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Chapter 5: Problem 61

A person rides a Ferris wheel that turns with constant angular velocity. Her weight is \(520.0 \mathrm{N}\). At the top of the ride her apparent weight is \(1.5 \mathrm{N}\) different from her true weight. (a) Is her apparent weight at the top \(521.5 \mathrm{N}\) or \(518.5 \mathrm{N} ?\) Why? (b) What is her apparent weight at the bottom of the ride? (c) If the angular speed of the Ferris wheel is \(0.025 \mathrm{rad} / \mathrm{s},\) what is its radius?

Short Answer

Expert verified
Answer: The apparent weight of the person at the top of the Ferris wheel ride is 518.5 N, while at the bottom, it is 521.5 N. The radius of the Ferris wheel is approximately 8.35 meters.

Step by step solution

01

(a) Determine the apparent weight at the top

We know that the apparent weight at the top of the ride is different from the true weight by 1.5 N. We need to determine if it will be greater or lesser than the true weight. Since the person is at the top of the Ferris wheel as it moves, the centripetal force acts in the downward direction, while the gravitational force (weigth) acts in the upward direction. Hence, the apparent weight will be lesser than the true weight at the top of the ride. So, the apparent weight at the top of the ride is 518.5 N.
02

(b) Calculate the apparent weight at the bottom of the ride

At the bottom of the Ferris wheel, both the gravitational force (weight) and the centripetal force act in the same direction. We can use the formula for the apparent weight to find the apparent weight at the bottom of the ride. Apparent Weight = True Weight + Centripetal Force We first need to find the centripetal force that acts on the person when they are at the top of the ride: Centripetal Force = True Weight - Apparent Weight at the top Centripetal Force = 520 N - 518.5 N Centripetal Force = 1.5 N Now we can find the apparent weight at the bottom of the ride: Apparent Weight at bottom = True Weight + Centripetal Force Apparent Weight at bottom = 520 N + 1.5 N Apparent Weight at bottom = 521.5 N
03

(c) Find the radius of the Ferris wheel

We will use the formula for centripetal force to find the radius of the Ferris wheel: Centripetal Force = Mass × (Radius × Angular velocity)^2 We can start by finding the mass of the person from their true weight: Mass = Weight / Gravitational Acceleration Mass = 520 N / 9.81 m/s^2 Mass ≈ 53 kg Now we can rearrange the centripetal force formula to solve for the radius: Radius = (Centripetal Force / (Mass × Angular velocity^2))^(1/2) Plugging in the values: Radius = (1.5 N / (53 kg × (0.025 rad/s)^2))^0.5 Radius ≈ 8.35 m So, the radius of the Ferris wheel is approximately 8.35 meters.

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Most popular questions from this chapter

A child pushes a merry-go-round from rest to a final angular speed of 0.50 rev/s with constant angular acceleration. In doing so, the child pushes the merry-go-round 2.0 revolutions. What is the angular acceleration of the merry- go-round?
Your car's wheels are \(65 \mathrm{cm}\) in diameter and the wheels are spinning at an angular velocity of 101 rad/s. How fast is your car moving in kilometers per hour (assume no slippage)?
Grace is playing with her dolls and decides to give them a ride on a merry-go- round. She places one of them on an old record player turntable and sets the angular speed at 33.3 rpm. (a) What is their angular speed in rad/s? (b) If the doll is \(13 \mathrm{cm}\) from the center of the spinning turntable platform, how fast (in \(\mathrm{m} / \mathrm{s}\) ) is the doll moving?
An elevator cable winds on a drum of radius \(90.0 \mathrm{cm}\) that is connected to a motor. (a) If the elevator is moving down at $0.50 \mathrm{m} / \mathrm{s},$ what is the angular speed of the drum? (b) If the elevator moves down \(6.0 \mathrm{m},\) how many revolutions has the drum made?
A biologist is studying plant growth and wants to simulate a gravitational field twice as strong as Earth's. She places the plants on a horizontal rotating table in her laboratory on Earth at a distance of \(12.5 \mathrm{cm}\) from the axis of rotation. What angular speed will give the plants an effective gravitational field \(\overrightarrow{\mathrm{g}}_{\mathrm{eff}},\) whose magnitude is \(2.0 \mathrm{g} ?\) \([\) Hint: Remember to account for Earth's gravitational field as well as the artificial gravity when finding the apparent weight. \(]\)
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