/*! 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 75 To form a pendulum, a \(0.092 \m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

To form a pendulum, a \(0.092 \mathrm{~kg}\) ball is attached to one end of a rod of length \(0.62 \mathrm{~m}\) and negligible mass, and the other end of the rod is mounted on a pivot. The rod is rotated until it is straight up, and then it is released from rest so that it swings down around the pivot. When the ball reaches its lowest point, what are (a) its speed and (b) the tension in the rod? Next, the rod is rotated until it is horizontal, and then it is again released from rest. (c) At what angle from the vertical does the tension in the rod equal the weight of the ball? (d) If the mass of the ball is increased, does the answer to (c) increase, decrease, or remain the same?

Short Answer

Expert verified
(a) 3.48 m/s, (b) 3.44 N, (c) \(48.2^\circ\), (d) Angle remains the same.

Step by step solution

01

Identify the given variables and constants

We have a ball of mass, \( m = 0.092 \text{ kg} \), attached to a rod of length, \( L = 0.62 \text{ m} \). The ball is released from a height corresponding to the vertical position. The acceleration due to gravity is \( g = 9.8 \text{ m/s}^2 \).
02

Using Conservation of Energy to find speed at the lowest point

The ball starts with potential energy when it is at the topmost position and kinetic energy at the lowest point. Initially, all the energy is potential, \( U_i = mgh \) where \( h = L \). At the lowest point, it is all kinetic energy, \( K_f = \frac{1}{2} mv^2 \). Equating potential and kinetic energy, we get:\[ mgh = \frac{1}{2} mv^2 \].Solving for \( v \): \[ gh = \frac{1}{2} v^2 \implies v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.62} \approx 3.48 \text{ m/s}. \]
03

Calculate the tension in the rod at the lowest point

At the lowest point, the tension in the rod has to overcome the gravitational force and provide the centripetal force. Therefore, the tension can be given by \[ T = mg + \frac{mv^2}{L}. \]Substitute the known values:\[ T = 0.092 \times 9.8 + \frac{0.092 \times (3.48)^2}{0.62}. \]Compute this to get:\( T \approx 1.72 + 1.72 = 3.44 \text{ N}. \)
04

Understanding the scenario for tension equal to weight

In this scenario, the ball is released from the horizontal position, and the tension equals the weight when the net force toward the center of the circle equals zero. This can be expressed through \[ T = mg \Rightarrow mg = \frac{mv^2}{L}\cos\theta + mg\cos\theta, \]simplifying to \[ g \cos\theta = \frac{v^2}{L}. \]
05

Use energy conservation for the horizontal starting position

Initially, the height is \( L \sin(90) = L \) and final height is \( L \cos\theta \). Using the conservation of energy:\( mgL = mgh_f + \frac{1}{2}mv^2 \) gives us\( mgL = mgL\cos\theta + \frac{1}{2}mv^2 \),solving for \( v^2 \)\( v^2 = 2gL(1-\cos\theta) \).
06

Set up the equation for angle \(\theta \)

We equate this with our previous equation for \(g\cos\theta = \frac{v^2}{L} \) giving us\[ g \cos\theta = \frac{2gL(1-\cos\theta)}{L} \]\[ \cos\theta = 2(1-\cos\theta) \]\[ 3\cos\theta = 2 \Rightarrow \cos\theta = \frac{2}{3} \].Thus, \( \theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ \).
07

Effect of mass increase on the angle

The equation derived for \( \theta \) was independent of mass, that means the angle \( \theta \) is determined by geometry and gravity alone.\ Consequently, the answer to part (c) does not change with the mass of the ball.

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.

Conservation of Energy
In the world of physics, the conservation of energy principle tells us that energy cannot be created or destroyed, but can only change forms. When we talk about a pendulum, we're looking at the trade-off between potential and kinetic energy.
When the ball at the end of our pendulum starts at the highest point, it has maximum potential energy and zero kinetic energy—since it's not moving. This potential energy is calculated using the formula, \( U = mgh \), where \( m \) is the mass, \( g \) is the gravitational acceleration, and \( h \) is the height from which it is released.
  • At the top, energy is stored as gravitational potential.
  • As the pendulum swings down, potential energy is converted into kinetic energy \( K = \frac{1}{2}mv^2 \).
  • At the lowest point, all energy has been turned into kinetic, making the pendulum ball move at its greatest speed.
This simple conversion is paramount in solving pendulum problems, as it helps us find the speed at the lowest point using the equation: \( mgh = \frac{1}{2}mv^2 \). Solving for \( v \), we get \( v = \sqrt{2gh} \). Thus, knowing \( h = 0.62 \text{ m} \), we find \( v \approx 3.48 \text{ m/s} \).
Tension in the Rod
The tension in the rod is essential to keep the ball moving in a circular path. At any given moment, especially at the lowest point of the swing, this tension must handle a couple of forces:
  • The gravitational force tugging the ball downward, \( mg \).
  • The centripetal force needed to keep the ball moving in its curved path.
The formula to determine the tension at the lowest point of the swing is: \[ T = mg + \frac{mv^2}{L} \]Where \( v = 3.48 \text{ m/s} \) is the speed at the bottom we calculated using kinetic energy principles, and \( L = 0.62 \text{ m} \) is the length of the rod.
Using these values, we find that the tension in the rod is approximately \( T = 3.44 \text{ N} \). The rod must exert this amount of force to counteract downward gravity while making sure the ball continues in its circular motion.
Centripetal Force
Centripetal force is the "center-seeking" force that ensures any body undergoing circular motion does not move away from its circular path. It's the force that pulls the ball inwards, maintaining its path around the pivot.
When the ball swings to its lowest point, the centripetal force is provided by the tension in the rod, less the gravitational pull.
  • The total centripetal force required is determined by the equation: \( F_c = \frac{mv^2}{L} \).
  • This force is part of the total tension in the rod: \[ T = F_g + F_c = mg + \frac{mv^2}{L} \].
Here, we are using the speed \( v \) found previously to see how this inward force fits into the overall motion of the pendulum. Without adequate tension providing centripetal force, the ball would not be able to maintain its curved path; instead, it would attempt to move in a straight line due to inertia.
Gravitational Force
Gravitational force is the ever-present force that pulls objects toward each other; in our pendulum problem, it's the force pulling the ball downwards towards Earth's center. This force acts constantly on the pendulum, contributing significantly to its motion.
The gravitational force that acts on the ball can be described simply using the formula:
\( F_g = mg \)
  • This force plays the role of both initial energy source—as the pendulum drops—and as a constant force being countered at every point of the swing.
  • It directly affects the tension in the rod, as tension must be high enough to counteract this force, especially at the lowest point where both tension and gravity act downwardly.
It's important to understand that while the pendulum appears to just swing back and forth, behind this movement resides a fascinating interplay between gravitational force and the structural tension managing and balancing these natural forces.

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 certain spring is found \(n o t\) to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \(52.8 x+38.4 x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x=0.500 \mathrm{~m}\) to \(x=1.00 \mathrm{~m} .\) (b) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is stretched by an amount \(x=1.00 \mathrm{~m} .\) If the particle is then released from rest, what is its speed at the instant the stretch in the spring is \(x=0.500 \mathrm{~m} ?\) (c) Is the force exerted by the spring conservative or nonconservative? Explain.

Figure 8 - 34 shows a thin rod, of length \(L=2.00 \mathrm{~m}\) and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass \(m=5.00 \mathrm{~kg}\) is attached to the other end. The rod is pulled aside to angle \(\theta_{0}=30.0^{\circ}\) and released with initial velocity \(\vec{v}_{0}=0 .\) As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and (b) what is the change in the gravitational potential energy of the ball-Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released? (d) Do the magnitudes of the answers to (a) through (c) increase, decrease, or remain the same if angle \(\theta_{0}\) is increased?

A block of mass \(m=2.0 \mathrm{~kg}\) is dropped from height \(h=40 \mathrm{~cm}\) onto a spring of spring constant \(k=1960 \mathrm{~N} / \mathrm{m}\) (Fig. 8 -39). Find the maximum distance the spring is compressed.

A \(50 \mathrm{~g}\) ball is thrown from a window with an initial velocity of \(8.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) above the horizontal. Using energy methods, determine (a) the kinetic energy of the ball at the top of its flight and (b) its speed when it is \(3.0 \mathrm{~m}\) below the window. Does the answer to (b) depend on either (c) the mass of the ball or (d) the initial angle?

A 1500 kg car begins sliding down a \(5.0^{\circ}\) inclined road with a speed of \(30 \mathrm{~km} / \mathrm{h} .\) The engine is turned off, and the only forces acting on the car are a net frictional force from the road and the gravitational force. After the car has traveled \(50 \mathrm{~m}\) along the road, its speed is \(40 \mathrm{~km} / \mathrm{h} .\) (a) How much is the mechanical energy of the car reduced because of the net frictional force? (b) What is the magnitude of that net frictional force?

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.