Question

(a) Calculate the angular momentum (in kg-m/s) of an ice skater spinning at 6.00 rav/s given his moment of inertia is 0.470 kg m? kg-m/s (b) He reduces his rate of spin (his angular velocity) by extending wis arms and increasing his moment of inertia Find the value of his moment of inertia (in kg) ir his angular velocity drops to 2.05 rev/s. kgim² (c) Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque (in Nm) was exerted if this takes 18.0 seconds? N-m

Answer 1

(a) Given the angular frequency of rotation of ice skater f = 6rev/s and the moment of inerti of the ice skater I = 0.470kgm^2. The angular velocity of the ice skater is given by,

$\omega = 2\pi f = 2\pi \times 6=37.7rad/s$

The angular momentum of the ice skater is given by,

$L=I\omega = 0.470\times 37.7=17.72kgm^2/s$

So the angular momentum of the ice skater is 17.72kgm^2/s.

(b) Now the angular velocity drops to $\omega'$ = 2.05rps. Now,

$\omega'=2\pi \times 2.05=12.88rad/s$

Let I' be the new moment of inertia. Now by law of conservation of angular momentum,

I'w=1 - Iw

$I'= \frac{I\omega}{\omega'}=\frac{17.72}{12.88}=1.37kgm^{2}$

So the new moment of inertia is 1.37kgm^2.

(c) Now the ice skater slows his angular velocity to 3 rev/s. Therefore,

$\omega'=2\pi\times 3= 18.85rad/s$

Now the angular acceleration $\alpha$ after t = 18s is,

$\omega'= \omega-\alpha t$

$\alpha =\frac{ \omega-\omega'}{t}=\frac{37.7-18.85}{18}=1.05rad/s^{2}$

Now the torque is given by,

$\tau=I\alpha=0.470\times 1.05=0.49Nm$

So the torque exerted is 0.49Nm.