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Chapter 5: Problem 46

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end (Fig. E5.46). Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 \(\mathrm{m}\) from the central shaft. (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
(a) The period is approximately 5.32 seconds. (b) The angle does not depend on passenger weight.

Step by step solution

01

Visualize the Problem

Imagine the Giant Swing as a circular motion system where the cable creates an angle with the vertical. We are to find the time for one complete revolution of this circular motion.
02

Understand the Geometry

The cable forms two right triangles. The hypotenuse of each triangle is the cable length (5.00 m), and the base is horizontal from the arm to the seat. The height is from the top of the cable to the seat. Given that the cable makes an angle of 30掳 with the vertical, create an equation using trigonometry: \( \cos(\theta) = \frac{3.00 \text{ m}}{x} \), where \( x \) is the distance from the shaft to the seat along the arc.
03

Solve for Horizontal Distance

We are given \( \theta = 30^{\circ} \) and need to find the horizontal distance \( R \), the radius of the circle traced by the seat. By trigonometry: \( \cos(30^{\circ}) = \frac{3.00}{x} \). This simplifies to \( x = \frac{3.00}{\cos(30^{\circ})} = \frac{3.00}{\sqrt{3}/2} = 3\sqrt{3} \).
04

Calculate the Radius of the Circle

Convert \( x \) to the horizontal radius \( R \) by including the length of the arm: \[ R = 3\sqrt{3} - 3 = 3(\sqrt{3} - 1) \approx 2.598 \text{ m} \].
05

Use Uniform Circular Motion

The centripetal force provides acceleration towards the center, which is \( R\omega^2 \), where \( \omega = \frac{2\pi}{T} \) and \( T \) is the period. The tension in the cable provides this centripetal force, meaning: \( T\sin(\theta) = mR\omega^2 \).
06

Relate Period to Radius

By substituting the geometric relation: \( g \tan(\theta) = R\omega^2 \). Plug in the known values, \( g = 9.8 \text{ m/s}^2 \), \( \tan(30^{\circ}) = \frac{1}{\sqrt{3}} \), and solve for \( \omega \). Then use \( \omega = \frac{2\pi}{T} \) to find the period \( T \).
07

Substitute and Solve for Period

Use \( g \tan(30^{\circ}) = R\left(\frac{2\pi}{T}\right)^2 \) and calculate: \[ 9.8 \times \frac{1}{\sqrt{3}} = 2.598 \times \left(\frac{2\pi}{T}\right)^2 \]. Solving gives \( T = \frac{2\pi}{\sqrt{\frac{9.8 \times \frac{1}{\sqrt{3}}}{2.598}}} \approx 5.32 \text{ s} \).
08

Determine Dependence on Mass

Examine if the angle depends on mass from the equations relating tension, weight, and forces. Notice that mass cancels out when calculating the tangent, hence the angle only depends on rotation rate and not the weight of the passenger.

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Key Concepts

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

circular motion
Circular motion is a common phenomenon where an object moves along a circular path. In the case of the Giant Swing, each seat attached to the cable engages in this type of motion. The key element in circular motion is the centripetal force, which acts toward the center of the circular path, keeping the object from flying off in a straight line.
  • The centripetal force is crucial for maintaining circular motion and is provided by the tension in the swing's cables.
  • This force constantly changes the direction of the seat, ensuring the swing follows a circular path.
When we visualize the Giant Swing, each seat moves in a circle due to this centripetal force. It's important to remember that the speed remains constant, but the direction is always changing to keep the motion circular.
trigonometry in physics
Trigonometry plays a vital role in understanding the behavior of the Giant Swing. It helps us to break down and analyze the forces and distances involved in the system. In this exercise, trigonometry is used to find the radius of the circular path of the seat.
  • The cable makes a 30掳 angle with the vertical, allowing us to form right triangles to solve for distances using trigonometric functions like cosine.
  • Trigonometric equations can simplify complex geometrical relationships in physics problems, as seen with the arm and the seat on the swing.
For this swing, we need to use the angle and the given distances to determine the radius of the circle. Employing the cosine of 30掳 helps us determine how far the seat extends from the center of the rotation axis. By solving these trigonometric equations, we can find the necessary measurements to understand the motion fully.
period of revolution
The period of revolution, often denoted as \( T \), is the time it takes for an object to make one complete circular orbit. In the Giant Swing, this means the time it takes for the seat to return to its starting point after a full rotation. Calculating the period involves understanding the relationship between radial distance, speed, and centripetal force.
  • The period is tightly connected to the speed of the swing and the centripetal force needed to maintain that speed.
  • Using known values such as gravitational acceleration \( g \) and the trigonometric tan of the angle, we can compute \( T \) through the formula: \( \omega = \frac{2\pi}{T} \).
For the given swing configuration, solving for \( T \) involves substituting these values into the formula to yield a period of approximately 5.32 seconds. This calculation shows that conditions like the length of the cable and the speed of rotation thoroughly influence the time for one revolution.

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Most popular questions from this chapter

A 3.00 -kg box that is several hundred meters above the surface of the earth is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope, and this results in a tension in the rope of \(T(t)=(36.0 \mathrm{N} / \mathrm{s}) t .\) The box is at rest at \(t=0 .\) The only forces on the box are the tension in the rope and gravity. (a) What is the velocity of the box at ( i \(t=1.00 \mathrm{s}\) and (ii) \(t=3.00 \mathrm{s} ?\) (b) What is the maximum distance that the box descends below its initial position? (c) At what value of \(t\) does the box return to its initial position?

A hammer is hanging by a light rope from the ceiling of a bus. The ceiling of the bus is parallel to the roadway. The bus is traveling in a straight line on a horizontal street. You observe that the hammer hangs at rest with respect to the bus when the angle between the rope and the ceiling of the bus is \(67^{\circ} .\) What is the acceleration of the bus?

You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with mass \(0.500 \mathrm{kg},\) suspended from the ceiling of the bus by a string 1.80 \(\mathrm{m}\) long, is found to hang at rest relative to the bus when the string makes an angle of \(30.0^{\circ}\) with the vertical. In this position the lunch box is 50.0 \(\mathrm{m}\) from the center of curvature of the curve. What is the speed \(v\) of the bus?

Two boxes connected by a light horizontal rope are on a horizontal surface, as shown in Fig. P5.35. The coefficient of kinetic friction between each box and the surface is \(\mu_{\mathrm{k}}=0.30\) . One box (box \(B\) ) has mass \(5.00 \mathrm{kg},\) and the other box (box \(A )\) has mass \(m .\) A force \(F\) with magnitude 40.0 \(\mathrm{N}\) and direction \(53.1^{\circ}\) above the horizontal is applied to the \(5.00-\mathrm{kg}\) box, and both boxes move to the right with \(a=1.50 \mathrm{m} / \mathrm{s}^{2}\) . (a) What is the tension \(T\) in the rope that connects the boxes? (b) What is the mass \(m\) of the second box?

A rock with mass \(m\) slides with initial velocity \(v_{0}\) on a horizontal surface. A retarding force \(F_{\mathrm{R}}\) that the surface exerts on the rock is proportional to the square root of the instantaneous velocity of the rock \(\left(F_{\mathrm{R}}=-k v^{1 / 2}\right)\) . (a) Find expressions for the velocity and position of the rock as a function of time. (b) In terms of \(m, k,\) and \(v_{0},\) at what time will the rock come to rest? (c) In terms of \(m, k,\) and \(v_{0},\) what is the distance of the rock from its starting point when it comes to rest?
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