Conservation of Angular Momentum — Question 1

Question 1

Pulling inward conserves angular momentum while increasing energy

A skater initially has inertia $4.0\,\mathrm{kg\,m^2}$ and angular speed $2.0\,\mathrm{rad/s}$. By pulling inward, the skater reduces inertia to $1.6\,\mathrm{kg\,m^2}$. External torque is negligible.

1. Find final angular speed.

2. Find the change in rotational energy and identify its source.

Question 1 - Solution

Pulling inward conserves angular momentum while increasing energy

(a) Derivation and calculation

Negligible external angular impulse gives $$L=I_i\omega_i=4.0(2.0)=8.00\ \mathrm{kg\,m^2/s}=I_f\omega_f.$$ Thus $$\boxed{\omega_f=\frac{I_i\omega_i}{I_f}=\frac8{1.6}=5.00\ \mathrm{rad/s}}.$$

(b) Result and interpretation

The rotational energies are $$K_i=\tfrac12(4)(2)^2=8.00\ \mathrm J,\qquad K_f=\tfrac12(1.6)(5)^2=20.0\ \mathrm J.$$ Therefore $$\boxed{\Delta K=20-8=12.0\ \mathrm J}.$$ Internal muscular work supplies this increase. Fixed angular momentum implies $K=L^2/(2I)$, so reducing inertia raises rotational energy.

Physical check

The final momentum $1.6(5)=8$ is unchanged even though the rotational energy increases.