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Chapter 11: Problem 51

a wheel is rotating freely at angular speed 800 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels? (b) What fraction of the original rotational kinetic energy is lost?

Short Answer

Expert verified
(a) 267 rev/min; (b) 2/3 of the original energy is lost.

Step by step solution

01

Define the Known Variables

The initial angular speed of the first wheel is 800 revolutions per minute (rev/min), which we need to convert to radians per second. Since 1 revolution = \(2\pi\) radians, and there are 60 seconds in a minute, we have: \(\omega_1 = 800 \times \frac{2\pi}{60} \) radians/second. Let's denote the rotational inertia of the first wheel as \(I_1\), and the second wheel as \(I_2 = 2I_1\).
02

Apply Conservation of Angular Momentum

When the second wheel is coupled to the shaft, the total system's angular momentum before and after must be equal. Before coupling, the system's angular momentum is \( I_1 \cdot \omega_1 \). After coupling, the total angular momentum is \((I_1 + I_2) \cdot \omega_f\). So, \( I_1 \cdot \omega_1 = (I_1 + 2I_1) \cdot \omega_f\).
03

Solve for Final Angular Speed

Using the momentum equation from Step 2, we can solve for the final angular speed \( \omega_f\): \( \omega_f = \frac{I_1 \cdot \omega_1}{3I_1} = \frac{\omega_1}{3} \). Substituting \(\omega_1 = 800 \times \frac{2\pi}{60}\) rad/s, \(\omega_f = \frac{800 \times \frac{2\pi}{60}}{3}\) rad/s.
04

Calculate Initial Rotational Kinetic Energy

The initial kinetic energy of the system is \( KE_i = \frac{1}{2} I_1 \omega_1^2 \).
05

Calculate Final Rotational Kinetic Energy

The final kinetic energy of the system is \( KE_f = \frac{1}{2} (I_1 + 2I_1) \omega_f^2 = \frac{1}{2} \times 3I_1 \times \left(\frac{\omega_1}{3}\right)^2 \).
06

Determine the Fraction of Kinetic Energy Lost

The fraction of kinetic energy lost is \( \frac{KE_i - KE_f}{KE_i} \). Substitute the expressions for \( KE_i \) and \( KE_f \) to calculate this fraction.

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Key Concepts

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

Rotational Inertia
Rotational inertia, also known as moment of inertia, is an object's resistance to changes in its rotational motion. Think of it as the equivalent of mass in linear motion. The larger an object's rotational inertia, the more torque it requires to increase its angular velocity.

It's essential in understanding how objects spin. For example, in our exercise, we have two wheels where the second one has twice the rotational inertia of the first. This means that the second wheel is more resistant to changes in its spin than the first wheel.

The formula for small objects or individual points can be defined as:
  • For an object like a thin hoop, rotational inertia is given by \(I = m imes r^2\), where \(m\) is the mass and \(r\) is the radius.
  • Complex shapes need their own specific formulas, often based on symmetry.
Understanding these properties helps us predict how the system's overall rotation will behave once both wheels are spinning together.
Rotational Kinetic Energy
Rotational kinetic energy is the energy due to the rotation of an object and is part of its total kinetic energy. The formula for rotational kinetic energy is given by:
\[ KE = \frac{1}{2} I \omega^2 \] where \(I\) is the rotational inertia, and \(\omega\) is the angular velocity.

In the initial situation, only the first wheel is rotating, so its kinetic energy depends on its inertia and angular velocity. After coupling the second wheel, the system's total kinetic energy is shared by both wheels.

This kinetic energy is subject to change when the two wheels combine. In our original exercise, the conservation of angular momentum causes the final rotational kinetic energy to differ from the initial energy, leading to a fraction of energy loss. The mathematical solution provided shows this energy difference, illustrating how the total kinetic energy before coupling is different from after. By knowing the initial and final kinetic energies, we can determine what fraction of this energy is lost due to the coupling.
Angular Velocity
Angular velocity refers to how fast an object is rotating. It is the rate of change of the angular position of a rotating body and is usually measured in radians per second. It reflects both how much rotation occurs and the direction of this rotation.

In our exercise, the first wheel has an angular speed of 800 revolutions per minute initially. When calculating angular speed in radians per second, we must convert these units appropriately. This conversion is crucial in applying the conservation of angular momentum formula.
  • Angular speed: \( \omega = \frac{revolutions}{minute} \times \frac{2\pi}{60} \)
  • For our problem, \( \omega_1 = 800 \times \frac{2\pi}{60} \text{ radians/second} \).
When a second wheel with twice the rotational inertia is attached, we apply conservation principles to find the system's new angular velocity. This fundamental change highlights how angular velocity interplays with inertia to maintain momentum in a system, showcasing core physics principles in rotational dynamics.

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Most popular questions from this chapter

A \(1000 \mathrm{~kg}\) car has four \(10 \mathrm{~kg}\) wheels. When the car is moving, what fraction of its total kinetic energy is due to rotation of the wheels about their axles? Assume that the wheels are uniform disks of the same mass and size. Why do you not need to know the radius of the wheels?

A horizontal vinyl record of mass \(0.10 \mathrm{~kg}\) and radius \(0.10 \mathrm{~m}\) rotates freely about a vertical axis through its center with an angular speed of \(4.7 \mathrm{rad} / \mathrm{s}\) and a rotational inertia of \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2}\) Putty of mass \(0.020 \mathrm{~kg}\) drops vertically onto the record from above and sticks to the edge of the record. What is the angular speed of the record immediately afterwards?

A sanding disk with rotational inertia \(1.2 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\) is attached to an electric drill whose motor delivers a torque of magnitude \(16 \mathrm{~N} \cdot \mathrm{m}\) about the central axis of the disk. About that axis and with the torque applied for \(33 \mathrm{~ms}\), what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?

A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved \(1.0 \mathrm{~m}\) up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

A man stands on a platform that is rotating (without friction) with an angular speed of \(1.2 \mathrm{rev} / \mathrm{s} ;\) his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If by moving the bricks the man decreases the rotational inertia of the system to \(2.0\) \(\mathrm{kg} \cdot \mathrm{m}^{2}\), what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy?
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