A student sits on a freely rotating stool holding dumbbells, each of mass `5.0 kg` (Fig). When his arms are extended horizontally (Fig a), the dumbbells are `1.0 m` from the axis of rotation and the student rotate with an angular speed of `1.0 rad//s`. The moment of inertia of the student plus stool is `5.0 kg.m^2` and is assumed to be constant. The student pulls the dumbbells inward horizontally to a position `0.50 m` from the rotationa are (Fig.) The new angular speed of the student is.
.

Options

A. `1.5 rad//s`
B. `2.5 rad//s`
C. `2.0 rad//s`
D. `1.25 rad//s`

Answer

C

Explanation

( c) The total angular momentum of the system of the student, the stool, and the weight about the axis of rotation is given by
`I_(total) = I_(weights) + I_(student) = 2(mr^2) + 5.0 kg.m^2`
Before : `r = 1.0 m`
Thus, `I_i = 2(5.0 kg)(1.0 m)^2 + 5.0 kg.m^2 = 15 kg.m^2`
After : `r = 0.50 m`
Thus, `I_i = 2(5.0 kg)(0.50 m)^2 + 5.0 kg.m^2 = 7.50 kg.m^2`
We now use conservation of angular momentum
`I_f omega_f = I_i omega_i`
or `omega_f = ((I_i)/(I_f)) omega_i = ((15.0)/(7.5))(1.0 rad//s) = 2 rad//s`.