Helper

    OK, I think I figured this out.

    When I first read this, I interpreted it as meaning that when the two disks came into contact then (kinetic) friction between the disks would start to rotate the disk M2 until the two disks had the same periferal velocity. (At which point rolling friction would take over.) Had that been the case, then ω2 = 2*ω1. I thought the "rotational" bit must be a typo!!

    Hiowever it seems that in addition to the kinetic friction between the disks, there is also some (internal) friction slowing disk2 down until ω2 = ω1.

    My mistake - sorry 'bout that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rotational collision
  1. Collision and rotation (Replies: 6)

Loading...