Question

1. determine the rotational inertia of a bowling ball with a mass of 5 kg and a radius of 10.1 cm.

1,020 kg·cm²
765 kg·cm²
510 kg·cm²
204 kg·cm²

Explanation:

Step1: Recall solid sphere inertia formula

The rotational inertia of a solid sphere is $I = \frac{2}{5}MR^2$

Step2: Substitute given values

$M=5\ \text{kg}$, $R=10.1\ \text{cm}$
$I = \frac{2}{5} \times 5 \times (10.1)^2$

Step3: Calculate the result

First simplify $\frac{2}{5} \times 5 = 2$, then $(10.1)^2=102.01$, so $I=2\times102.01=204.02\approx204\ \text{kg·cm}^2$

Answer:

204 kg·cm²