Ch 05: Applying Newton's Laws

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 05: Applying Newton's Laws

Problem 54a

Chapter 5, Problem 54a

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of $100$ m. Its name comes from its $60$ arms, each of which can function as a second hand (so that it makes one revolution every $60.0$ s). Find the speed of the passengers when the Ferris wheel is rotating at this rate.

Verified step by step guidance

Step 1: Understand the problem. The Ferris wheel has a diameter of 100 m, which means its radius is half of the diameter, or 50 m. The wheel completes one revolution every 60.0 seconds. We need to calculate the speed of the passengers, which is the tangential speed.

Step 2: Recall the formula for tangential speed. Tangential speed (v) is given by the equation:

v = r ω

, where

r

is the radius of the circular path and

ω

is the angular velocity.

Step 3: Calculate the angular velocity. Angular velocity

ω

is defined as the rate of change of the angle per unit time. For one complete revolution, the angle is

2 π

radians, and the time taken is 60.0 seconds. Use the formula:

ω = \frac{2 π}{t}

, where

t

is the time for one revolution.

Step 4: Substitute the values into the angular velocity formula. Plug in

60.0

seconds for

t

to find

ω

. This will give the angular velocity in radians per second.

Step 5: Use the tangential speed formula. Substitute the radius

50

m and the calculated angular velocity

ω

into the formula

v = r ω

to find the tangential speed of the passengers.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Speed

Centripetal speed refers to the constant speed of an object moving in a circular path. It is calculated by the formula v = 2πr/T, where v is the speed, r is the radius of the circle, and T is the period of rotation. In the case of the Cosmo Clock 21, the radius is half the diameter, and the period is the time taken for one complete revolution.

Radius of the Ferris Wheel

The radius of a circle is the distance from the center to any point on its circumference. For the Cosmo Clock 21, with a diameter of 100 m, the radius is 50 m. This measurement is crucial for calculating the speed of the passengers as they move along the circular path of the Ferris wheel.

Period of Rotation

The period of rotation is the time it takes for an object to complete one full revolution around a circular path. In this scenario, the Ferris wheel completes one revolution every 60 seconds. This time interval is essential for determining the speed of the passengers, as it directly influences the calculation of centripetal speed.

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Textbook Question

The 'Giant Swing' at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable $5.00$ m long, and the upper end of the cable is fastened to the arm at a point $3.00$ m from the central shaft (Fig. E $5.50$ ). Find the time of one revolution of the swing if the cable supporting a seat makes an angle of $30.0 degrees$ with the vertical.

In another version of the 'Giant Swing' (see Exercise $5.50$ ), the seat is connected to two cables, one of which is horizontal (Fig. E $5.51$ ). The seat swings in a horizontal circle at a rate of $28.0$ rpm (rev/min). If the seat weighs $255$ N and an $825$ -N person is sitting in it, find the tension in each cable.

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates 'artificial gravity' at the outside rim of the station. If the diameter of the space station is $800$ m, how many revolutions per minute are needed for the 'artificial gravity' acceleration to be $9.80$ m/s²?

1613

views

Textbook Question

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of $100$ m. Its name comes from its $60$ arms, each of which can function as a second hand (so that it makes one revolution every $60.0$ s). A passenger weighs $882$ N at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel?

820

views

Textbook Question

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of $100$ m. Its name comes from its $60$ arms, each of which can function as a second hand (so that it makes one revolution every $60.0$ s). What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?

875

views

Textbook Question

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of $100$ m. Its name comes from its $60$ arms, each of which can function as a second hand (so that it makes one revolution every $60.0$ s). What then would be the passenger's apparent weight at the lowest point?

980

views