Angular momentum: six guided questions

the axle can hold the centre in place by exerting a force. Ask whether that force has a moment about the axle’s own axis.

Choose disk plus lump, including both objects before and after contact. Their mutual contact interaction is internal. The disk alone receives an external contact torque from the lump, so the disk’s individual angular momentum generally changes.

The axle can exert a force while contributing zero torque about its own axis. With negligible external angular impulse about $z$, the combined system’s total $L_z$ is conserved. Total linear momentum need not be conserved because the axle can exert a force impulse. Kinetic energy generally decreases when the lump sticks.

Check: a correct claim names the combined system, the $z$-axis and the contact interval. It does not claim that all forces vanish.

use $L_z=m(xv_y-yv_x)$. The horizontal component is radial at this position.

The sign is negative, corresponding to clockwise angular momentum. The magnitude is 0.30 kg·m²/s. Although the speed is 5.0 m/s, only its perpendicular component contributes here. Using $$\begin{aligned}mvR&=(0.20)(5.0)(0.50)\\&=0.50\,\mathrm{kg\cdot m^2/s}\end{aligned}$$ would incorrectly treat the whole velocity as tangential.

Check: changing $v_x$ while keeping this position and $v_y$ fixed leaves $L_z$ unchanged. The radial component contributes zero.

calculate the initial angular momentum, then add the angular impulse with its sign.

The wheel still rotates counterclockwise. Its angular momentum is not conserved over this interval. The negative torque reduces the positive angular momentum.

Check: $$\begin{aligned}\alpha&=-0.10/0.20\\&=-0.50\,\mathrm{rad/s^2},\end{aligned}$$ giving $$\begin{aligned}\omega_f&=3.0-0.50(2.0)\\&=2.0\,\mathrm{rad/s}.\end{aligned}$$ Under that same constant torque it would first reach zero after 6.0 s, longer than the stated interval.

add two $mr^2$ terms to the platform-and-actuator inertia. Conserve $L_z$ before evaluating energy.

The kinetic energy increases by 2.70 J. The internal actuator supplies energy as it pulls the masses inward. With a lossless mechanism and no other endpoint energy changes, its stored energy decreases by that amount. Conservation of angular momentum did not require conservation of kinetic energy.

Check: $I_f\omega_f=(0.625)(4.8)=3.00$. Also $K_f/K_i=1.60=I_i/I_f$. The total mass did not change; its distance from the axis did.

include the lump’s $mR^2$ in the final inertia. Use angular momentum to find $\omega_f$; do not start by setting $K_i=K_f$.

For disk plus lump about $z$,

Kinetic energy decreases by 0.60 J in the inelastic collision. This does not conflict with angular-momentum conservation about $z$.

Check: $I_f\omega_f=0.40\,\mathrm{kg\cdot m^2/s}$. The final clay speed is $R\omega_f=0.50\,\mathrm{m/s}$. Adding the disk energy 0.15 J and the clay energy 0.05 J reproduces the total 0.20 J.

sum the two initial angular momenta with their signs, add the given angular impulse and divide by the final combined inertia.

The disk starts with negative angular velocity, so

The final rotation is counterclockwise. Ignoring the motor would give $0.10/0.45=2/9\,\mathrm{rad/s}$, about +0.22 rad/s. The supplied external angular impulse changes the answer substantially. Its nonzero value prevents setting $L_{z,f}=L_{z,i}$, regardless of the collision’s duration.

Check: $$\begin{aligned}(0.45)(2/15)-0.10\\=-0.040\,\mathrm{kg\cdot m^2/s},\end{aligned}$$ exactly the given angular impulse. An angular impulse gives an angular-momentum change; its units are not joules, so it is not by itself a value for the motor’s work.