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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) Angular frequency is 3.52 rad/s. (b) Total mechanical energy is 0.0203 J. (c) Speed at lowest point is 0.411 m/s.

Step by step solution

01

Calculate Angular Frequency

To find the angular frequency \( \omega \) of the pendulum's motion, use the formula for a simple pendulum in simple harmonic motion: \[ \omega = \sqrt{\frac{g}{L}} \] where \( g = 9.81 \frac{\text{m}}{\text{s}^2} \) is the acceleration due to gravity and \( L = 0.79 \text{ m} \) is the length of the pendulum.\[\omega = \sqrt{\frac{9.81}{0.79}} = \sqrt{12.4177} = 3.52 \text{ rad/s}.\]
02

Calculate Total Mechanical Energy

The total mechanical energy \( E \) in a simple pendulum system consists solely of potential and kinetic energy. At the maximum displacement (initial position), the energy is purely potential: \[ E = mgh. \] The height \( h \) is found from the length \( L \) and the angle \( \theta = 8.50^{\circ} \) using \( h = L(1 - \cos\theta) \). Convert \( \theta \) to radians: \( \theta = \frac{8.50\pi}{180} \).\[ \theta = 0.1484 \text{ rad}. \ h = 0.79(1 - \cos 0.1484) = 0.79(1 - 0.9891) \approx 0.00864 \text{ m}. \]Calculate the energy:\[E = 0.24 \times 9.81 \times 0.00864 = 0.0203 \text{ J}.\]
03

Calculate the Bob's Speed

At the lowest point, all the mechanical energy is kinetic. The speed \( v \) can be found using:\[ E = \frac{1}{2}mv^2. \]Substitute the known values for mass \( m \) and energy \( E \):\[0.0203 = \frac{1}{2} \times 0.24 \times v^2. \v^2 = \frac{2 \times 0.0203}{0.24} \approx 0.1692.\]Thus, \[ v = \sqrt{0.1692} \approx 0.411 \text{ m/s}. \]

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Key Concepts

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

Angular Frequency
Angular frequency is a key concept in understanding the motion of pendulums. It represents how quickly an object oscillates back and forth. For a simple pendulum, like the one in this exercise, the motion can be described using the formula for angular frequency \[\omega = \sqrt{\frac{g}{L}}. \]Here, \(g\) is the acceleration due to gravity, typically \(9.81 \frac{\text{m}}{\text{s}^2}\), and \(L\) is the length of the pendulum. Knowing the angular frequency helps in predicting the time period and the dynamics of the pendulum. It tells us how many oscillations the pendulum makes in a given time. In this problem, the pendulum with a length of \(0.79\, \text{m}\) has an angular frequency of approximately \(3.52\, \text{rad/s}\). This means that the pendulum completes \(3.52\) radians of oscillation every second, which is fundamental to understand the pendulum’s entire motion.
  • Angular frequency is directly related to the length of the pendulum.
  • Shorter pendulums have higher angular frequencies.
  • Angular frequency is measured in radians per second \((\text{rad/s})\).
Mechanical Energy
Mechanical energy in the context of pendulum motion is the sum of its kinetic and potential energies. It remains constant if we assume there are no non-conservative forces like air resistance acting on the pendulum. At the initial position, when the pendulum is at its maximum angle, its potential energy is maximized and kinetic energy is zero because the pendulum is momentarily stationary. As the pendulum swings down, potential energy is converted into kinetic energy. The lowest point of the swing is where kinetic energy is maximized, since this is the point of greatest speed, and potential energy is minimized.For this problem, the total mechanical energy \(E\) can be calculated using the potential energy at maximum displacement \[E = mgh,\]where \(m\) is the mass of the pendulum bob, \(g\) is the acceleration due to gravity, and \(h\) is the height relative to the lowest point. The energy calculated for our pendulum is approximately \(0.0203\, \text{J}\).
  • Mechanical energy remains constant in an ideal pendulum.
  • Total energy is a blend of potential and kinetic energies.
  • Conserved mechanical energy allows us to calculate velocities and positions.
Pendulum Motion
Pendulum motion is a classic example of simple harmonic motion (SHM), characterized by its back-and-forth oscillation about an equilibrium position. In SHM, the restoring force that brings the pendulum back to its equilibrium is proportional to the displacement from that position. In our scenario, when the pendulum is displaced by an angle of \(8.50^{\circ}\), it begins to move back and forth. The key points of interest in pendulum motion are its amplitude (the maximum angular displacement), period (the time taken to complete one full oscillation), and the phase of the motion which describes its position at any given time.At the equilibrium position, the pendulum's speed is at its maximum and all the potential energy is converted into kinetic energy. This was evident in our problem when calculating the bob's speed: we used the total mechanical energy, knowing it was entirely kinetic at the lowest point, to find the speed approximately \(0.411\, \text{m/s}\).
  • Pendulum demonstrates repetitive and predictable SHM.
  • The potential energy is maximum at the peak of the swing.
  • The pendulum's velocity is highest as it passes through equilibrium.

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Most popular questions from this chapter

A 15.0 -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of \(5.00 \mathrm{m} / \mathrm{s}\) in \(0.500 \mathrm{s}\). In the process, the spring is stretched by \(0.200 \mathrm{m} .\) The block is then pulled at a constant speed of \(5.00 \mathrm{m} / \mathrm{s}\), during which time the spring is stretched by only \(0.0500 \mathrm{m} .\) Find \((\mathrm{a})\) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

A pen contains a spring with a spring constant of \(250 \mathrm{N} / \mathrm{m}\). When the tip of the pen is in its retracted position, the spring is compressed \(5.0 \mathrm{mm}\) from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional \(6.0 \mathrm{mm}\). How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.

A block of mass \(m=0.750 \mathrm{kg}\) is fastened to an unstrained horizontal spring whose spring constant is \(k=82.0 \mathrm{N} / \mathrm{m} .\) The block is given a displacement of \(+0.120 \mathrm{m},\) where the \(+\) sign indicates that the displacement is along the \(+x\) axis, and then released from rest. (a) What is the force (magnitude and direction) that the spring exerts on the block just before the block is released? (b) Find the angular frequency \(\omega\) of the resulting oscillatory motion. (c) What is the maximum speed of the block? (d) Determine the magnitude of the maximum acceleration of the block.

Astronauts on a distant planet set up a simple pendulum of length \(1.2 \mathrm{m} .\) The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

A \(1.00 \times 10^{-2}\) -kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is \(124 \mathrm{N} / \mathrm{m}\). The block is shoved parallel to the spring axis and is given an initial speed of \(8.00 \mathrm{m} / \mathrm{s},\) while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

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