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Q6.3-23 PE

Expert-verifiedFound in: Page 222

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.**

- The centripetal force required by her is \({\rm{482}}{\rm{.79 N}}\).
- The centripetal force required by her is \({\rm{17}}{\rm{.35 N}}\).
- 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 \({\rm{0}}{\rm{.08}}\).

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

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.

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 \({\rm{482}}{\rm{.79 N}}\).

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 \({\rm{17}}{\rm{.35 N}}\).

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 \({\rm{0}}{\rm{.08}}\).

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