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Q6.3-23 PE

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College Physics (Urone)
Found in: Page 222
College Physics (Urone)

College Physics (Urone)

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

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Short Answer

(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 \({\rm{0}}{\rm{.08}}\).
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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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

Most popular questions for Physics Textbooks

Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s2. Who do you agree with and why?

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that \(\theta \) (as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is, \(\theta = {\tan ^{ - 1}}\,{v^2}/rg\)

(b) Calculate \(\theta \) for a \(12.0{\rm{ m}}/{\rm{s}}\) turn of radius \(30.0{\rm{ m}}\) (as in a race).

Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle \(\theta \), the speed \(v\), and the radius of curvature \(r\) of the turn similar to that for the ideal banking of roadways.

Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.

Microwave ovens rotate at a rate of about 6 rev/min. What is this in revolutions per second? What is the angular velocity in radians per second?

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