Question

A skater has rotational inertia 4.2 kg.m2 with his fists held to his chest and 5.7 kg.m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they’re essentially on his ro- tation axis, how fast will he be spinning?

Answer 1

Answer:

6.13428 rev/s

Explanation:

$I_f$ = Final moment of inertia = 4.2 kgm²

I = Moment of inertia with fists close to chest = 5.7 kgm²

$\omega_i$ = Initial angular speed = 3 rev/s

$\omega_f$ = Final angular speed

r = Radius = 76 cm

m = Mass = 2.5 kg

Moment of inertia of the skater is given by

$I_i=I+2mr^2$

In this system the angular momentum is conserved

$L_f=L_i\\\Rightarrow I_f\omega_f=I_i\omega_i\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{(5.7+2\times 2.5\times 0.76^2)3}{4.2}\\\Rightarrow \omega_f=6.13428\ rev/s$

The rotational speed will be 6.13428 rev/s