Question

A student sits on a rotating stool holding two 5 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.66 rad/sec. The moment of inertia of the student plus the stool is 8 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.31 m from the rotation axis. Calculate the final angular speed of the student.

Answer 1

Answer:$\omega _f=1.185\ rad/s$

Explanation:

Given

mass of objects $m=5\ kg$

Initially mass is at $r=0.9\ m$

Initial angular speed $\omega_i=0.66\ rad/s$

Moment of inertia of student and  stool is $I_s=8\ kg-m^2$

Finally masses are at a distance of $r_f=0.31\ m$ from axis

$I_i=I_p+I_m$

$I_i=8+2\times 5\times (0.9)^2$

$I_i=16.1\ kg-m^2$

Final moment of inertia of the system

$I_f=I_s+I_m$

$I_f=8+2\times 5\times (0.31)^2$

$I_f=8+0.961=8.961\ kg-m^2$

As there is no external torque therefore moment of inertia is conserved

$I_i\omega _i=I_f\omega _f$

$\omega _f=\frac{16.1}{8.96}\times 0.66$

$\omega _f=1.796\times 0.66$

$\omega _f=1.185\ rad/s$