Question

A student sits on a rotating stool holding two 2.8-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg · m2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.34 m from the rotation axis. (a) Find the new angular speed of the student.

Answer 1

Answer:

Explanation:

Given

mass of objects m is 2.8 kg

objects is $r=1 m$ away from axis of rotation

angular  velocity $\omega _1=0.75 rad/s$

moment of inertia of stool and student is$=3 kg-m^2$

New distance of objects $r'=0.34 m$

conserving Angular momentum

Initial Angular momentum $L_1=I\omega _1$

$I_1=3+2\times 2.8\times 1^2$

$I_1=3+5.6=8.6 kg-m^2$

$L_1=8.6\times 0.75=6.45$

For Second case

$I_2=3+2\times 2.8\times 0.34^2=3.32 kg-m^2$

$L_2=3.32\times \omega _2$

$L_1=L_2$

$6.45=3.32\times \omega _2$

$\omega _2=1.94 rad/s$