Question

A 7.0 kg bowling ball has a moment of inertia of 2.8x10-2 kg m2, and a radius of 0.10 m. If it rolls down the lane at an angular speed of 40 rad/s without slipping, determine its angular momentum.

Answer 1

Hi there!

Angular momentum is equivalent to:

$\large\boxed{L = I\omega}$

L = angular momentum (kgm²/s)

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Plug in the given values for moment of inertia and angular speed:

$L = (0.028)(40) = \boxed{1.12 kgm^2/s}$