Answer to Question #270779 in Physics for Miyo

Question

Angular Momentum

Solve the problems.

1. A small particle of mass 0.20 kg is being whirled in a horizontal circle at the end of a 20 m long string at a constant speed of 1.0 m/s. Determine ito angular momentum about its axis of rotation.

2. A bowling ball has a mans of 5.5 kg and a radius of 12.0 cm. It is released so that it rolls down the alley at a rate of 12 rev/s. Find the magnitude of its angular momentum

Answer 1

1) Let's first find the angular velocity of the particle:

$$v=r\omega,$$ $$\omega=\dfrac{v}{r}=\dfrac{1.0\ \dfrac{m}{s}}{20\ m}=0.05\ \dfrac{rad}{s}.$$

Finally, we can find angular momentum of the particle about its axis of rotation:

$$L=I\omega=mr^2\omega,$$ $$L=0.20\ kg\times(20\ m)^2\times0.05\ \dfrac{rad}{s}=4.0\ \dfrac{kg\times m^2}{s}.$$

2) Let's first convert rev/s to rad/s:

$$\omega=12\ \dfrac{rev}{s}\times\dfrac{2\pi\ rad}{1\ rev}=75.4\ \dfrac{rad}{s}.$$

Finally, we can find angular momentum of the bowling ball about its axis of rotation:

$$L=I\omega=\dfrac{2}{5}mr^2\omega,$$ $$L=\dfrac{2}{5}\times5.5\ kg\times(0.12\ m)^2\times75.4\ \dfrac{rad}{s}=2.39\ \dfrac{kg\times m^2}{s}.$$