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

Short Answer

Expert verified

Therefore the angular speed of the baseball91rpm.

Step by step solution

01

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=Iω

Here, Lis angular momentum, /is moment of inertia, and ωis 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=25MR2

02

Step :2 Radius of the ball

Substitute 25MR2for /in the equation L=lω

L=25MR2ω

Re arrange this equation ω

ω=52LMR2

Radius of the ball

R=d2

=22cm2

=(11cm)0.01mcm

=0.11m

03

Step :3 Substitution

Substitute 5kgforM,0.11mforR, and0.23kgm2/sforLinω=52LMd2,

ω=520.23kgm2/s(5kg)(0.11m)2

=(9.5rad/s)1rev2πrad60s1min

=90.7rev/min

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