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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 \(5.00 \mathrm{~m}\) long, and the upper end of the cable is fastened to the arm at a point \(3.00 \mathrm{~m}\) from the central shaft (Fig. E5.50). (a) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of \(30.0^{\circ}\) with the vertical. (b) Does the angle depend on the weight of the passenger for a given rate of revolution?

Short Answer

Expert verified
The time of one revolution of the swing is \(2 \pi \times 5 cos(30)\) seconds and does not depend on the weight of the passenger.

Step by step solution

01

Finding the Radius

First, we need to find the radius of the circular path. For this, let's use some trigonometry with the given angle and the length of the cable from where it's fastened to the arm. The provided angle with the vertical is \(30.0^{\circ}\) and the hypotenuse (cable length) is \(5.00 \mathrm{~m}\). So we can apply the cosine rule, \(cos \theta = \frac{adjacent}{hypotenuse}\), to find the adjacent side, which in this case is the radius (R) of the circular path, \(R = 5cos(30)\).
02

Deriving the formula for the Time period

The speed (v) of the swing can be represented as \(v = \omega R\), where \( \omega \) is the angular velocity and can be represented as \(\omega = \frac{2 \pi}{T}\), \( T \) being the time period of one revolution. Substitute the value of \( \omega \) in the first equation to get: \( v = R \frac{2 \pi}{T}\) or \( T = R \frac{2 \pi}{v} \)
03

Solving for unknown

We know that the giant swing moves at a constant speed, but we don't know the exact value. However, we can still get the value for \(T\) as \(\frac{v}{v} = 1\). Based on this logic, \( T = R \frac{2 \pi}{v} = 2 \pi R\)
04

Inserting values into the formula

We substitute the value of \(R\) that we got in step 1 into the equation to get the time period of one revolution. Hence, \(T = 2 \pi \times 5 cos(30)\)
05

Evaluating the weight factor

One important aspect to note here is that the speed at which the passenger swings does not depend on their weight. The only force that would be affecting them as they swing is the gravitational pull. So, no the time period, and hence the angle, does not depend on the weight of the passenger for a given rate of revolution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Path Radius Calculation
Understanding the radius of a circular path is crucial in circular motion physics problems. In the Giant Swing example, the radius represents the horizontal distance from the central shaft to the point where the cable is fastened to the arm.

To find this distance, we use trigonometry. We're given the cable length as the hypotenuse and the angle it makes with the vertical. Using the cosine rule, \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\), we calculate the adjacent side, which is in fact our radius (R). This step forms the foundation of solving related circular motion problems since many other quantities, such as angular velocity and time period of revolution, depend on the radius.
Trigonometry in Physics
Trigonometry is not just an abstract mathematical concept but a practical tool in physics, especially in analyzing situations involving angles, such as the case with the Giant Swing.

The use of trigonometric functions, like cosine in our exercise, allows us to relate different sides of a right triangle to angles. In physics, these sides often represent vectors or paths, essential for understanding motion dynamics. While we used the cosine rule in our example, other functions like sine and tangent are equally important in different scenarios involving inclined planes, pendulum swings, and orbital paths.
Angular Velocity
Angular velocity, symbolized by \(\omega\), is a measure of how fast an object rotates or revolves around a central point. It is the angle covered per unit of time, usually expressed in radians per second. Unlike linear velocity, angular velocity is concerned with rotation rather than straight-line motion.

In our exercise, we derived angular velocity from the time period of one revolution using \(\omega = \frac{2\pi}{T}\), where \(2\pi\) radians is the full angle for one rotation. This relationship is central to understanding rotational dynamics and is applicable across a range of circular motion scenarios from simple playground rides to complex planetary orbits.
Time Period of Revolution
The time period of revolution, denoted by T, is the time it takes for an object to complete one full rotation around a circular path. It's an important concept in determining the motion of objects in circular paths, like the seats of the Giant Swing.

In calculating the time period, we first needed the radius, which we found using trigonometry. Then, through understanding angular velocity, we could express the time period as \(T = \frac{2\pi R}{v}\), relating both the radius and the speed of the object in the circular path. For a constant speed, the time period becomes a simple product of the radius and \(2\pi\), which showcases how pivotal these foundational concepts are in solving complex physical phenomena.

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Most popular questions from this chapter

A box with mass \(10.0 \mathrm{~kg}\) moves on a ramp that is inclined at an angle of \(55.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is \(\mu_{\mathrm{k}}=0.300 .\) Calculate the magnitude of the acceleration of the box if you push on the box with a constant force \(F=120.0 \mathrm{~N}\) that is parallel to the ramp surface and (a) directed down the ramp, moving the box down the ramp; (b) directed up the ramp, moving the box up the ramp.

Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g\) 's) assumed to be constant, during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) How long did this force last?

DATA You are an engineer working for a manufacturing company. You are designing a mechanism that uses a cable to drag heavy metal blocks a distance of \(8.00 \mathrm{~m}\) along a ramp that is sloped at \(40.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between these blocks and the incline is \(\mu_{\mathrm{k}}=0.350 .\) Each block has a mass of \(2170 \mathrm{~kg}\). The block will be placed on the bottom of the ramp, the cable will be attached, and the block will then be given just enough of a momentary push to overcome static friction. The block is then to accelerate at a constant rate to move the \(8.00 \mathrm{~m}\) in \(4.20 \mathrm{~s}\). The cable is made of wire rope and is parallel to the ramp surface. The table gives the breaking strength of the cable as a function of its diameter; the safe load tension, which is \(20 \%\) of the breaking strength; and the mass per meter of the cable: $$ \begin{array}{cccc} \begin{array}{c} \text { Cable Diameter } \\ \text { (in.) } \end{array} & \begin{array}{c} \text { Breaking Strength } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Safe Load } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Mass per Meter } \\ (\mathbf{k g} / \mathbf{m}) \end{array} \\ \hline \frac{1}{4} & 24.4 & 4.89 & 0.16 \\ \frac{3}{8} & 54.3 & 10.9 & 0.36 \\ \frac{1}{2} & 95.2 & 19.0 & 0.63 \\ \frac{5}{8} & 149 & 29.7 & 0.98 \\ \frac{3}{4} & 212 & 42.3 & 1.41 \\ \frac{7}{8} & 286 & 57.4 & 1.92 \\ 1 & 372 & 74.3 & 2.50 \\ \hline \end{array} $$ (a) What is the minimum diameter of the cable that can be used to pull a block up the ramp without exceeding the safe load value of the tension in the cable? Ignore the mass of the cable, and select the diameter from those listed in the table. (b) You need to know safe load values for diameters that aren't in the table, so you hypothesize that the breaking strength and safe load limit are proportional to the cross-sectional area of the cable. Draw a graph that tests this hypothesis, and discuss its accuracy. What is your estimate of the safe load value for a cable with diameter \(\frac{9}{16}\) in.? (c) The coefficient of static friction between the crate and the ramp is \(\mu_{\mathrm{s}}=0.620,\) which is nearly twice the value of the coefficient of kinetic friction. If the machinery jams and the block stops in the middle of the ramp, what is the tension in the cable? Is it larger or smaller than the value when the block is moving? (d) Is the actual tension in the cable, at its upper end, larger or smaller than the value calculated when you ignore the mass of the cable? If the cable is \(9.00 \mathrm{~m}\) long, how accurate is it to ignore the cable's mass?

Apparent Weight. A \(550 \mathrm{~N}\) physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is \(850 \mathrm{~kg}\). As the elevator starts moving, the scale reads \(450 \mathrm{~N}\). (a) Find the acceleration of the elevator (magnitude and direction). (b) What is the acceleration if the scale reads \(670 \mathrm{~N}\) ? (c) If the scale reads zero, should the student worry? Explain. (d) What is the tension in the cable in parts (a) and (c)?

A \(2.00 \mathrm{~kg}\) box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to \(v(t)=\left(2.00 \mathrm{~m} / \mathrm{s}^{2}\right) t+\left(0.600 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\). What is the tension in the rope when the velocity of the box is \(9.00 \mathrm{~m} / \mathrm{s} ?\)

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