Angular momentum and velocity

  • #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?
 

Answers and Replies

  • #2
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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.
 
  • #3
Astronuc
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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.
 
  • #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...
 
  • #5
Astronuc
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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].
 
  • #6
Wow I feel dumb :rofl:
Thanks
 
  • #7
Astronuc
Staff Emeritus
Science Advisor
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Don't feel dumb - just be careful. :wink: :biggrin:
 

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