Question

In the swing carousel amusement park ride, riders sit in chairs that are attached by a chain to a large rotating drum as shown in (Figure 1). As the carousel turns, the riders move in a large circle with the chains tilted out from the vertical. In one such carousel, the riders move in a 15-m-radius circle and take 7.9 s to complete one revolution.

What is the angle of the chains, as measured from the vertical?

Answer 1

Answer:$\theta =44.068^{\circ}$

Explanation:

Given

time taken to complete the circle=7.9 s

radius of circle(r)=15 m

velocity of rider is given by $=\frac{2\pi r}{t}$

$v=\frac{2\pi 15}{7.9}=11.93 m/s$

Let us suppose T is the tension in the chain and $\theta$is the angle which chain makes with vertical

Therefore $T\sin \theta =\frac{mv^2}{r}-1$

$T\cos \theta=mg$ --2

Divide 1 & 2 we get

$tan\theta =\frac{v^2}{rg}$

$tan\theta =0.968$

$\theta =44.068^{\circ}$

Answer 2

The angle the chains measures from the vertical is 44.1⁰.

The given parameters;

• radius of the circle = 15 m
• time of motion for a complete revolution, t = 7.9 s

• let the angle the chain makes with the vertical = θ

The vertical component of the tension on the rope;

$Tcos(\theta) = mg$

The horizontal component of the tension on the chain;

$Tsin(\theta) = ma_c$

The coefficient of friction is calculated as;

$\mu_k = \frac{Tsin(\theta)}{Tcos(\theta) } = \frac{ma_c}{mg} \\\\tan(\theta) = \frac{a_c}{g}$

where;

$a_c$ is the centripetal acceleration of the riders

The velocity of the riders is calculated as;

$v = \frac{2\pi r}{t} \\\\v = \frac{2\pi \times 15}{7.9} \\\\v = 11.93 \ m/s$

The centripetal acceleration is calculated as;

$a_c = \frac{v^2}{r} \\\\a_c = \frac{(11.93)^2}{15} =9.488 \ m/s^2$

The angle the chain measures from the vertical is calculated as;

$tan(\theta) = \frac{a_c}{g} \\\\tan(\theta) = \frac{9.488}{9.8} \\\\tan(\theta) = 0.968\\\\\theta = tan^{-1} (0.968)\\\\\theta = 44.1^0$

Thus, the angle the chains measures from the vertical is 44.1⁰.

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