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Chapter 11: Problem 68

A brass wire is \(1.40 \mathrm{~m}\) long and has a cross-sectional area of \(6.00 \mathrm{~mm}^{2}\). A small steel ball with mass \(0.0800 \mathrm{~kg}\) is attached to the end of the wire. You hold the other end of the wire and whirl the ball in a vertical circle of radius \(1.40 \mathrm{~m}\). What speed must the ball have at the lowest point of its path if its fractional change in length of the brass wire at this point from its unstretched length is \(2.0 \times 10^{-5}\) ? Treat the ball as a point mass.

Short Answer

Expert verified
The speed must be \( \approx 13.2 \mathrm{~m/s} \) at the lowest point of its path.

Step by step solution

01

Hooke's Law calculation

The fractional change in length due to the tension can be described using Hooke's law in the form \( \Delta L = \frac{T}{YA} \), where \( Y \) is the Young's modulus of brass \( = 9.0脳10^{10} \mathrm{~Nm^{-2}} \), \( A \) is the cross-sectional area \( = 6.00 脳 10^{-6} \mathrm{~m^{2}} \) and \( T \) is the tension in the wire. Rearranging for T, gives us \( T = \Delta L \times YA \). The given fractional change in length, \( \Delta L \) is \( 2.0 \times 10^{-5} \). Substituting the values into the equation gives us \( T = 2.0 \times 10^{-5} \times 9.0 脳 10^{10} \mathrm{~Nm^{-2}} \times 6.00 脳 10^{-6} = 10.8 \mathrm{~N} \).
02

Equation of Motion for the ball

The forces acting on the ball at the lowest point in its circular path are gravity and the tension from the wire. The ball travels in a circular path, so the net force acting on it is centripetal. The equation of motion thus becomes \( T = m (g + \frac{v^{2}}{r}) \), where \( m \) is the mass of the ball \( = 0.0800 \mathrm{~kg} \), \( g \) is the acceleration due to gravity \( = 9.8 \mathrm{~m/s^{2}} \), \( v \) is the speed of the ball, and \( r \) is the radius of the circle \( = 1.40 \mathrm{~m} \).
03

Solving for Speed

Substituting \( T = 10.8 \mathrm{~N} \) from step 1 into the equation from step 2 gives \( 10.8 \mathrm{~N} = 0.0800 \mathrm{~kg}(9.8 \mathrm{~m/s^{2}} + \frac{v^{2}}{1.40 \mathrm{~m}}) \). This equation can be re-arranged to solve for \( v = \sqrt{ ((\frac{T}{m}) - g) \times r} \). Substituting the known values gives \( v = \sqrt{((135 - 9.8) \times 1.40) = \sqrt{175} \mathrm{~m/s}} \)

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Key Concepts

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

Understanding Hooke's Law
Hooke's Law is a principle that describes the behavior of springs and other elastic objects. It dictates that the force needed to extend or compress a spring by some distance is proportional to that distance. This can be represented by the formula:
\[ F = k \times \triangle L \]
Where \( F \) is the force applied, \( k \) is the spring constant (a measure of the stiffness of the spring), and \( \triangle L \) is the change in length from the spring's original length. In the case of the circular motion problem involving a brass wire and a steel ball, Hooke's Law helps us calculate the tension in the wire when it is stretched by a particular amount due to the weight and motion of the ball. The wire behaves like a spring, and the tension in it is equivalent to the force that would be applied to a spring. So Hooke's Law transforms into the context of materials, allowing us to calculate the change in length (\( \triangle L \) as a result of a specific tension (\( T \). By rearranging the formula, we can solve for the tension necessary to produce the observed change in length when the ball is at the lowest point of its path.
Decoding Young's Modulus
Young's modulus, also known as the elastic modulus, is a measure of the stiffness of a solid material. It signifies how much a material will deform under a given load and can be calculated using the equation:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
Where 'Stress' is the force causing the deformation divided by the area to which the force is applied (\( \frac{F}{A} \)), and 'Strain' is the fractional change in length (\( \frac{\triangle L}{L_0} \)) with respect to the original length (\( L_0 \)). In our circular motion exercise, we apply Young's modulus to find out the tension in the wire when it stretches due to the centripetal force acting on the steel ball. The modulus tells us how much tension is needed for the wire to stretch by a certain amount, which is key to understanding how materials behave when they are part of a dynamic system like the rotating ball on a wire.
Centripetal Force in Circular Motion
Centripetal force is indispensable for understanding circular motion in physics. It is the force that keeps an object moving in a circular path and it's always directed towards the center of the circle. The magnitude of the centripetal force (\( F_c \)) required to keep an object moving along a circular path of radius \( r \) at speed \( v \) is given by:
\[ F_c = \frac{mv^2}{r} \]
In this formula, \( m \) symbolizes the mass of the object and \( v \) is its velocity at any given point in the circular path. Our exercise focuses on a ball being whirled around on a wire, which experiences this very force. As the ball is at the lowest point of its vertical circle, the tension in the wire provides the required centripetal force for the circular motion, alongside the gravitational force acting on the ball. Understanding how to calculate this force is crucial for solving problems involving circular motion, as seen in the step-by-step solution for the brass wire and steel ball scenario. By analyzing the forces at play and utilizing centripetal force calculations, we can determine the speed needed for the ball to maintain its circular trajectory.

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Most popular questions from this chapter

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 door \(1.00 \mathrm{~m}\) wide and \(2.00 \mathrm{~m}\) high weighs \(330 \mathrm{~N}\) and is supported by two hinges, one \(0.50 \mathrm{~m}\) from the top and the other \(0.50 \mathrm{~m}\) from the bottom. Each hinge supports half the total weight of the door. Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

A 3.00-m-long, \(190 \mathrm{~N}\), uniform rod at the \(\mathrm{zoo}\) is held in a horizontal position by two ropes at its ends (Fig. E11.21). The left rope makes an angle of \(150^{\circ}\) with the rod, and the right rope makes an angle \(\theta\) with the horizontal. A \(90 \mathrm{~N}\) howler monkey (Alouatta seniculus) hangs motionless \(0.50 \mathrm{~m}\) from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\). First make a free-body diagram of the rod.

A claw hammer is used to pull a nail out of a board (see Fig. \(\mathrm{P} 11.45\) ). The nail is at an angle of \(60^{\circ}\) to the board, and a force \(\overrightarrow{\boldsymbol{F}}_{1}\) of magnitude \(400 \mathrm{~N}\) applied to the nail is required to pull it from the board. The hammer head contacts the board at point \(A,\) which is \(0.080 \mathrm{~m}\) from where the nail enters the board. A horizontal force \(\vec{F}_{2}\) is applied to the hammer handle at a distance of \(0.300 \mathrm{~m}\) above the board. What magnitude of force \(\overrightarrow{\boldsymbol{F}}_{2}\) is required to apply the required \(400 \mathrm{~N}\) force \(\left(F_{1}\right)\) to the nail? (Ignore the weight of the hammer.)

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?
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