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85P

Expert-verifiedFound in: Page 326

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?**

- The angular speed of the merry-go-round is $\mathrm{\omega}=\frac{\mathrm{mRv}}{\mathrm{I}+{\mathrm{MR}}^{2}}$.
- The linear speed of the girl is $\mathrm{v}=\mathrm{R\omega}=\frac{\left({\mathrm{mR}}^{2}\mathrm{v}\right)}{\mathrm{I}+{\mathrm{MR}}^{2}}$.

Mass of girl is M

Merry go round of radius is R

Rotational inertia is I

**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.

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

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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