Question

How fast, in rpm, would a $5.0kg22cm$diameter bowling ball have to spin to have an angular momentum of $0.23{\mathrm{kgm}}^2/\mathrm{s}$?

Short answer

Therefore the angular speed of the baseball$91\mathrm{rpm}$.

Step-by-step solution

Step : 1 Introduction

Determine the angular velocity of the ball using the equations for angular momentum and moment of inertia of a spherical object.

The magnitude of angular momentum is expressed as,

$L=I\omega$

Here, $L$is angular momentum, $/$is moment of inertia, and $\omega$is angular speed

Consider the bouling ball as a spherical object then the expression for its moment of inertia is,

Here, $M$ is mass of the baseball and $R$ is its radius.

$I=\frac{2}{5}M{R}^2$

Step :2 Radius of the ball

Substitute $\frac{2}{5}M{R}^2$for $/$in the equation $L=l\omega$

$L=\left(\frac{2}{5}M{R}^2\right)\omega$

Re arrange this equation $\omega$

$\omega =\frac{5}{2}\left(\frac{L}{M{R}^2}\right)$

Radius of the ball

$R=\frac{d}{2}$

$=\frac{22\mathrm{cm}}{2}$

$=\left(11\mathrm{cm}\right)\left(\frac{0.01\mathrm{m}}{\mathrm{cm}}\right)$

$=0.11\mathrm{m}$

Step :3 Substitution

Substitute $5\mathrm{kg}\text{for}M,0.11\mathrm{m}\text{for}R\text{, and}0.23\mathrm{kg}\cdot {\mathrm{m}}^2/\mathrm{s}\text{for}L\text{in}\omega =\frac{5}{2}\frac{L}{M{d}^2}\text{,}$

$\omega =\frac{5}{2}\frac{\left(0.23\mathrm{kg}\cdot {\mathrm{m}}^2/\mathrm{s}\right)}{\left(5\mathrm{kg}\right)(0.11\mathrm{m}{)}^2}$

$=(9.5\mathrm{rad}/\mathrm{s})\left(\frac{1\mathrm{rev}}{2\pi \mathrm{rad}}\right)\left(\frac{60\mathrm{s}}{1\min }\right)$

$=90.7\mathrm{rev}/\min$