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Chapter 16: Problem 34

34\. Demolition Ball In Fig. \(16-37\), a \(2500 \mathrm{~kg}\) demolition ball swings from the end of a crane. The length of the swinging segment of cable is \(17 \mathrm{~m}\). (a) Find the period of the swinging, assuming that the system can be treated as a simple pendulum. (b) Does the period depend on the ball's mass?

Short Answer

Expert verified
(a) The period is approximately 8.27 seconds. (b) No, the period does not depend on the ball's mass.

Step by step solution

01

Understanding the Simple Pendulum Model

To solve this problem, recognize that a swinging demolition ball attached to a crane can be approximated as a simple pendulum. The period of a simple pendulum depends on the length of the cable and the acceleration due to gravity, but not on the mass of the ball.
02

Applying the Formula for the Period

The period of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \]where: - \( T \) is the period, - \( L \) is the length of the pendulum ( L = 17 \text{ meters} ), and - \( g \) is the acceleration due to gravity (\( g \approx 9.8 \text{ meters per second squared} \)).
03

Calculating the Period

Substitute the values for \( L \) and \( g \) into the formula: \[ T = 2\pi \sqrt{\frac{17}{9.8}} \]Calculate the inside term first:\[ \frac{17}{9.8} \approx 1.7347 \]Now take the square root:\[ \sqrt{1.7347} \approx 1.317 \]Finally, multiply by \(2\pi\): \[ T = 2\pi \times 1.317 \approx 8.27 \, \text{seconds} \]
04

Answering Part (b)

The period of a simple pendulum, as shown by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), does not depend on the mass of the pendulum. Therefore, the mass of the ball does not affect the period.

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Key Concepts

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

Period Calculation
The period of a simple pendulum is the time it takes for the pendulum to swing back and forth once. This is a crucial concept when solving problems related to pendulums. The period is given by the formula: \ \( T = 2\pi \sqrt{\frac{L}{g}} \). Here, \( T \) represents the period, \( L \) is the length of the pendulum, and \( g \) stands for the acceleration due to gravity.

To calculate the period:
  • First, substitute the values into the formula
  • Solve the fraction \( \frac{L}{ g} \)
  • Take the square root of the result
  • Finally, multiply by \( 2\pi \)
In our example: \( L = 17 \) m and \( g = 9.8 \) m/s虏, calculations show \ \( T \approx 8.27 \) seconds. Thus, the demolition ball takes about 8.27 seconds for a complete swing.
Pendulum Physics
A simple pendulum consists of a mass (also known as the bob) attached to a string of a certain length that swings under the influence of gravity. When released, the pendulum swings in a regular, periodic motion. This motion is governed by the laws of physics.
Key Points to Remember:
  • The pendulum's motion is a type of harmonic motion
  • The period does not depend on the mass of the bob
  • The amplitude of the swing (initial displacement) should be small for the simple pendulum formula to be accurate
This approach assumes negligible air resistance and a small angular displacement. As long as these conditions are met, the simple pendulum model is a good approximation.
Acceleration Due to Gravity
Gravity plays a vital role in the pendulum's motion. The acceleration due to gravity, denoted by \( g \), is approximately \( 9.8 \, m/s^2 \) on the surface of the Earth. This constant determines how fast an object accelerates when dropped.

In the context of a pendulum:
  • \( g \) appears in the denominator of the formula for the period
  • A higher value of \( g \) would result in a shorter period
  • A lower value of \( g \) would lengthen the period


For instance, if you took the demolition ball to a planet where \( g \) was different, the period would change accordingly. Thus, understanding \( g \) is crucial for accurately calculating the period of a pendulum.

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Most popular questions from this chapter

11\. Automobile Spring An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of \(3.00 \mathrm{~Hz}\). (a) What is the spring constant of each spring if the mass of the car is \(1450 \mathrm{~kg}\) and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging \(73.0 \mathrm{~kg}\) each, ride in the car? (Again, consider an even distribution of mass.)

10\. Speaker Diaphragm A loudspeaker diaphragm is oscillating in simple harmonic motion with a frequency of \(440 \mathrm{~Hz}\) and \(\mathrm{a}\) maximum displacement of \(0.75 \mathrm{~mm}\). What are (a) the angular frequency, (b) the maximum speed and (c) the magnitude of the maximum acceleration?

46\. Angular Amplitude For a simple pendulum, find the angular amplitude \(\Theta\) at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by \(1.0 \%\). (See "Trigonometric Expansions" in Appendix E.)

49\. Mechanical Energy Find the mechanical energy of a blockspring system having a spring constant of \(1.3 \mathrm{~N} / \mathrm{cm}\) and an oscillation amplitude of \(2.4 \mathrm{~cm}\).

58\. Lightly Damped The amplitude of a lightly damped oscillator decreases by \(3.0 \%\) during each cycle. What fraction of the mechanical energy of the oscillator is lost in each full oscillation?
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