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Chapter 15: Problem 38

A \(95 \mathrm{~kg}\) solid sphere with a \(15 \mathrm{~cm}\) radius is suspended by a vertical wire. A torque of \(0.20 \mathrm{~N} \cdot \mathrm{m}\) is required to rotate the sphere through an angle of \(0.85\) rad and then maintain that orientation. What is the period of the oscillations that result when the sphere is then released?

Short Answer

Expert verified
The period of oscillation is approximately 3.779 seconds.

Step by step solution

01

Understand the Problem

We have a sphere suspended by a wire and a torque is applied to rotate it. We need to find the period of the oscillations when it is released.
02

Find the Moment of Inertia

For a solid sphere, the moment of inertia is given by \(I = \frac{2}{5} m r^2\). Here, \(m = 95\, \text{kg}\) and \(r = 0.15\, \text{m}\). Substitute these values to find \(I\):\[ I = \frac{2}{5} \times 95 \times (0.15)^2 \approx 0.85575\, \text{kg} \cdot \text{m}^2 \]
03

Relate Torque and Angle to Calculate Spring Constant

The torque \(\tau\) is related to the spring constant \(k_s\) and the angle \(\theta\) by \(\tau = k_s \theta\). Given \(\tau = 0.20\, \text{N} \cdot \text{m}\) and \(\theta = 0.85\, \text{rad}\), solve for \(k_s\):\[ k_s = \frac{0.20}{0.85} \approx 0.235\, \text{N} \cdot \text{m/rad} \]
04

Use Formula to Find the Period of Oscillations

The period of oscillations \(T\) for torsional oscillation is given by \(T = 2\pi \sqrt{\frac{I}{k_s}}\). Substitute the known values of \(I\) and \(k_s\):\[ T = 2\pi \sqrt{\frac{0.85575}{0.235}} \approx 3.779\, \text{seconds} \]
05

Finalize the Solution

Conclude that the period of the oscillations of the sphere when released is approximately 3.779 seconds.

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Key Concepts

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

Understanding Moment of Inertia
Moment of inertia is a core concept in rotational dynamics. It describes how mass is distributed in an object and its resistance to rotational motion. Think of it as similar to mass in linear motion 鈥� it shows how hard it is to start, stop, or change the speed of rotation. For a solid sphere, the formula to find the moment of inertia (\( I \)) is given by:\[I = \frac{2}{5} m r^2\]In this formula:
  • \( m \) represents the mass of the object. For our sphere, it is 95 kg.

  • \( r \) is the radius of the sphere, which is 0.15 meters in this example.
To find the moment of inertia, we plug these values into the equation. This gives us \( I \approx 0.85575\, \text{kg} \cdot \text{m}^2 \). Knowing this value helps us understand how the object's mass is distributed with regard to its axis of rotation.
Exploring the Spring Constant
The spring constant, denoted as \( k_s \), measures the stiffness of a spring or a torsional system. In the context of torsional oscillations, it helps us understand how much torque is needed to achieve a certain rotation angle. It's similar to how a stiffer spring requires more force to be compressed or stretched.We use the relation between torque and angle to find \( k_s \) for the sphere, described by:\[ \tau = k_s \theta \]Here:
  • \( \tau \) is the applied torque, 0.20 \( \text{N} \cdot \text{m} \) in this problem.

  • \( \theta \) is the angle in radians the sphere rotates, which is 0.85 rad.
By rearranging the formula, \( k_s \) is found to be approximately 0.235 \( \text{N} \cdot \text{m/rad} \). This indicates the rotational stiffness of the system.
Understanding Torque in Rotational Movement
Torque is the rotational equivalent of linear force. It measures how much a force acting on an object causes it to rotate, and is calculated as the product of force and the lever arm distance from the pivot point at which the force is applied.For torsional oscillations, torque drives the motion, influencing how objects like the sphere transition between potential energy and kinetic energy in a cycle. The formula for torque in the context of this exercise relates to the torsional spring constant and the angle of twist:\[ \tau = k_s \theta\]In this formula:
  • \( \tau \) is the torque, essential for causing rotational movement.

  • \( k_s \) is the spring constant, reflecting the system's stiffness.

  • \( \theta \) is the angle in radians that the sphere is rotated.
Understanding torque is crucial because it allows us to calculate the energy dynamics of the system and directly influences the period of torsional oscillations when the sphere is set into motion.

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Most popular questions from this chapter

An oscillator consists of a block of mass \(0.500 \mathrm{~kg}\) connected to a spring. When set into oscillation with amplitude \(35.0 \mathrm{~cm}\), the oscillator repeats its motion every \(0.500 \mathrm{~s}\). Find the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring.

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?

The balance wheel of an old-fashioned watch oscillates with angular amplitude \(\pi\) rad and period \(0.500 \mathrm{~s}\). Find (a) the maximum angular speed of the wheel, (b) the angular speed at displacement \(\pi / 2 \mathrm{rad}\), and \((\mathrm{c})\) the magnitude of the angular acceleration at displacement \(\pi / 4\) rad.

The scale of a spring balance that reads from 0 to \(15.0 \mathrm{~kg}\) is \(12.0 \mathrm{~cm}\) long. A package suspended from the balance is found to oscillate vertically with a frequency of \(2.00 \mathrm{~Hz}\). (a) What is the spring constant? (b) How much does the package weigh?

A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to \(1.00 \mu \mathrm{m} .\) (a) At what frequency is the magnitude \(a\) of the diaphragm's acceleration equal to \(g\) ? (b) For greater frequencies, is \(a\) greater than or less than \(g\) ?
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