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Fundamentals Of Physics
Found in: Page 326
Fundamentals Of Physics

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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Short Answer

A girl of mass M stands on the rim of a frictionless merry-go-round of radius R and rotational inertia I that is not moving. She throws a rock of mass m horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is v. Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

  1. The angular speed of the merry-go-round is ω=mRvI+MR2.
  2. The linear speed of the girl is v=Rω=mR2vI+MR2.
See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Data

Mass of girl is M

Merry go round of radius is R

Rotational inertia is I

Step 2: Determining the concept

Using the formula L=Iω and v=rω, find the angular speed of the merry-go-round and the linear speed of the girl.

Formulae are as follow:

L=Iωv=Rω

Where, ω is angular frequency, L is angular momentum, I is moment of inertia, v is velocity and R is radius.

Step 3: (a) Determining the angular speed of the merry-go-round

Initially, the merry-go-round is not moving; therefore the initial angular momentum of the system is zero.

The final angular momentum the girl and merry-go-round is,

L=I+MR2ω

The final angular momentum associated with the thrown rock is -mRv.

Therefore, according to the conservation of the angular momentum,

0=I+MR2ω-mRvω=mRvI+MR2

Hence, the angular speed of the merry-go-round is ω=mRvI+MR2.

Step 4: (b) Determining the linear speed of the girl

The linear speed of the girl is,

v=Rω

Hence, the linear speed of the girl is v=Rω=mR2vI+MR2.

Therefore, using the formula for the angular momentum and relationship between linear and angular velocity, the expression for linear and angular velocity can be found.

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