Question

how fast, in rpm , would a 5.6 kg , 20- cm -diameter bowling ball have to spin to have an angular momentum of 0.22 kgm2/s ?

Answer 1

The angular velocity in rpm of the bowling ball is equal to 71.32 rpm.

Given the following data:

Mass = 5.6 kg

Diameter = 20 cm to m = 0.2 m

Angular momentum = 0.22 kgm2/s

Radius = Diameter/2 = 0.22/2 = 0.11 m