Conservation of Angular Momentum — Question 3

Question 3

Vertical deposition brings mass but no axial angular momentum

A horizontal disk has $I=1.0\,\mathrm{kg\,m^2}$ and $\omega=4\,\mathrm{rad/s}$. A $1.0\,\mathrm{kg}$ lump falls vertically and sticks at radius $0.50\,\mathrm m$. The axle supplies negligible vertical-axis torque.

1. Find the final angular speed.

2. Find the loss of azimuthal rotational energy; explain why this is not the total three-dimensional impact loss.

Question 3 - Solution

Vertical deposition brings mass but no axial angular momentum

(a) Derivation and calculation

Incoming momentum is vertical, so $$(\vec r\times\vec p_{\mathrm{incoming}})_z=0.$$ Axial angular momentum conservation gives $$I_i\omega_i=(I_i+mR^2)\omega_f,$$ $$\boxed{\omega_f=\frac{1(4)}{1+1(.50)^2}=3.20\ \mathrm{rad/s}}.$$

(b) Result and interpretation

The azimuthal rotational energies are $$K_{{\mathrm{rot}},i}=\tfrac12(1)(4)^2=8.00\ \mathrm J,$$ $$K_{{\mathrm{rot}},f}=\tfrac12(1.25)(3.20)^2=6.40\ \mathrm J.$$ Thus $\boxed{\Delta K_{\mathrm{rot,loss}}=1.60\ \mathrm J}$. Stopping the downward motion involves additional energy that depends on the unspecified vertical impact speed; this is not the full three-dimensional impact-loss calculation.

Physical check

The final axial momentum is $1.25(3.2)=4$, matching the initial disk momentum.