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Q6.3-23 PE

Expert-verified
Found in: Page 222

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) A $$22.0{\rm{ kg}}$$ child is riding a playground merry-go-round that is rotating at $$40.0{\rm{ rev}}/{\rm{min}}$$. What centripetal force must she exert to stay on if she is $$1.25{\rm{ m}}$$ from its centre? (b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at $$3.00{\rm{ rev}}/\min$$ if she is $$8.00{\rm{ m}}$$ from its center? (c) Compare each force with her weight.

1. The centripetal force required by her is $${\rm{482}}{\rm{.79 N}}$$.
2. The centripetal force required by her is $${\rm{17}}{\rm{.35 N}}$$.
3. The ratio of the centripetal force to the weight of the answer in part (a) is$${\rm{2}}{\rm{.24}}$$, and the ratio of the centripetal force to the weight of the answer in part (b) is $${\rm{0}}{\rm{.08}}$$.
See the step by step solution

## Step 1: Centripetal force

Centripetal force is a force that is always orthogonal to the motion of the body and directed towards the center of the circular path.

## Step 2: (a) Centripetal force when she ${\bf{1}}.{\bf{25}}{\rm{ }}{\bf{m}}$ from centre

The angular velocity of the merry-go-round when it rotates at $${\rm{40}}{\rm{.0 rev/min}}$$ is, where $$N$$ is the rate of revolution of the merry-go-round.

$\omega = 2\pi N\ Here, $$N$$ is the rate of revolution of the merry-go-round. Substitute $${\rm{40}}{\rm{.0 rev/min}}$$ for $$N$$, $$\begin{array}{c}\omega = 2\pi \times \left( {40.0{\rm{ rev}}/\min } \right)\\ = 2\pi \times \left( {40.0{\rm{ rev}}/\min } \right) \times \left( {\frac{{1{\rm{ }}\min }}{{60{\rm{ s}}}}} \right)\\ = 4.19{\rm{ rad}}/{\rm{s}}\end{array}$$ The centripetal force is, $$F = m{\omega ^2}r$$ Here, \[m$ is the mass of the child, $$\omega$$ is the angular velocity of the merry-go-round, and $$r$$ is the distance of the child from the center of the merry-go-round.

Substitute $${\rm{22}}{\rm{.0}}\,{\rm{kg}}$$ for $m$, $${\rm{4}}{\rm{.19 rad/s}}$$ for $$\omega$$, and $${\rm{1}}{\rm{.25 m}}$$ for $$r$$,

$$\begin{array}{c}F = \left( {22.0{\rm{ kg}}} \right) \times {\left( {4.19{\rm{ rad}}/{\rm{s}}} \right)^2} \times \left( {1.25{\rm{ m}}} \right)\\ = 482.79{\rm{ N}}\end{array}$$

Hence, the centripetal force required by her is $${\rm{482}}{\rm{.79 N}}$$.

## Step 3: (b) Centripetal force for stay on amusement park

When the merry-go-round is rotating at , the angular velocity of the merry-go-round is,

$\omega = 2\pi N$

Here, $$N$$ is the rate of revolution of the merry-go-round.

Substitute $${\rm{3}}{\rm{.00 rev/min}}$$ for $$N$$,

$$\begin{array}{c}\omega = 2\pi \times \left( {3.00{\rm{ rev}}/\min } \right)\\ = 2\pi \times \left( {3.00{\rm{ rev}}/\min } \right) \times \left( {\frac{{1{\rm{ }}\min }}{{60{\rm{ s}}}}} \right)\\ = 0.314{\rm{ rad}}/{\rm{s}}\end{array}$$

The centripetal force is,

$$F = m{\omega ^2}r$$

Here, a$m$ is the mass of the child, $$\omega$$ is the angular velocity of the merry-go-round, and $$r$$ is the distance of the child from the center of the merry-go-round.

Substitute $${\rm{22}}{\rm{.0}}\,{\rm{kg}}$$ for $m$, $${\rm{0}}{\rm{.314 rad/s}}$$ for $$\omega$$, and $${\rm{8}}{\rm{.00 m}}$$ for $$r$$,

$$\begin{array}{c}F = \left( {22.0{\rm{ kg}}} \right) \times {\left( {0.314{\rm{ rad}}/{\rm{s}}} \right)^2} \times \left( {8.00{\rm{ m}}} \right)\\ = 17.35{\rm{ N}}\end{array}$$

Hence, the centripetal force required by her is $${\rm{17}}{\rm{.35 N}}$$.

## Step 4: (c) Ratio of centripetal force to the weight of the child

The child's weight is,

$$W = mg$$

Here, $$m$$ is the mass of the child, and $$g$$ is the acceleration due to gravity.

Substitute $${\rm{22}}{\rm{.0Kg}}$$ for $${\rm{m}}$$, and $${\rm{9}}{\rm{.8 m/}}{{\rm{s}}^{\rm{2}}}$$ for $$g$$,

$$\begin{array}{c}W = \left( {22.0{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\\ = 215.6{\rm{ N}}\end{array}$$

The ratio of the centripetal force in part (a) to the weight is,

$$\eta = \frac{F}{W}$$

Substitute $${\rm{482}}{\rm{.79 N}}$$ for $$F$$, and $${\rm{215}}{\rm{.6 N}}$$ for $$W$$,

$$\begin{array}{c}\eta = \frac{{482.79{\rm{ N}}}}{{215.6{\rm{ N}}}}\\ = 2.24\end{array}$$

The ratio of the centripetal force in part (b) to the weight is,

$$\eta = \frac{F}{W}$$

Substitute $${\rm{17}}{\rm{.35 N}}$$ for $$F$$, and $${\rm{215}}{\rm{.6 N}}$$ for $$W$$,

$$\begin{array}{c}\eta = \frac{{17.35{\rm{ N}}}}{{215.6{\rm{ N}}}}\\ = 0.08\end{array}$$

Hence, the ratio of the centripetal force to the weight of the answer in part (a) is $${\rm{2}}{\rm{.24}}$$, and the ratio of the centripetal force to the weight of the answer in part (b) is $${\rm{0}}{\rm{.08}}$$.