Question

An ice skater has a moment of inertia of 5.0 kg · m2when her arms are outstretched, and at this time she is spinning at 3.0 rev/s. If she pulls in her arms and decreases her moment of inertia to 2.0 kg · m2, how fast will she be spinning?

Answer 1

Answer:

Explanation:

Given

Initial Moment of Inertia $I_1=5 kg-m^2$

initial Spin $N_1=3 rev/s$

$\omega _1=2\pi N_1=2\pi 3=6\pi rad/s$

Final Moment Moment of Inertia $I_2=2 kg-m^2$

Conserving Angular momentum

$L_1=L_2$

$I_1\omega _1=I_2\omega _2$

$5\times 6\pi=2\times \omega _2$

$\omega _2=15\pi rad/s$

$N_2=\frac{\omega _2}{2\pi }$

$N_2=\frac{15\pi }{2\pi}=7.5 rev/s$