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Chapter 12: Problem 39

A wheel is rotating freely at rotational 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 rotational 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
The final rotational speed is 88.9 rev/min, and \( \frac{2}{3} \) of the original kinetic energy is lost.

Step by step solution

01

- Identify Known Quantities

The initial wheel has a rotational speed of 800 rev/min and the second wheel is initially at rest. Let's denote the rotational inertia of the first wheel as I and the second wheel is 2I.
02

- Calculate Initial Angular Momentum

Initial angular momentum (\textbf{L_initial}) is given by the product of rotational inertia and angular velocity. Angular velocity in radians per second can be found by converting rev/min to rad/s: \(\text{Angular velocity} = 800 \text{ rev/min} \times \frac{2\text{蟺 rad}}{1 \text{rev}} \times \frac{1 \text{min}}{60 \text{s}} = \frac{80蟺}{3} \text{ rad/s}\).Thus, \(L_initial = I \times \frac{80蟺}{3}\).
03

- Horizon: The Second Wheel

The second wheel is initially at rest, so its initial angular momentum is zero. \(L_{initial, combined} = I \times \frac{80蟺}{3} + 2I \times 0 = I \times \frac{80蟺}{3}\).
04

- Calculate Final Angular Momentum

When the second wheel is coupled to the shaft, the total rotational inertia becomes \(I + 2I = 3I\). Let the final angular speed be \(蠅_f\). Conservation of angular momentum tells us that the initial angular momentum equals the final angular momentum: \(I \times \frac{80蟺}{3}\) = \(3I \times 蠅_f\).
05

- Solve for Final Angular Speed

Solving the equation \(I \times \frac{80蟺}{3} = 3I \times 蠅_f\), we find: \(蠅_f = \frac{80蟺}{9}\). Converting back to rev/min: \(蠅_f = \frac{80蟺}{9} \text{ rad/s} \times \frac{60 \text{s}}{1 \text{min}} \times \frac{1 \text{rev}}{2蟺 \text{rad}} = \frac{800}{9} \text{ rev/min}\).
06

- Calculate Initial Kinetic Energy

The initial rotational kinetic energy is given by \(\frac{1}{2}I蠅^2\). Substituting the initial values: \(K_{initial} = \frac{1}{2}I \left(\frac{80蟺}{3}\right)^2 = \frac{1}{2}I \times \frac{6400蟺^2}{9}\).
07

- Calculate Final Kinetic Energy

The final rotational kinetic energy is given by \(\frac{1}{2} \times 3I \times 蠅_f^2\). Substituting the final values: \(K_{final} = \frac{1}{2} \times 3I \left(\frac{80蟺}{9}\right)^2 = \frac{1}{2} \times 3I \frac{6400蟺^2}{81} = \frac{9600蟺^2I}{81}\).
08

- Calculate Fraction of Energy Lost

To get the fraction of energy lost: \(\frac{K_{initial} - K_{final}}{K_{initial}}\). Substituting the values: \(\frac{3200蟺^2I/9 - 9600蟺^2I/81}{3200蟺^2I/9} = \frac{3200蟺^2I/9 (1 - (3/9))}{3200蟺^2I/9} = 1 - \frac{1}{3} = \frac{2}{3}\). Therefore, \(\frac{2}{3}\) of the initial kinetic energy is lost.

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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 the moment of inertia, plays a similar role in rotational motion to mass in linear motion. It quantifies how difficult it is to change the rotational speed of an object around an axis. The larger the rotational inertia, the harder it is to change the object's rotation. For the given problem, the first wheel has a rotational inertia denoted as I. The second wheel has twice this rotational inertia, so it is represented as 2I. When these wheels are combined, their total rotational inertia becomes the sum of both, which is 3I.
Angular Velocity
Angular velocity describes how fast an object rotates or revolves relative to another point, measured in radians per second (rad/s) or revolutions per minute (rev/min). Initially, the first wheel spins at 800 rev/min. This can be converted into angular velocity in rad/s using the formula: \(800 \times \frac{2\pi\text{ rad}}{1 \text{ rev}} \times \frac{1 \text{min}}{60 \text{s}} = \frac{80\pi}{3} \text{ rad/s}\). The second wheel initially is at rest, so its angular velocity is 0 rad/s. After couping both wheels, the speed of rotation changes due to the increased rotational inertia, maintaining the conservation of angular momentum.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation, calculated by the equation \(\frac{1}{2}I\omega^2\), where I is the moment of inertia and \(\omega\) is angular velocity. Initially, the first wheel's kinetic energy is \(\frac{1}{2}I\left(\frac{80\pi}{3}\right)^2\). After the second wheel is coupled, the final kinetic energy of the combined system becomes \(\frac{1}{2} \times 3I \left(\frac{80\pi}{9}\right)^2\). Notably, the change in kinetic energy is a key aspect in understanding energy losses.
Energy Loss Calculation
Energy loss occurs when kinetic energy is converted into other forms of energy, often due to friction or inelastic coupling. Here, to calculate the fraction of the original kinetic energy lost, we compare the initial and final kinetic energies. The formula for energy loss fraction is: \(\frac{K_{initial} - K_{final}}{K_{initial}}\). The initial kinetic energy is higher than the final kinetic energy due to the increased rotational inertia of the combined system. Therefore, \(\frac{2}{3}\) of the initial energy is lost.
Moment of Inertia
The moment of inertia (I) is the rotational equivalent of mass in linear motion. It depends on both the object's mass and the distribution of that mass relative to the axis of rotation. For a single rigid body, moment of inertia is calculated with different formulas, depending on the geometry of the object. In this exercise, the initial wheel's moment of inertia is represented as I, and the secondary wheel's moment of inertia is 2I due to having twice the rotational inertia. Understanding and calculating the moment of inertia is crucial in solving rotational dynamics problems and predicting how objects will behave under rotational forces.

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Most popular questions from this chapter

What is the torque about the origin on a jar of jalape帽o peppers located at coordinates \((3.0 \mathrm{~m},-2.0 \mathrm{~m}, 4.0 \mathrm{~m})\) due to (a) force \(\vec{F}_{A}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}+(5.0 \mathrm{~N}) \hat{\mathrm{k}},(\mathrm{b})\) force \(\vec{F}_{B}=\) \((-3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}-(5.0 \mathrm{~N}) \hat{\mathrm{k}}\), and \((\mathrm{c})\) the vector sum of \(\vec{F}_{A}\) and \(\vec{F}_{B} ?\) (d) Repeat part (c) about a point with coordinates \((3.0 \mathrm{~m},\), \(2.0 \mathrm{~m}, 4.0 \mathrm{~m}\) ) instead of about the origin.

A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g}\). Its axle radius is \(3.2 \mathrm{~mm}\), and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its translational acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) translational speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) rotational speed?

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A uniform disk of mass \(10 \mathrm{~m}\) and radius \(3.0 r\) can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass \(m\) and radius \(r\) lies on top of the larger disk, concentric with it. Initially the two disks rotate together with a rota- tional velocity of \(20 \mathrm{rad} / \mathrm{s}\). Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their rotational velocity about the center of the larger disk? (b) What is the ratio \(K_{2} / K_{1}\) of the new kinetic energy of the two- disk system to the system's initial kinetic energy?

A \(4.0 \mathrm{~kg}\) particle moves in an \(x y\) plane. At the instant when the particle's position and velocity are \(\vec{r}=\) \((2.0 \mathrm{~m}) \hat{\mathrm{i}}+(4.0 \mathrm{~m}) \hat{\mathrm{j}}\) and \(\vec{v}=(-4.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\), the force on the particle is \(\vec{F}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}\). At this instant, determine (a) the particle's rotational momentum about the origin, (b) the particle's rotational momentum about the point \(x=0, y=4.0 \mathrm{~m},(\mathrm{c})\) the torque acting on the particle about the origin, and (d) the torque acting on the particle about the point \(x=0.0 \mathrm{~m}, y=4.0 \mathrm{~m}\).
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