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Physics For Scientists & Engineers
Found in: Page 175
Physics For Scientists & Engineers

Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

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Illustration

Short Answer

Figure P6.57 shows a photo of a swing ride at an amusement park. The structure consists of a horizontal, rotating, circular platform of diameter D from which seats of mass m are suspended at the end of mass less chains of length d. When the system rotates at constant speed, the chains swing outward and make an angle θ with the vertical. Consider such a ride with the following parameters: D = 8.00 m, d = 2.50 m, m = 10.0 kg , and θ = 28.0° (a) What is the speed of each seat? (b) Draw a diagram of forces acting on the combination of a seat and a 40.0 - kg child and (c) find tension in the chain.

a) The speed of each seat is 5.19 m/s .

b) The diagram of forces acting on the combination of a seat is shown in step 2.

c) The tension in the chain is 555N.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Centrifugal force

The expression for the centrifugal force is given by,

F=mv2r

Here m is the mass of the object, v is the speed and r is the radius, F is the centrifugal force. Centrifugal force is also known by the name of centripetal force.

Calculate the speed of each seat

(a) The below figure shows the direction of forces and direction.

Here F is the centrifugal force, D is the diameter, d is the length of the chain, m is the mass of the seats, M is the mass of the child, T is the tension, θis the angle, Tcosθis the vertical component of the tension and Tsinθ is the horizontal component of the tension.

From the above figure, the radius is equal to

r=dsinθ+D2

Substitute 8.0 m for D, 2.50 m for d, 28o for θ in the above equation.

r=2.50 sin 28°+8.02=5.17 m

From the above figure, the expression for the centrifugal force is

F=M+mv2r ....... (1)

Substitute 5.17 m for r, 40kg for M, 10kg for m in the above equation.

F=40+10v25.17 =9.67v2

Now balance the force in horizontal and vertical direction.

Horizontal direction

Tsinθ=F

Substitute M+mv2r for F from equation (1) into the above equation.

Tsinθ=M+mv2r ...... (2)

Vertical direction

Tcosθ=M+mg ....... (3)

The ratio of equation (2) and equation (3) is

TsinθTcosθ=M+mv2rM+mg tanθ=v2rg v=rgtanθ

Substitute role="math" localid="1663678381997" 5.17 m for r, 9.8 m/s2 for g and 28° for θin the above equation.

v= 5.179.8 tan 28° =5.19 m/s

Therefore, the speed of each seat is 5.19 m/s.

Drawing the diagram of forces acting on the combination of a seat

(b)

The total mass is equivalent to the mass of child and seat.

Mtotal=M+m

Substitute 40kg for M and 10 kg for m in the above equation.

Mtotal=M+m =40+10 kg =50 kg

The total weight of combination, i.e seat and child is,

W=Mtotalg

Substitute 50 kg for Mtotal and 9.8 m/s.

W= 509.8 =490 N

The horizontal component of the tension is T sin 28° and the vertical component of the tension is T cos 28°.

The below diagram shows the direction of the forces acting on the combination of a seat.

Calculating the tension in the chain

(c)

From equation (2).

Tsinθ=M+mv2r

As, T sin θ=F and F=9.67v2.

Tsinθ=9.67v2

On solving equation

T=9.67v2sin θ

Substitute 28° for θ and 5.19 m/s for v in the above equation.

T=9.675.192sin 28° =555 N

Therefore, the tension in the chain is 555 N.

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