Conservation of Angular Momentum — Question 5

Question 5

Walking speed relative to a platform differs from ground speed

A person modeled as a $50\,\mathrm{kg}$ point mass walks on the rim of a radius-$2\,\mathrm m$ platform of inertia $100\,\mathrm{kg\,m^2}$. The system starts at rest with negligible external axial torque. The person reaches angular speed $0.60\,\mathrm{rad/s}$ relative to the platform.

1. Find the platform angular velocity and the person's ground-frame angular velocity.

2. Find the final kinetic energy and its source.

Question 5 - Solution

Walking speed relative to a platform differs from ground speed

(a) Derivation and calculation

Let $\omega$ be the platform rate and $\omega_p$ the person ground-frame rate. The relative-speed condition is $$\omega_p-\omega=.60.$$ With person inertia $mR^2=50(2)^2=200$ kg m$^2$, conservation from rest gives $$100\omega+200(\omega+.60)=0.$$ Therefore $\boxed{\omega=-.400\ \mathrm{rad/s}}$ and $\boxed{\omega_p=+.200\ \mathrm{rad/s}}$.

(b) Result and interpretation

The final kinetic energy is $$\boxed{K=\tfrac12(100)(-.40)^2+\tfrac12(200)(.20)^2=8+4=12.0\ \mathrm J}.$$ The person supplies this energy by internal work. Using $.60$ rad/s as the ground-frame speed would violate the specified relation $\omega_p-\omega=.60$.

Physical check

The final momenta are $-40$ and $+40$ kg m$^2$/s, preserving the initial zero total.