/*! 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 88 A thin, light wire is wrapped ar... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin, light wire is wrapped around the rim of a wheel, as shown in Fig. E9.49. The wheel rotates about a stationary horizontal axle that passes through the center of the wheel. The wheel has radius 0.180 \(\mathrm{m}\) and moment of inertia for rotation about the axle of \(I=0.480 \mathrm{kg} \cdot \mathrm{m}^{2}\) . A small block with mass 0.340 \(\mathrm{kg}\) is suspended from the free end of the wire. When the system is released from rest, the block descends with constant acceleration. The bearings in the wheel at the axle are rusty, so friction there does \(-6.00 \mathrm{J}\) of work as the block descends 3.00 \(\mathrm{m}\) . What is the magnitude of the angular velocity of the wheel after the block has descended 3.00 \(\mathrm{m} ?\)

Short Answer

Expert verified
The angular velocity of the wheel is approximately 4.08 rad/s.

Step by step solution

01

Determine the block's gravitational force

The block's mass is given as \(0.340 \, \text{kg}\). The force due to gravity acting on it is its weight, calculated by multiplying its mass by the acceleration due to gravity (\(g = 9.81 \, \text{m/s}^2\)). The gravitational force (\(F_g\)) is:\[F_g = m \cdot g = 0.340 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 3.33 \, \text{N}\]
02

Calculate work done by gravity over the descent

The work done by gravity when the block descends by \(3.00 \, \text{m}\) is given by the formula \(W = F_g \cdot h\), where \(h\) is the height.\[W_g = 3.33 \, \text{N} \times 3.00 \, \text{m} = 9.99 \, \text{J}\]
03

Consider the work-energy principle

The total work done on the system is equal to the change in mechanical energy. Friction does negative work of \(-6.00 \, \text{J}\) on the system.The total mechanical work done on the wheel by gravity is the sum of the work done by gravity and the work done by friction (\(W_f = -6.00 \, \text{J}\)):\[W_{\text{total}} = W_g + W_f = 9.99 \, \text{J} - 6.00 \, \text{J} = 3.99 \, \text{J}\]
04

Relate work done to rotational energy

The work done (\(3.99 \, \text{J}\)) is transformed into rotational kinetic energy of the wheel. The formula for rotational kinetic energy (given an angular velocity \(\omega\)) is:\[K_r = \frac{1}{2} I \omega^2\]Setting \(W_{\text{total}}\) equal to \(K_r\), we have:\[3.99 \, \text{J} = \frac{1}{2} \times 0.480 \, \text{kg} \cdot \text{m}^2 \times \omega^2\]
05

Solve for angular velocity

Rearranging to solve for \(\omega\):\[\omega^2 = \frac{2 \times 3.99}{0.480}\]Calculate \(\omega\):\[\omega^2 = \frac{7.98}{0.480} = 16.625\]\[\omega = \sqrt{16.625} \approx 4.08 \, \text{rad/s}\]

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Key Concepts

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

Moment of Inertia
Moment of inertia is crucial in rotational dynamics as it describes how much an object resists rotational acceleration about an axis. Imagine it as the rotational counterpart of mass in linear motion. It depends on the distribution of mass within the object and how far each piece of mass is from the axis of rotation. For a rotating wheel in this exercise, the moment of inertia is given as 0.480 kg·m². This value indicates how difficult it is to change its rotational speed. A higher moment of inertia would mean more force is needed to rotate the wheel. Interestingly, in our problem, knowing the wheel's moment of inertia helps link the rotational motion to the forces applied. This is especially important when calculating change in angular velocity when a force, such as gravity, acts through the wrapped wire. For students, remember this: when an object has a large moment of inertia, it means that most of its mass is far from the axis, making it harder to spin.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or spins around a particular axis. It tells us how quickly the wheel in our problem is turning after the block has descended a certain distance. It is measured in radians per second (rad/s). After releasing the block, the tension in the wire applies a torque that results in angular acceleration. The exercise calculates this down to a final angular velocity of approximately 4.08 rad/s once the block has descended 3 meters. To find angular velocity, we must balance energy input and output. The wheel speeds up due to gravitational work done minus losses, like friction. Once the interplay of forces is understood through the work-energy principle, angular velocity emerges as a key indicator of motion transformations in rotational systems.
Work-Energy Principle
In rotational dynamics, the work-energy principle plays a fundamental role by stating that the work done on an object is equal to the change in its kinetic energy. For our wheel and block system, this concept helped us analyze the energy transitions from one form to another.As the block descends, gravitational force does work by transferring energy which is partially used to overcome frictional forces at the axle (-6.00 J) and the rest is converted into rotational energy. The equation looks like this:\[W_{ ext{total}} = W_{ ext{gravity}} + W_{ ext{friction}} = 9.99 ext{ J} - 6.00 ext{ J} = 3.99 ext{ J}\]This remaining energy (3.99 J) transforms into the rotational kinetic energy of the wheel as calculated in step 4. It’s important to remember that understanding friction's impact is vital here as it directly absorbs some of the potential work available to speed up rotation.
Rotational Kinetic Energy
Rotational kinetic energy is the energy possessed by an object due to its rotation. It is the rotational equivalent of linear kinetic energy and can be expressed with the formula:\[K_r = \frac{1}{2} I \omega^2\]In the wheel system, the total work done on the wheel, which is 3.99 J, transforms into rotational kinetic energy. Here, the initial energy from the gravitational pull (minus friction) results in an increase in this rotational energy.By setting the work done equal to rotational kinetic energy, you can solve for angular velocity, showing how energy conservation principles extend into rotational systems:\[3.99 ext{ J} = \frac{1}{2} \times 0.480 \text{ kg} \cdot \text{m}^2 \times \omega^2\]Understanding this equation highlights how energy is distributed in rotational motion, demonstrating the intricate balance of forces and motion. Knowing rotational kinetic energy helps in predicting how systems like this one respond to various forces.

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Most popular questions from this chapter

A uniform disk of radius \(R\) is cut in half so that the remaining half has mass \(M\) (Fig. E9.34a).(a) What is the moment of inertia of this half about an axis perpendicular to its plane through point \(A\) ? (b) Why did your answer in part (a) come out the same as if this were a complete disk of mass \(M ?\) (c) What would be the moment of inertia of a quarter disk of mass \(M\) and radius \(R\) about an axis perpendicular to its plane passing through point \(B(\) Fig. \(E 9.34 b) ?\)

A compound disk of outside diameter 140.0 \(\mathrm{cm}\) is made up of a uniform solid disk of radius 50.0 \(\mathrm{cm}\) and area density 3.00 \(\mathrm{g} / \mathrm{cm}^{2}\) surrounded by a concentric ring of inner radius \(50.0 \mathrm{cm},\) outer radius \(70.0 \mathrm{cm},\) and area density 2.00 \(\mathrm{g} / \mathrm{cm}^{2} .\) Find the moment of inertia of this object about an axis perpendicular to the plane of the object and passing through its center.

The angle \(\theta\) through which a disk drive turns is given by \(\theta(t)=a+b t-c t^{3},\) where \(a, b,\) and \(c\) are constants, \(t\) is in seconds, and \(\theta\) is in radians. When \(t=0, \theta=\pi / 4\) rad and the angular velocity is 2.00 \(\mathrm{rad} / \mathrm{s}\) , and when \(t=1.50 \mathrm{s},\) the angular acceleration is 1.25 \(\mathrm{rad} / \mathrm{s}^{2}\) . (a) Find \(a, b,\) and \(c,\) including their units. (b) What is the angular acceleration when \(\theta=\pi / 4\) rad? (c) What are \(\theta\) and the angular velocity when the angular acceleration is 3.50 \(\mathrm{rad} / \mathrm{s}^{2} ?\)

Energy is to be stored in a 70.0 -kg flywheel in the shape of a uniform solid disk with radius \(R=1.20 \mathrm{m.}\) To prevent structural failure of the flywheel, the maximum allowed radial acceleration of a point on its rim is 3500 \(\mathrm{m} / \mathrm{s}^{2} .\) What is the maximum kinetic energy that can be stored in the flywheel?

A uniform 2.00 -m ladder of mass 9.00 kg is leaning against a vertical wall while making an angle of \(53.0^{\circ}\) with the floor. A worker pushes the ladder up against the wall until it is vertical. What is the increase in the gravitational potential energy of the ladder?

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