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!

Angular momentum and velocity

  1. Nov 19, 2005 #1
    A large horizontal circular platform (M=81.1 kg, r=3.43 m) rotates about a frictionless vertical axle. A student (m=56.3 kg) walks slowly from the rim of the platform toward the center. The angular velocity w of the system is 4.90 rad/s when the student is at the rim. Find w (in rad/s) when the student is 1.67 m from the center.
    [tex] L_o= L_f [/tex]
    [tex] (1/2)MR^2 * \omega = (MR^2)(\omega) [/tex]
    [tex] (1/2) (81.1)(3.43^2) = (56.3)(1.67^2)(\omega) [/tex]
    Solving for omega gave me 6.1 rad/s which wasn't right.
    Can someone help me?
     
  2. jcsd
  3. Nov 19, 2005 #2
    The student is part of the system. So the initial rotational inertia and thus angular momentum has to take the student's rotational inertia into account. Then the student will move giving rise to a new rotational inertia for the system.
     
  4. Nov 19, 2005 #3

    Astronuc

    User Avatar

    Staff: Mentor

    The disc and the student represent a composite system, so one must consider the angular momentum of the disc and the student.

    The disc has moment of inertia 1/2mr2 and the student has moment of inertia msrs2, and if the student rotates, one must also consider that effect as well, but one would need the effective diameter of the student.

    See this discussion for a composite system and superposition of moments of inertia - http://hyperphysics.phy-astr.gsu.edu/hbase/tdisc.html#icomp

    Assume conservation of angular momentum applies as you did.
     
  5. Nov 19, 2005 #4
    Ok so I tried doing,
    [tex] L_o= (1/2)MR^2 + MR^2 * \omega[/tex]
    [tex] (1/2)(81.1)(3.43^2) +(56.3)(3.43^2) *4.90 [/tex]
    [tex] L_o= 3284 [/tex]
    [tex] L_f= (1/2)MR^2 + MR^2 *\omega[/tex]
    [tex] (1/2)(81.1)(3.43^2) +(56.3)(1.67^2) *\omega[/tex]
    Solving for omega gave me 5.04 rad/s, which isn't right...
     
  6. Nov 19, 2005 #5

    Astronuc

    User Avatar

    Staff: Mentor

    Make sure you group the terms correctly.

    Try

    [tex] L_o= ((1/2)MR^2 + MR^2) * \omega[/tex]

    Remember L = I x [itex]\omega[/itex] and I = [itex]\Sigma_i\,I_i[/itex].
     
  7. Nov 19, 2005 #6
    Wow I feel dumb :rofl:
    Thanks
     
  8. Nov 19, 2005 #7

    Astronuc

    User Avatar

    Staff: Mentor

    Don't feel dumb - just be careful. :wink: :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Angular momentum and velocity
Loading...