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30 August, 20:27

An amusement park ride consists of a rotating circular platform 8.03 m in diameter from which 10 kg seats are suspended at the end of 3.76 m massless chains. When the system rotates, the chains make an angle of 28.6 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2.

1. What is the speed of each seat?

2. If a child of mass 26.2 kg sits in a seat, what is the tension in the chain (for the same angle) ?

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  1. 30 August, 21:03
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    a) 5.57 m/s

    b) 403 N

    Explanation:

    We will find the radius, r of the motion using the formula

    r = radius of the platform + horizontal swing distance

    r = diameter/2 + Lsinθ, where L is the length of the massless chain and θ is the displaced angle

    r = 8.03/2 + 3.76 sin28.6

    r = 4.015 + 1.8

    r = 5.82 m

    a) To get the speed, we use the formula

    tanθ = v2/rg, if we make v subject of formula, we have

    v = √ (rgtanθ), where

    v = speed of the seats

    g = 9.8m/s²

    v = √ (5.82 * 9.8 * tan28.6)

    v = √ (5.82 * 9.8 * 0.545)

    v = √31.08

    v = 5.57 m/s

    b) To find the tension, we use the formula

    Tcosθ = mg, making T subject of formula, we have

    T = mg/cosθ, where T = tension in the chain and m = 10 kg + 26.2 kg = 36.2 kg and θ = 28.6°

    T = 36.2 * 9.8 / cos 28.6

    T = 354.76 / 0.88

    T = 403 N
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