Question

A horizontal circular platform (M=92.1kg, r=3.07m) rotates about a frictionless vertical axle. A student (m=92.3kg) walks slowly from the rim of the platform toward the center. The angular velocity w of the system is 4.1rad/s when the student is at the rim. Find w when the student is 1.47m from the center. Use "rad/s" as units.

Answer 1

  this is a conservation of angular momentum problem, where the angular momentum is the same when the student is at the rim as when the student is 1.47 m from the center

this occurs because the student exerts no torques on the platform, so we have
ang momentum before = ang momentum after

the angular momentum of the system is I w, where I is the moment of inertia and w is the angular velocity

the platform has moment of inertia 1/2 MR^2, and the student, a point mass, contributes an amount mr^2 to the moment of inertia where m, r are the mass and distance of the student from the center, so at the beginning we have

Ibefore = 1/2 MR^2 + mR^2 = R^2(1/2 M + m) =3.07^2*(0.5*92.1 + 92.3) = 1304 kg-m^2

I after = 1/2 MR^2 + mr^2 = 1/2 * 92.1kg*(3.07m)^2 + 92.3kg*(1.47m)^2 = 633.5 kg-m^2

so:

1304kgm^2 * 4.1 rad/s = 633.5 kg-m^2 * wafter

wafter = 8.4 rad/sec.........Answer

Hope it will help you.

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