Question

A `60.0-kg` woman stands at the western rim of a horizontal turntable having a moment of inertia of `50 kg.m^2` and radius of `2.0 m`. The turntable is initially at rest and free to rotate abot a frictionless, vertical axle through its centre. The woman then starts walking around the rim clockwise` ("as viewed from above the system")` at constant speed of `1.50 m//s` relative to the Earth. The final angular velocity of the woman and the turntable systems.
A. `0.36 rad//s ("counterclockwise")`
B. `1.8 rad//s ("counterclockwise")`
C. `3.6 rad//s ("clockwise")`
D. `0.36 rad//s("clockwise")`

Answer 1

Correct Answer - A
(a) From conservation of angularmomentum for the system of the woman and the turntable, we have `L_f = L_i = 0`,
So `L_f = I_(woman) omega_("woman") + I_("table") omega_("table") = 0`
and `omega_("table") = (-(I_(woman))/(I_("table"))) omega_(woman)`
=`((m_(woman) r^2)/(I_("table")))(v_(woman)/(r))`
=`(m_(woman) rv_(woman))/(I_("table"))`
`omega_("table") = -(60.0 kg(2.00 m)(1.50 m//s))/(500 kg .m^2)`
=` -0.360 rad//s`
or `omega_("table") = 0.360 rad//s ("counterclockwise")`.