/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 A flywheel (i.e., a massive disk... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 10

A flywheel (i.e., a massive disk capable of rotating about its central axis) has a moment of inertia of \(3.8 \mathrm{~kg} \cdot \mathrm{m}^{2}\). What constant torque is required to increase the wheel's frequency from \(2.0 \mathrm{rev} / \mathrm{s}\) to \(5.0 \mathrm{rev} / \mathrm{s}\) in \(6.0\) revolutions? Neglect friction. Given $$\begin{array}{lll} \theta=12 \pi \mathrm{rad} & \omega_{i}=4.0 \pi \mathrm{rad} / \mathrm{s} & \text { and } & \omega_{f}=10 \pi \mathrm{rad} / \mathrm{s} \end{array}$$ we can write Work done on wheel = Change in \(\mathrm{KE}_{r}\) of wheel $$\begin{aligned} \tau \theta &=\frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} I \omega_{i}^{2} \\ (\tau)(12 \pi \mathrm{rad}) &=\frac{1}{2}\left(3.8 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)\left[\left(100 \pi^{2}-16 \pi^{2}\right)(\mathrm{rad} / \mathrm{s})^{2}\right] \end{aligned}$$ which leads to \(\tau=42 \mathrm{~N} \cdot \mathrm{m}\). Notice in all of these problems that radians and seconds must be used.

Short Answer

Expert verified
A torque of 42 N路m is required.

Step by step solution

01

Identify given values and convert units

Given values are \(I = 3.8 \, \text{kg} \cdot \text{m}^2\), \(\omega_i = 4.0\pi \, \text{rad/s}\) (since \(2.0\, \text{rev/s} = 4.0\pi\, \text{rad/s}\)), \(\omega_f = 10\pi \, \text{rad/s}\) (since \(5.0\, \text{rev/s} = 10\pi\, \text{rad/s}\)), \(\theta = 6.0 \, \text{rev} = 12\pi \, \text{rad}\).
02

Apply the work-energy principle

The work done by torque \(\tau\) during rotation is equal to the change in rotational kinetic energy. The rotational kinetic energy is given by \(\Delta KE_r = \frac{1}{2} I \omega_f^2 - \frac{1}{2} I \omega_i^2\).
03

Substitute values into the equation

Substitute values into \((\tau)(12\pi) = \frac{1}{2} \times 3.8 \, \text{kg} \cdot \text{m}^2 \times [(10\pi)^2 - (4\pi)^2] \, \text{rad}^2/\text{s}^2\). This becomes \[\tau (12\pi) = \frac{1}{2} \times 3.8 \times (100\pi^2 - 16\pi^2)\, \text{N} \cdot \text{m} \].
04

Calculate the expression inside the bracket

Calculate \(100\pi^2 - 16\pi^2 = 84\pi^2\).
05

Solve for \(\tau\)

Now, plug \(84\pi^2\) back into the equation: \((\tau)(12\pi) = \frac{1}{2} \times 3.8 \times 84\pi^2\). Simplify to solve for \(\tau\): \[12\pi\tau = 159.6\pi^2\, \text{N} \cdot \text{m}\]. Therefore, solve for \(\tau\): \[\tau = \frac{159.6\pi}{12}\, \text{N} \cdot \text{m} = 42\, \text{N} \cdot \text{m}\].

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy is the energy an object possesses due to its rotation. Just like linear kinetic energy, which is given by \( \frac{1}{2}mv^2 \), the rotational kinetic energy of an object can be expressed as \( KE_r = \frac{1}{2} I \omega^2 \). Here, \( \omega \) is the angular velocity, which tells us how fast something is spinning.
Imagine a playground merry-go-round: as it spins faster, it gains more rotational kinetic energy. This energy depends not only on the speed of rotation but also on the mass distribution around the central axis, quantified by the moment of inertia \( I \). In our flywheel problem, the change in rotational kinetic energy helped us calculate the required torque.Understanding rotational kinetic energy can also help in designing efficient motors and understanding machinery dynamics. Using the correct units like radians helps ensure calculations work smoothly, as seen in this exercise.
Torque Calculation
Torque is crucial in understanding rotational motion, much like force is for linear motion. It measures how much a force acting on an object causes that object to rotate. In simple terms, it's the rotational effect of a force applied to a point at some distance from the axis of rotation. The formula for torque is given by \( \tau = r \times F \), where \( r \) is the distance from the axis and \( F \) is the force applied.
In the exercise, instead of dealing with force directly, we had a change in angular velocity. We used the relationship between torque, angular displacement \( \theta \), and rotational kinetic energy to find the needed torque to spin up the flywheel. The approach used the calculated work from the change in rotational kinetic energy which equals \( \tau \theta \). This demonstrates the interplay between physical quantities in rotational dynamics.Understanding torque is essential in mechanics, as it helps us design and predict the behavior of systems involving rotational motion, like engines or even simple tools like wrenches.
Work-Energy Principle
The work-energy principle in physics states that the work done on an object is equal to the change in its kinetic energy. For rotational motion, the principle holds as well, relating the work done by a torque to the change in rotational kinetic energy.
In the problem, the work done by the torque \( \tau \) was related to the change in rotational kinetic energy, symbolized by \( \Delta KE_r \). This relationship simplifies complex rotational motions to a straightforward energy calculation. By calculating the work, we could determine how much torque was necessary to increase the flywheel's angular speed within a given number of revolutions.This principle is important for both determining energy requirements in mechanical systems and understanding how energy converts during motion. In practical applications, it helps engineers and scientists calculate the energy transfers necessary for desired outcomes.

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Most popular questions from this chapter

A horizontal disk with a moment of inertia \(I_{1}\) is rotating freely at an angular speed of \(\omega_{1}\) when a second, nonrotating disk with a moment of inertia \(I_{2}\) is dropped on it (Fig. \(10-11\) ). The two then rotate as a unit. Find the final angular speed. Ignore the central rod. From the law of conservation of angular momentum, $$\begin{array}{l} \text { Angular momentum before }=\text { Angular momentum after }\\\ I_{1} \omega_{1}+I_{2}(0)=l_{1} \omega+I_{2} \omega\\\ \text { Solving this equation leads to } \quad \omega=\frac{l_{1} 4_{1}}{l_{1}+I_{2}} \end{array}.$$

Picture a rigid rod of length \(L\) having negligible mass. It has two identical tiny spheres both of mass \(m\), one at each end of the rod. Determine the moment of inertia about an axis perpendicular to the rod passing through its center, in terms of \(m\) and \(L\). [Hint: \(\mathrm{A}\) single point mass \((m)\) has a moment of inertia about an axis a distance \(r\) away of \(m r^{2}\). Here \(L\) is \(\left.\operatorname{not} r .\right]\)

A 6.0-kg bowling ball \(\left(I=2 M r^{2} / 5\right)\) starts from rest and rolls, without sliding, down a gradual slope until it reaches a point 80 \(\mathrm{cm}\) lower than its starting point. How fast is it then moving? Ignore friction losses. Do you actually need the mass? [Hint: Why were you not given \(r ?]\)

A 20 -kg solid disk \(\left(I=\frac{1}{2} M r^{2}\right)\) rolls on a horizontal surface at the rate of \(4.0 \mathrm{~m} / \mathrm{s}\). Compute its total KE. [Hint: Do you really need \(r ?]\)

A large roller in the form of a uniform cylinder is pulled by a tractor to compact earth; it has a \(1.80\) -m diameter and weighs 10 kN. If frictional losses can be ignored, what average horsepower must the tractor provide to accelerate the cylinder from rest to a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) in a horizontal distance of \(3.0 \mathrm{~m}\) ? The power required is equal to the work done by the tractor divided by the time it takes. The tractor does the following work: $$\text { Work }=(\Delta \mathrm{KE})_{r}+(\Delta \mathrm{KE})_{t}=\frac{1}{2} I \omega_{f}^{2}+\frac{1}{2} m v_{f}^{2}$$ We have \(v_{f}=4.0 \mathrm{~m} / \mathrm{s}, \omega_{f}=v_{f} / r=4.44 \mathrm{rad} / \mathrm{s}\), and \(m=10000 / 9.81\) \(=1019 \mathrm{~kg}\). The moment of inertia of the cylinder is $$I=\frac{1}{2} m r^{2}=\frac{1}{2}(1019 \mathrm{~kg})(0.90 \mathrm{~m})^{2}=413 \mathrm{~kg} \cdot \mathrm{m}^{2}$$ Substituting these values, the work required tums out to be \(12.23\) kJ. We still need the time taken to do this work. Because the roller went \(3.0 \mathrm{~m}\) with an average velocity \(v_{a v}=\frac{1}{2}(4+0)=2.0 \mathrm{~m} / \mathrm{s}\), $$\begin{array}{l} t=\frac{s}{v_{\sigma e}}=\frac{3.0 \mathrm{~m}}{2.0 \mathrm{~m} / \mathrm{s}}=1.5 \mathrm{~s}\\\ \text { Then }\\\ \text { Power }=\frac{\text { Work }}{\text { Time }}=\frac{12230 \mathrm{~J}}{1.5 \mathrm{~s}}=(8150 \mathrm{w})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=11 \mathrm{hp} \end{array}$$
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