Question

A) What is the angular momentum of 0.25kg ball rotating on the end of a rope in a circle of radius 0.5m at and angular speed of 15.0 rad/s?

B) What if it rotates at 450.rpm

Answer 1

PART A)

Angular momentum is defined as

$L = I\omega$

here we know that

$I = mr^2$

now we have

m = 0.25 kg

r = 0.5 m

now we will have

$I = (0.25 kg)(0.5m)^2$

$I = 0.0625 kg m^2$

also we know that

$\omega = 15 rad/s$

now we have

$L = (0.0625)(15) = 0.94 kg m^2 rad/s$

Part B)

now the ball is rotated with speed 450 rpm

now we have

f = 450 rpm = 450/60 hz

so angular speed is given as

$\omega = 2\pi f$

$\omega = 47.12 rad/s$

now again by above formula

$L = I\omega$

$L = (0.0625)(47.12) = 2.94 kg m^2 rad/s$