/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 An amusement park ride consists ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An amusement park ride consists of airplane-shaped cars attached to steel rods (Fig. \(\mathbf{P} \mathbf{1 1 . 8 4}\) ). Each rod has a length of \(15.0 \mathrm{~m}\) and a cross-sectional area of \(8.00 \mathrm{~cm}^{2}\). The rods are attached to a frictionless hinge at the top, so that the cars can swing outward when the ride rotates. (a) How much is each rod stretched when it is vertical and the ride is at rest? (Assume that each car plus two people seated in it has a total weight of \(1900 \mathrm{~N}\).) (b) When operating, the ride has a maximum angular speed of 12.0 rev \(/\) min. How much is the rod stretched then?

Short Answer

Expert verified
The rod is stretched by 1.78 mm when it is vertical and at rest, and by approximately 52.8 mm when the ride is operating at maximum angular speed.

Step by step solution

01

Initial Elongation

First, calculate the initial elongation due to the weight. Use the formula for the elongation of a material under force, which is \( \Delta L = FL/EA \). F = 1900 N (given weight of the car and passengers), L = 15 m (given length of the steel rod), A = 8 cm² = 8 x \(10^{-4}\) m² (cross-sectional area of the steel rod converted to m²), E = 200 x \(10^9\) N/m² (Approximate modulus of elasticity of steel). Substituting the values, we get \( \Delta L_1 = (1900*15)/(200*10^9*8*10^{-4}) = 1.78 * 10^{-3} m \)
02

Elongation Due To Angular Speed

For the second part, calculate the additional elongation due to the operational speed. The centripetal force (Fc) is calculated by = \(mv^2/r = mrω^2\). Since, \(ω\) = 2πf and f = 12 revolution/minute = 12/60 Hz, we can substitute Fc into the formula of elongation to get \( \Delta L_2 = (mrω^2L)/EA = 15*(1900/9.8)*(4π^2(12/60)^2*15)/(200*10^9*8*10^{-4}) = 0.051 m \). The total elongation when the ride is running is the sum of the initial elongation and the elongation due to the operational speed.
03

Total Elongation

Sum both elongations obtained from Steps 1 and 2 to get the total elongation of the rod when the ride is in operation. \( \Delta L_{Total} = \Delta L_1 + \Delta L_2 = 1.78 * 10^{-3} + 0.051 = 0.0528 m \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Elongation of Material
Understanding the elongation of materials is crucial for fields like engineering and physics as it helps us predict how structures behave under various forces. At its most basic, elongation refers to the extent to which a material stretches or lengthens when a force is applied to it. This concept can be seen in everyday life, such as in a rubber band stretching when pulled or the slight lengthening of a bridge as vehicles pass over it.

In our exercise scenario, the elongation of the steel rods in an amusement park ride is calculated using a fundamental formula: \( \Delta L = \frac{F \cdot L}{E \cdot A} \). This equation considers the force (F) applied, the original length of the material (L), the modulus of elasticity (E) which is a property of the material, and the cross-sectional area (A). The calculated elongation \( \Delta L \) tells us how much the steel rods stretch both when the ride is at rest and in motion.

It's noteworthy to mention that real-life materials usually have limits to their elasticity, and if overstretched, they might reach a point of plastic deformation where they do not return to their original length. This factor is important in design safety.
Modulus of Elasticity
The modulus of elasticity, also known as Young's modulus, is a measure of the stiffness of a material. It reflects how much a material will deform under a certain amount of stress and is a fundamental property that dictates how well a material can resist changing shape when force is applied.

In our exercise, we deal with steel, which has a high modulus of elasticity, indicative of its stiffness and resistance to being deformed (approximately 200 x \(10^9\) N/m²). The modulus of elasticity is what we use in the equation \( \Delta L = \frac{F \cdot L}{E \cdot A} \) to understand the initial elongation due to the weight of the cars and passengers when the amusement park ride is at rest. Materials with high moduli of elasticity are often chosen in construction and manufacturing when minimal deformation is desirable.
Centripetal Force
Centripetal force plays a central role in any motion that takes place along a circular path. It's the force that acts on an object moving in a circular route, directed towards the center of the circle. This force is essential for understanding how an object behaves when it's undergoing circular motion, such as the cars in our amusement park ride example.

When calculating the additional elongation due to the ride's motion, we consider the centripetal force using the formula \( F_c = m \cdot r \cdot \omega^2 \), where \( m \) is the mass of the object, \( r \) is the radius of the circular path, and \( \omega \) is the angular speed. Here, the centripetal force contributes to the total force stretching the steel rods and is a result of the combined effects of mass, angular speed, and the radius of rotation. This calculation highlights how physical forces can have practical and predictable impacts on materials.
Angular Speed
Angular speed is the rate at which an object rotates around a circle or pivot point, described in terms of the angle traversed per unit of time. In physics, it offers a way to quantify the rotational movement; determining how fast an object is spinning or revolving. For circular motion, angular speed (\(\omega\)) is often expressed in radians per second, but can also be represented as revolutions per minute (rpm), as seen in our exercise.

When the amusement park ride operates at a maximum of 12 revolutions per minute, we convert this measure to the standard unit (Hz) and then use angular speed to calculate the centripetal force causing further elongation in the steel rods. This demonstrates the practical application of angular speed in calculating dynamics of rotational systems and contributes to determining total forces experienced by structural components in motion.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) In Fig. P11.64 a \(6.00-\mathrm{m}\) -long, uniform beam is hanging from a point \(1.00 \mathrm{~m}\) to the right of its center. The beam weighs \(140 \mathrm{~N}\) and makes an angle of \(30.0^{\circ}\) with the vertical. At the right-hand end of the beam a \(100.0 \mathrm{~N}\) weight is hung; an unknown weight \(w\) hangs at the left end. If the system is in equilibrium, what is \(w ?\) You can ignore the thickness of the beam. (b) If the beam makes, instead, an angle of \(45.0^{\circ}\) with the vertical, what is \(w ?\)

A \(0.120 \mathrm{~kg},\) 50.0-cm-long uniform bar has a small \(0.055 \mathrm{~kg}\) mass glued to its left end and a small \(0.110 \mathrm{~kg}\) mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

Two people carry a heavy electric motor by placing it on a light board \(2.00 \mathrm{~m}\) long. One person lifts at one end with a force of \(400 \mathrm{~N}\), and the other lifts the opposite end with a force of \(600 \mathrm{~N}\). (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{~N},\) with its center of gravity at its center, and the two people exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

You are a construction engineer working on the interior design of a retail store in a mall. A 2.00 -m-long uniform bar of mass \(8.50 \mathrm{~kg}\) is to be attached at one end to a wall, by means of a hinge that allows the bar to rotate freely with very little friction. The bar will be held in a horizontal position by a light cable from a point on the bar (a distance \(x\) from the hinge) to a point on the wall above the hinge. The cable makes an angle \(\theta\) with the bar. The architect has proposed four possible ways to connect the cable and asked you to assess them: $$ \begin{array}{lllll} \text { Alternative } & \text { A } & \text { B } & \text { C } & \text { D } \\\ \hline x(\mathrm{~m}) & 2.00 & 1.50 & 0.75 & 0.50 \\ \theta(\text { degrees }) & 30 & 60 & 37 & 75 \end{array} $$ (a) There is concern about the strength of the cable that will be required. Which set of \(x\) and \(\theta\) values in the table produces the smallest tension in the cable? The greatest? (b) There is concern about the breaking strength of the sheetrock wall where the hinge will be attached. Which set of \(x\) and \(\theta\) values produces the smallest horizontal component of the force the bar exerts on the hinge? The largest? (c) There is also concern about the required strength of the hinge and the strength of its attachment to the wall. Which set of \(x\) and \(\theta\) values produces the smallest magnitude of the vertical component of the force the bar exerts on the hinge? The largest? (Hint: Does the direction of the vertical component of the force the hinge exerts on the bar depend on where along the bar the cable is attached?) (d) Is one of the alternatives given in the table preferable? Should any of the alternatives be avoided? Discuss.

A uniform \(300 \mathrm{~N}\) trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.