Question

A single ladybug is standing near the edge of a rotating solid disk that has an angular speed of 2.95 rad/s. The ladybug begins to walk in toward the center of the disk. Given the following data, what is the initial angular momentum of the bug+disk system?

• Ladybug: mass = 0.05 kg; initial position r = 0.15 m; final position r = 0.03 m
• Disk: mass = 0.25 kg; R = 0.18 m

What is the new angular speed (in rad/s) when the ladybug moves to her final position?

Answer 1

(a) We can find the initial angular momentum of the bug+disk system as follows:

$$L_i=L_{bug,i}+L_{disk,i},$$ $$L_i=r_i^2m_{bug}\omega_i+\dfrac{1}{2}MR^2\omega_i,$$

$L_i=(0.15\ m)^2\cdot0.05\ kg\cdot2.95\ \dfrac{rad}{s}+\dfrac{1}{2}\cdot0.25\ kg\cdot(0.18\ m)^2\cdot2.95\ \dfrac{rad}{s},$

$$L_i=0.015\ \dfrac{kg\cdot m^2}{s}.$$

(b) Since there is the change in rotation of the disk when the ladybug moves to her final position, we can apply the Law of Conservation of Angular Momentum:

$$L_i=L_f,$$ $$L_f=(r_f^2m_{bug}+\dfrac{1}{2}MR^2)\omega_f.$$

From this equation we can find the new angular speed when the ladybug moves to her final position:

$$\omega_f=\dfrac{L_f}{r_f^2m_{bug}+\dfrac{1}{2}MR^2},$$ $$\omega_f=\dfrac{0.015\ \dfrac{kg\cdot m^2}{s}}{(0.03\ m)^2\cdot0.05\ kg+\dfrac{1}{2}\cdot0.25\ kg\cdot(0.18\ m)^2},$$ $$\omega_f=3.66\ \dfrac{rad}{s}.$$