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Chapter 14: Problem 58

A 2.50 -kg rock is attached at the end of a thin, very light rope 1.45 \(\mathrm{m}\) long. You start it swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after 10\(\frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

Short Answer

Expert verified
The system loses energy due to air resistance and friction, converting kinetic energy into thermal energy and sound.

Step by step solution

01

Calculate Initial Height

First, determine the vertical displacement when the rock is released from an angle of \(11^{\circ}\). The height \(h\) can be calculated using the length of the rope \(L = 1.45 \, \text{m}\) and the angle \(\theta = 11^{\circ}\). Using the relation: \[ h = L (1 - \cos\theta)\]Substituting the values gives:\[ h_1 = 1.45 \, (1 - \cos 11^{\circ})\]
02

Calculate Final Height

Similarly, determine the vertical displacement when the rock rises to an angle of \(4.5^{\circ}\) after 10.5 swings. Using the same formula as before, calculate:\[h_2 = 1.45 \, (1 - \cos 4.5^{\circ})\]
03

Calculate Initial Potential Energy

The initial potential energy \(E_1\) can be calculated using:\[E_1 = mgh_1\]where \(m = 2.50 \, \text{kg}\), \(g = 9.81 \, \text{m/s}^2\). Substitute \(h_1\) from Step 1.
04

Calculate Final Potential Energy

The final potential energy \(E_2\) has to be calculated similarly:\[E_2 = mgh_2\]using \(h_2\) from Step 2.
05

Calculate Energy Lost

The energy lost \(\Delta E\) can be found by the difference:\[\Delta E = E_1 - E_2\]Calculate \(\Delta E\) using the values obtained for \(E_1\) and \(E_2\).
06

Explain Energy Loss

The energy lost is typically due to friction and air resistance. As the rock swings through the air and the rope moves, they experience resistance that converts kinetic energy into thermal energy and sound, which disperses into the environment.

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Key Concepts

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

Potential Energy Calculation
When analyzing pendulum motion, potential energy plays an important role. In this exercise, we calculate the potential energy at two different points of the pendulum's swing: the initial release point and the subsequent highest point reached after several swings.
Let's review how these calculations are done:
  • **Initial Height Calculation**: When the pendulum is released, it starts from a certain height above the lowest point of its swing. This height can be calculated using the formula: \[ h = L (1 - \cos \theta) \]where \( L \) is the length of the pendulum and \( \theta \) is the angle with the vertical.
  • **Potential Energy Formula**: The potential energy at any point is given by the formula: \[ E = mgh \]where \( m \) is the mass of the pendulum bob, \( g \) is the acceleration due to gravity (\(9.81 \text{ m/s}^2\)), and \( h \) is the height we calculated.
  • **Initial and Final Energies**: By using the above formulas, we determine the initial potential energy at the start and the energy at the final height (after recalling the angles: \(11^{\circ}\) initially and \(4.5^{\circ}\) after swings).
Understanding and calculating these potential energies allow us to analyze how much energy the system had initially and how much it has after undergoing motion.
Mechanical Energy Loss
In real-world systems, energy is rarely conserved in totality due to factors such as friction and air resistance, both of which are prevalent in pendulum motion. In this particular exercise, we observed that the pendulum does not swing back to its original height after several oscillations.
Let's consider how mechanical energy loss occurs in this scenario:
  • **Energy Loss Calculation**: After determining both initial and final potential energy using the calculations in the earlier section, the energy lost by the system can be found by: \[ \Delta E = E_1 - E_2 \]where \(E_1\) and \(E_2\) are the energies at initial (highest) and final (lower) points, respectively.
  • **Cause of Energy Loss**: As the pendulum swings, air resistance acts on it, reducing its speed. Simultaneously, internal friction within the rope or at the pivot point converts some kinetic energy into heat.
  • **Energy Transformation**: These transformations essentially cause mechanical (kinetic and potential) energy to dissipate into other forms, like thermal energy and sound. This means the pendulum will continually lose energy to its surroundings, preventing it from reaching its initial height.
Recognizing the energy conversion from mechanical forms to less useful forms helps understand why mechanical energy does not seem conserved.
Pendulum Motion Analysis
The motion of a pendulum is a classic example of simple harmonic motion, yet in practice, real-life pendula encounter forces that complicate this simple model. In this example, we observe how the pendulum's amplitude diminishes over time.
Let's analyze the pendulum's motion:
  • **Ideal Pendulum Motion**: Ideally, a pendulum with no external forces should swing back to its original height, maintaining constant mechanical energy. This would mean constant swings back and forth.
  • **Real-World Deviations**: In reality, each swing dissipates some energy due to external forces such as air resistance and friction. These forces act against the pendulum's motion, reducing its amplitude over time.
  • **Analysis Over Time**: As recorded in the initial conditions, after 10.5 swings, the pendulum's amplitude reduces significantly. Observing the angle from \(11^{\circ}\) to \(4.5^{\circ}\), we see a loss of height, a direct indication of energy loss in each swing.
Understanding pendulum motion involves grasping the concept of how theoretical scenarios may not hold in real-world situations due to continuous mechanical energy conversion into other, less useful forms.

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Most popular questions from this chapter

CP A rocket is accelerating upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) from the launchpad on the earth. Inside a small, 1.50 -kg ball hangs from the ceiling by a light, 1.10-m wire. If the ball is displaced \(8.50^{\circ}\) from the vertical and released, find the amplitude and period of the resulting swings of this pendulum.

A 1.80 -kg connecting rod from a car engine is pivoted about a horizontal knife edge as shown in Fig. E14.53. The center of gravity of the rod was located by balancing and is 0.200 \(\mathrm{m}\) from the pivot. When the rod is set into small-amplitude oscillation, it makes 100 complete swings in 120 s. Calculate the moment of inertia of the rod about the rotation axis through the pivot.

CP A rifle bullet with mass 8.00 \(\mathrm{g}\) and initial horizontal velocity 280 \(\mathrm{m} / \mathrm{s}\) strikes and embeds itself in a block with mass 0.992 \(\mathrm{kg}\) that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0 \(\mathrm{cm}\) . After the impact, the block moves in SHM. Calculate the period of this motion.

CP SHM in a Car Engine. The motion of the piston of an automobile engine is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 \(\mathrm{m}\) and the engine runs at 4500 \(\mathrm{rev} / \mathrm{min}\) , compute the acceleration of the piston at the endpoint of its stroke. (b) If the piston has mass \(0.450 \mathrm{kg},\) what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the mid- point of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?

Don't Miss the Boat. While on a visit to Minnesota ("Land of \(10,000\) Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the \(1500-\mathrm{kg}\) boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 \(\mathrm{cm} .\) The boat takes 3.5 \(\mathrm{s}\) s to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 \(\mathrm{kg}\) ) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 \(\mathrm{cm}\) of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?
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