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95 An engineer has an odd-shaped \(10 \mathrm{~kg}\) object and needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant \(\kappa=0.50 \mathrm{~N} \cdot \mathrm{m} .\) If this torsion pendulum oscillates through 20 cycles in \(50 \mathrm{~s},\) what is the rotational inertia of the object?

Short Answer

Expert verified
The rotational inertia is approximately 0.0791 kg⋅m².

Step by step solution

01

Define the formula for period

The period of oscillation for a torsion pendulum is given by the formula \( T = 2\pi \sqrt{\frac{I}{\kappa}} \), where \( T \) is the period in seconds, \( I \) is the rotational inertia, and \( \kappa \) is the torsion constant.
02

Calculate the period

It is given that the pendulum completes 20 cycles in 50 seconds. Therefore, the period \( T \) is the total time divided by the number of cycles: \( T = \frac{50 \text{ s}}{20} = 2.5 \text{ s} \).
03

Rearrange the formula to find rotational inertia

By rearranging the formula \( T = 2\pi \sqrt{\frac{I}{\kappa}} \), we can solve for \( I \): \( I = \frac{\kappa T^2}{4\pi^2} \).
04

Substitute given values into the formula

Substitute \( T = 2.5 \text{ s} \) and \( \kappa = 0.50 \text{ N} \cdot \text{m} \) into the formula: \( I = \frac{0.50 \times (2.5)^2}{4\pi^2} \).
05

Calculate the rotational inertia

Calculate \( I \) using the substituted values: \[ I = \frac{0.50 \times 6.25}{4 \times \pi^2} = \frac{3.125}{39.478} \approx 0.0791 \text{ kg} \cdot \text{m}^2 \].

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Key Concepts

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

Understanding the Torsion Pendulum
A torsion pendulum is a fascinating type of pendulum that uses a twisting string or wire to oscillate. Unlike a traditional pendulum that swings back and forth under gravity, a torsion pendulum twists around a vertical axis. A classic example is a disk or bar suspended from a wire. When you twist it and let it go, the torsion in the wire causes it to rotate back and forth. The motion is similar to a swinging pendulum, but here it's rotational.
This unique device is used in clocks and scientific experiments to measure time periods precisely. It's vital for engineers and physicists to understand the principles behind its motion to apply them in practical situations.
Role of the Torsion Constant
The torsion constant, often denoted by the symbol \( \kappa \), is a crucial factor in the study of torsion pendulums. It measures the wire’s resistance to twisting. Think of \( \kappa \) as how stiff or flexible a wire is. A larger \( \kappa \) means the wire is stiffer and resists twisting more, whereas a smaller \( \kappa \) indicates a more pliable wire.
In the context of our problem, knowing the torsion constant allows us to determine the object’s rotational inertia by relating it to the period of oscillation. This constant helps engineers ensure that the wire is suitable for the specific application they are dealing with. Understanding \( \kappa \) ensures that the device behaves predictably under twisting motions, which is essential in precision engineering.
  • Stiffer wires have higher torsion constants and are less flexible.
  • More pliable wires have lower torsion constants and are easier to twist.
Calculating the Period of Oscillation
The period of oscillation, represented by \( T \), is the time taken for one complete cycle of motion in a system like a torsion pendulum. It's calculated by dividing the total time of several cycles by the number of cycles completed. In our exercise, 20 cycles took 50 seconds, so each cycle (or period) lasted 2.5 seconds.
Knowing the period is critical because it allows us to figure out other properties of the system, like its rotational inertia. The period depends on both the torsion constant and the object's rotational inertia. That's why it's essential for calculating how much the object will resist changes in its rotation. Precision in measuring \( T \) ensures accuracy in all subsequent calculations related to the pendulum's dynamics.
Concept of Oscillating Motion
Oscillating motion refers to any repetitive back and forth movement around a central value or point of equilibrium. In physics, oscillations can be linear, like a spring moving up and down, or rotational, like our torsion pendulum. Key characteristics of oscillating motion are its amplitude, period, and frequency.
For rotational or twisting motion, features like rotational inertia also play a significant role. Oscillations are vital in various technologies, from telling time with pendulum clocks to transmitting radios waves using oscillator circuits.
Understanding oscillations helps in designing efficient systems that rely on these movements, ensuring they function correctly under different conditions.
  • Amplitude is how far the object moves from its rest position.
  • The period is the time taken for one complete cycle of movement.
  • Frequency is how many cycles are completed in a specific time frame.

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Most popular questions from this chapter

Hanging from a horizontal beam are nine simple pendulums of the following lengths: \((\mathrm{a}) 0.10,(\mathrm{~b}) 0.30,(\mathrm{c}) 0.40,(\mathrm{~d}) 0.80,(\mathrm{e}) 1.2,(\mathrm{f}) 2.8,\) (g) \(3.5,\) (h) 5.0 , and (i) 6.2 m. Suppose the beam undergoes horizontal oscillations with angular frequencies in the range from \(2.00 \mathrm{rad} / \mathrm{s}\) to \(4.00 \mathrm{rad} / \mathrm{s} .\) Which of the pendulums will be (strongly) set in motion?

An oscillating block-spring system takes \(0.75 \mathrm{~s}\) to begin repeating its motion. Find (a) the period, (b) the frequency in hertz, and (c) the angular frequency in radians per second.

In Fig. 15-37, two blocks \((m=1.8 \mathrm{~kg}\) and \(M=10 \mathrm{~kg})\) and a spring \((k=200 \mathrm{~N} / \mathrm{m})\) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is \(0.40 .\) What amplitude of simple harmonic motion of the spring- blocks system puts the smaller block on the verge of slipping over the larger block?

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position \(y_{i}\) such that the spring is at its rest length. The object is then released from \(y_{i}\) and oscillates up and down, with its lowest position being \(10 \mathrm{~cm}\) below \(y_{i}\). (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is \(8.0 \mathrm{~cm}\) below the initial position? (c) An object of mass \(300 \mathrm{~g}\) is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) How far below \(y_{i}\) is the new equilibrium (rest) position with both objects attached to the spring?

A (hypothetical) large slingshot is stretched \(2.30 \mathrm{~m}\) to launch a \(170 \mathrm{~g}\) projectile with speed sufficient to escape from Earth \((11.2 \mathrm{~km} / \mathrm{s})\). Assume the elastic bands of the slingshot obey Hooke's law. (a) What is the spring constant of the device if all the elastic potential energy is converted to kinetic energy? (b) Assume that an average person can exert a force of \(490 \mathrm{~N}\). How many people are required to stretch the elastic bands?

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