Anonymous
6/15/2025, 2:40:10 AM
No.16698076
Consider two identical disks in a vacuum with a nonzero rotational inertia, [math]I_1=I_2=I[/math], rotating at the same, nonzero rate in opposite directions, [math]\omega_1=+\omega[/math] and [math]\omega_2=-\omega[/math]. The two disks drift towards one another along a thin, massless cable until they grind together and reach a common angular velocity, [math]\omega_1=\omega_2=\omega_f[/math]. Determine the final angular velocities and whether Newton’s Third Law holds for (a) the case where total angular momentum is conserved, and (b) the case where total rotational kinetic energy is conserved.
(a)
[math]L_1=+I\omega[/math]
[math]L_2=-I\omega[/math]
[math]L_{tot,i}=L_1+L_2=I\omega-I\omega=0[/math]
[math]L_{tot,f}=I\omega_f+I\omega_f=2I\omega_f[/math]
[math]2I\omega_f=0 \rightarrow \omega_f=0[/math]
[math]τ_1=Iα_1=I \frac{\Delta \omega_1}{\Delta t}=-I \frac{\omega}{\Delta t}[/math]
[math]τ_2=Iα_2=I \frac{\Delta \omega_2}{\Delta t}=+I \frac{\omega}{\Delta t}[/math]
[math]τ_2=-τ_1[/math]
This result is consistent with the Third Law.
(b)
[math]K_1=\frac{1}{2} I\omega^2[/math]
[math]K_2=\frac{1}{2} I\omega^2[/math]
[math]K_{tot,i}=K_1+K_2=\frac{1}{2} I\omega^2+\frac{1}{2} I\omega^2=I\omega^2[/math]
[math]K_{tot,f}=\frac{1}{2} I {\omega_f}^2+\frac{1}{2} I{\omega_f}^2=I{\omega_f}^2[/math]
[math]I{\omega_f}^2=I\omega^2 \rightarrow \omega_f=\pm \omega[/math]
Note that either branch implies that one of the disks must retain its initial angular velocity and the other must end up with the same angular speed but flip its direction. Choosing the positive branch and looking at the torque:
[math]τ_1=Iα_1=I \frac{\Delta \omega_1}{\Delta t}=0[/math]
[math]τ_2=Iα_2=I \frac{\Delta \omega_2}{\Delta t}=2I \frac{\omega}{\Delta t}[/math]
[math]τ_2 \neq -τ_1[/math]
This result is not consistent with the Third Law. Furthermore, since the net torque is not zero, the rotational work is not zero, which implies that rotational kinetic energy was not conserved.
(a)
[math]L_1=+I\omega[/math]
[math]L_2=-I\omega[/math]
[math]L_{tot,i}=L_1+L_2=I\omega-I\omega=0[/math]
[math]L_{tot,f}=I\omega_f+I\omega_f=2I\omega_f[/math]
[math]2I\omega_f=0 \rightarrow \omega_f=0[/math]
[math]τ_1=Iα_1=I \frac{\Delta \omega_1}{\Delta t}=-I \frac{\omega}{\Delta t}[/math]
[math]τ_2=Iα_2=I \frac{\Delta \omega_2}{\Delta t}=+I \frac{\omega}{\Delta t}[/math]
[math]τ_2=-τ_1[/math]
This result is consistent with the Third Law.
(b)
[math]K_1=\frac{1}{2} I\omega^2[/math]
[math]K_2=\frac{1}{2} I\omega^2[/math]
[math]K_{tot,i}=K_1+K_2=\frac{1}{2} I\omega^2+\frac{1}{2} I\omega^2=I\omega^2[/math]
[math]K_{tot,f}=\frac{1}{2} I {\omega_f}^2+\frac{1}{2} I{\omega_f}^2=I{\omega_f}^2[/math]
[math]I{\omega_f}^2=I\omega^2 \rightarrow \omega_f=\pm \omega[/math]
Note that either branch implies that one of the disks must retain its initial angular velocity and the other must end up with the same angular speed but flip its direction. Choosing the positive branch and looking at the torque:
[math]τ_1=Iα_1=I \frac{\Delta \omega_1}{\Delta t}=0[/math]
[math]τ_2=Iα_2=I \frac{\Delta \omega_2}{\Delta t}=2I \frac{\omega}{\Delta t}[/math]
[math]τ_2 \neq -τ_1[/math]
This result is not consistent with the Third Law. Furthermore, since the net torque is not zero, the rotational work is not zero, which implies that rotational kinetic energy was not conserved.