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Expert-verified Found in: Page 326 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # 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 $\mathrm{\omega }=\frac{\mathrm{mRv}}{\mathrm{I}+{\mathrm{MR}}^{2}}$.
2. The linear speed of the girl is $\mathrm{v}=\mathrm{R\omega }=\frac{\left({\mathrm{mR}}^{2}\mathrm{v}\right)}{\mathrm{I}+{\mathrm{MR}}^{2}}$.
See the step by step solution

## 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 ${\mathbf{L}}{\mathbf{=}}{\mathbf{I\omega }}$ and ${\mathbf{v}}{\mathbf{=}}{\mathbf{r\omega }}$, find the angular speed of the merry-go-round and the linear speed of the girl.

Formulae are as follow:

$\mathrm{L}=\mathrm{I\omega }\phantom{\rule{0ex}{0ex}}\mathrm{v}=\mathrm{R\omega }$

Where, $\mathrm{\omega }$ 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,

$\mathrm{L}=\left(\mathrm{I}+{\mathrm{MR}}^{2}\right)\mathrm{\omega }$

The final angular momentum associated with the thrown rock is $-\mathrm{mRv}$.

Therefore, according to the conservation of the angular momentum,

$0=\left(\mathrm{I}+{\mathrm{MR}}^{2}\right)\mathrm{\omega }-\mathrm{mRv}\phantom{\rule{0ex}{0ex}}\mathrm{\omega }=\frac{\mathrm{mRv}}{\left(\mathrm{I}+{\mathrm{MR}}^{2}\right)}$

Hence, the angular speed of the merry-go-round is $\mathrm{\omega }=\frac{\mathrm{mRv}}{\left(\mathrm{I}+{\mathrm{MR}}^{2}\right)}$.

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

The linear speed of the girl is,

$\mathrm{v}=\mathrm{R\omega }$

Hence, the linear speed of the girl is $\mathrm{v}=\mathrm{R\omega }=\frac{{\mathrm{mR}}^{2}\mathrm{v}}{\mathrm{I}+{\mathrm{MR}}^{2}}$.

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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