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The length of a simple pendulum is 0.79 \(\mathrm{m}\) and the mass of the particle (the bob) at the end of the cable is 0.24 \(\mathrm{kg}\) . The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

Short Answer

Expert verified
(a) \( \omega \approx 3.52 \text{ rad/s} \). (b) \( E \approx 0.027 \text{ J} \). (c) \( v \approx 0.33 \text{ m/s} \).

Step by step solution

01

Identify the formula for angular frequency

The angular frequency for a simple pendulum undergoing simple harmonic motion is given by the formula: \( \omega = \sqrt{\frac{g}{L}} \), where \( g \) is the acceleration due to gravity \( (9.81 \, \text{m/s}^2) \) and \( L \) is the length of the pendulum.
02

Calculate the angular frequency

Using the given pendulum length \( L = 0.79 \, \text{m} \), substitute into the formula for angular frequency: \( \omega = \sqrt{\frac{9.81}{0.79}} \). This simplifies to \( \omega \approx 3.52 \, \text{rad/s} \).
03

Determine the formula for total mechanical energy

The total mechanical energy \( E \) of a pendulum in simple harmonic motion is given by \( E = \frac{1}{2} k A^2 \), where \( k \) is the effective spring constant and \( A \) is the amplitude. For a pendulum, \( k = \frac{mg}{L} \) and \( A = L\sin(\theta_0) \), with \( m = 0.24 \, \text{kg} \) and \( \theta_0 = 8.50^{\circ} \).
04

Calculate the total mechanical energy

Convert \( \theta_0 = 8.50^{\circ} \) to radians: \( \theta_0 = \frac{8.50 \times \pi}{180} \) radians. Hence, the amplitude \( A = 0.79 \cdot \sin(8.50^{\circ}) \). \( k = \frac{0.24 \times 9.81}{0.79} \approx 2.97 \, \text{N/m} \). Now calculate \( E = \frac{1}{2} \times 2.97 \times (0.79 \cdot \sin(8.50^{\circ}))^2 \approx 0.027 \, \text{J} \).
05

Apply conservation of energy to find speed at lowest point

At the lowest point, all the mechanical energy is kinetic: \( E = \frac{1}{2}mv^2 \). Use this to solve for speed, \( v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 0.027}{0.24}} \approx 0.33 \, \text{m/s} \).

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Key Concepts

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

simple pendulum
A simple pendulum is a fascinating example of simple harmonic motion. It consists of a weight, or bob, which is attached to a fixed point by a string or rod of negligible mass.
The bob swings back and forth through an arc under the influence of gravity. The motion is periodic and predictable.
  • The restoring force is due to gravity, which acts to bring the pendulum back to its lowest point.
  • The pendulum exhibits simple harmonic motion for small angles (usually less than 15 degrees) due to the linear relationship between the restoring force and displacement.
  • The period of a pendulum does not depend on its mass but only its length and the acceleration due to gravity.
This type of motion is used extensively in clocks and other timing devices. It's a fantastic demonstration of periodic motion that is affected only by the length of the pendulum and gravity, making it a simple yet effective tool for measuring time.
angular frequency
Angular frequency, often represented by the Greek letter \( \omega \), is a measure that describes how quickly the pendulum oscillates. It's the rate of rotation for any periodic motion and is directly connected to the concept of the simple pendulum.
For a pendulum, the formula for angular frequency is \( \omega = \sqrt{\frac{g}{L}} \), where \( g \) represents the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \), and \( L \) is the length of the pendulum.
  • Angular frequency provides insight into how fast the pendulum goes back and forth.
  • Higher angular frequency means faster oscillation, which happens when the pendulum is shorter.
  • Angular frequency is measured in radians per second, reflecting its cyclic nature.
Understanding angular frequency helps us predict how long it will take the pendulum to complete one full oscillation. It is part of the broader understanding of periodic systems and essential for examining more complex harmonic systems.
mechanical energy
Mechanical energy in a simple pendulum consists of both potential and kinetic forms. When the pendulum is lifted to its highest point, potential energy is at its maximum. As it swings down, this energy transforms into kinetic energy.
The total mechanical energy of the pendulum remains constant if we neglect air resistance and friction, which aligns with the principles of conservation of energy.
  • Potential energy depends on the height of the bob and is highest at the extreme points of the swing.
  • Kinetic energy is greatest at the lowest point of the swing, where speed is maximized.
  • Total mechanical energy does not change, albeit the distribution between potential and kinetic varies during the motion.
This constant energy exchange in a frictionless scenario illustrates a fundamental concept in physics, showing how energy is conserved in an isolated system, enhancing our understanding of oscillatory motion.
conservation of energy
Conservation of energy is a critical concept that applies perfectly to the motion of a simple pendulum. It states that in a closed system, without external forces doing work, the total energy remains constant.
In the context of a pendulum, this means all potential energy is converted to kinetic energy and vice versa.
  • At the highest points, all the pendulum’s energy is potential (gravitational).
  • At the lowest point, energy is purely kinetic as the pendulum moves fastest.
  • Throughout the swing, the total energy, which is the sum of kinetic and potential energies, does not change.
This principle allows us to calculate variables like the pendulum’s speed at various points using energy relations rather than forces. It underscores a vital aspect of physics: energy transformation without loss in ideal conditions is a timeless trait across various systems.

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Most popular questions from this chapter

An 86.0 -kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{N} / \mathrm{m}\) . He accidentally slips and falls freely for 0.750 \(\mathrm{m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

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