Answer to Question #280736 in Physics for Phia

Question

A 4.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s. Its angular momentum about the center of the circle is:

a: 12 kg.m2 /s

b. 24 kg.m2 /s

c. 48 kg.m2 /s

d. 6 kg.m2/s

Answer 1

Given:

$m=4.0\:\rm kg$

$r=0.50\:\rm m$

$\omega=12\:\rm rad/s$

The angular momentum is given by

$$L=I\omega=mr^2\omega$$

$$L=4.0*0.50^2*12=12\:\rm kg\cdot m^2/s$$