Question

A bowling ball has a mass of 5.5 kg and a radius of 12.0 cm. It is released so that

Answer 1

$L = 2.4 \; \text{kg} \cdot \text{m}^{2} \cdot \text{s}^{-1}$

Explanation

The angular momentum of a rolling body is the product of the body's moment of inertia and its angular velocity.

$L = I \cdot \omega$

where

• $L$ is the angular momentum of a rolling body;
• $I$ is the body's moment of inertia; And
• $\omega$ is the body's angular moment.

What's the moment of inertia of this bowling ball?

Assuming that the ball is a solid sphere. For a solid sphere,

$I = \dfrac{2}{5} \; m \cdot r^2$.

where

• $I$ is the moment of inertia of the sphere;
• $m$ is the mass of the sphere; and
• $r$ is the radius of the sphere.

$r = 12.0 \; \text{cm} = 0.120 \; \text{m}$ for this sphere. $m = 5.5 \; \text{kg}$.

$I = \dfrac{2}{5} \; m \cdot r^{2}\\\phantom{I} = \dfrac{2}{5} \times 5.5 \times 0.120^{2}\\\phantom{I} = 0.0317 \; \text{kg}\cdot m^{2}$

What's the angular momentum of this bowling ball?

$\omega = 12 \; \text{rev} \cdot s^{-1} = 12 \times 2\; \pi \; \text{rad} \cdot \text{s}^{-1} = 75.4 \; \text{rad}\cdot \text{s}^{-1}$.

$L = I \cdot \omega = 0.0317 \times 75.4 = 2.4 \; \text{kg}\cdot \text{m}^{2} \cdot \text{s}^{-1}$.