Chapter 12: Q.45 (page 332)

How fast, in rpm, would a $5.0 k g 22 c m$ diameter bowling ball have to spin to have an angular momentum of $0.23 {kgm}^{2} / s$ ?

Short Answer

Expert verified

Therefore the angular speed of the baseball $91 rpm$ .

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,

$L = I ω$

Here, $L$ is angular momentum, $/$ is moment of inertia, and $ω$ 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 $L = l ω$

$L = (\frac{2}{5} M R^{2}) ω$

Re arrange this equation $ω$

$ω = \frac{5}{2} (\frac{L}{M R^{2}})$

Radius of the ball

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

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

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

$= 0.11 m$

Step :3 Substitution

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

$ω = \frac{5}{2} \frac{(0.23 kg \cdot m^{2} / s)}{(5 kg) (0.11 m)^{2}}$

$= (9.5 rad / s) (\frac{1 rev}{2 π rad}) (\frac{60 s}{1 \min})$

$= 90.7 rev / \min$

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How fast, in rpm, would a $5.0 k g 22 c m$ diameter bowling ball have to spin to have an angular momentum of $0.23 {kgm}^{2} / s$ ?

Short Answer

Step by step solution

Step : 1 Introduction

Step :2 Radius of the ball

Step :3 Substitution

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How fast, in rpm, would a 5.0kg22cmdiameter bowling ball have to spin to have an angular momentum of 0.23kgm2/s?

Short Answer

Step by step solution

Step : 1 Introduction

Step :2 Radius of the ball

Step :3 Substitution

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

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Study anywhere. Anytime. Across all devices.

How fast, in rpm, would a $5.0 k g 22 c m$ diameter bowling ball have to spin to have an angular momentum of $0.23 {kgm}^{2} / s$ ?