Conservation of Angular Momentum — Question 2

Question 2

Tangential capture preserves axial momentum but dissipates energy

A stationary turntable has inertia $0.20\,\mathrm{kg\,m^2}$ about a frictionless vertical axle. A $0.50\,\mathrm{kg}$ particle moving tangentially at $3\,\mathrm{m/s}$ sticks at radius $0.40\,\mathrm m$. Neglect external angular impulse about the axle.

1. Find common angular speed just after capture.

2. Find kinetic energy lost in the horizontal capture.

Question 2 - Solution

Tangential capture preserves axial momentum but dissipates energy

(a) Derivation and calculation

The incoming tangential particle supplies axial momentum $$L_i=muR=.50(3)(.40)=.600\ \mathrm{kg\,m^2/s}.$$ After sticking, $$I_f=I_0+mR^2=.20+.50(.40)^2=.280\ \mathrm{kg\,m^2},$$ $$\boxed{\omega_f=L_i/I_f=.600/.280=15/7=2.143\ \mathrm{rad/s}}.$$

(b) Result and interpretation

The horizontal kinetic energies are $$K_i=\tfrac12mu^2=2.25\ \mathrm J,\qquad K_f=\frac{L_i^2}{2I_f}=\frac9{14}\ \mathrm J.$$ Hence $$\boxed{K_{\mathrm{lost}}=\frac94-\frac9{14}=\frac{45}{28}=1.607\ \mathrm J}.$$ The axle may supply a linear impulse while contributing no angular impulse about its own axis.

Physical check

The capture conserves axial angular momentum but dissipates kinetic energy through the sticking interaction.