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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. \( \Delta L_{Total} = \Delta L_1 + \Delta L_2 = 1.78 * 10^{-3} + 0.051 = 0.0528 m \).

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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.

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Most popular questions from this chapter

You are to use a long, thin wire to build a pendulum in a science mu- seum. The wire has an unstretched length of \(22.0 \mathrm{~m}\) and a cir- cular cross section of \(\begin{array}{lll}\text { diameter } & 0.860 & \mathrm{~mm}\end{array}\) it is made of an alloy that has a large breaking stress. One end of the wire will be attached to the ceiling, and a \(9.50 \mathrm{~kg}\) metal sphere will be attached to the other end. As the pendulum swings back and forth, the wire's maximum angular displacement from the vertical will be \(36.0^{\circ}\). You must determine the maximum amount the wire will stretch during this motion. So, before you attach the metal sphere, you suspend a test mass (mass \(m\) ) from the wire's lower end. You then measure the increase in length \(\Delta l\) of the wire for several different test masses. Figure \(\mathbf{P} 1 \mathbf{1} .86,\) a graph of \(\Delta l\) versus \(m\) shows the results and the straight line that gives the best fit to the data. The equation for this line is \(\Delta l=(0.422 \mathrm{~mm} / \mathrm{kg}) m\) (a) Assume that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2},\) and use Fig. \(\mathrm{P} 11.86\) to calculate Young's modulus \(Y\) for this wire. (b) You remove the test masses, attach the \(9.50 \mathrm{~kg}\) sphere, and release the sphere from rest, with the wire displaced by \(36.0^{\circ} .\) Calculate the amount the wire will stretch as it swings through the vertical. Ignore air resistance.

A steel cable with cross-sectional area \(3.00 \mathrm{~cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8} \mathrm{~Pa}\). Find the maximum upward acceleration that can be given a \(1200 \mathrm{~kg}\) elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

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