/*! 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 62 A uniform hollow disk has two pi... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform hollow disk has two pieces of thin, light wire wrapped around its outer rim and is supported from the ceiling (Fig. P10.62). Suddenly one of the wires breaks, and the remaining wire does not slip as the disk rolls down. Use energy conservation to find the speed of the center of this disk after it has fallen a distance of 2.20 \(\mathrm{m} .\)

Short Answer

Expert verified
The speed of the disk's center after falling 2.20 m is approximately 5.38 m/s.

Step by step solution

01

Identify the Initial and Final States

The initial state is when the disk is hanging stationary at the height before any of the wires break. The final state is when the disk has fallen a distance of 2.20 meters, with the disk rolling without slipping.
02

Write the Conservation of Energy Equation

The conservation of energy states that total energy in the initial state should equal total energy in the final state. Initially, the disk has potential energy and no kinetic energy. Finally, the disk will have translational kinetic energy, rotational kinetic energy, and less potential energy. The equation is:\[U_i + K_i = U_f + K_f\]where \(U_i\) is the initial potential energy, \(K_i\) is the initial kinetic energy (which is zero), \(U_f\) is the final potential energy, and \(K_f\) is the sum of translational and rotational kinetic energies.
03

Express Energies in Terms of Known Variables

Initially, the only energy is potential energy: \(U_i = mgh\), where \(h = 2.20\, \text{m}\). The final energies are: \(U_f = mg(h-2.20)\), translational kinetic energy: \(K_{trans} = \frac{1}{2} mv^2\), and rotational kinetic energy: \(K_{rot} = \frac{1}{2} I \omega^2\), with \(I = \frac{1}{2} mR^2\) for a hollow disk.
04

Relate Angular and Linear Velocities

For rolling without slipping, \(v = R \omega\). Substituting \(\omega = \frac{v}{R}\) into the rotational kinetic energy equation gives:\[K_{rot} = \frac{1}{2} \times \frac{1}{2} mR^2 \left(\frac{v}{R}\right)^2 = \frac{1}{4} mv^2\].
05

Solve for the Final Speed

Using energy conservation, we set up the equation:\[mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 + mg(h-2.20)\]Simplifying, we have:\[mg \times 2.20 = \frac{3}{4} mv^2\]Cancel \(m\) and solve for \(v\):\[v = \sqrt{\frac{4gh}{3}}\]Substitute \(g \approx 9.81\, \text{m/s}^2\) and \(h = 2.20\, \text{m}\) to find \(v\).
06

Calculate the Final Speed

Substituting the values, we get:\[v = \sqrt{\frac{4 \times 9.81 \times 2.20}{3}}\approx \sqrt{28.9}\approx 5.38\, \text{m/s}\].

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Key Concepts

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

Potential Energy
Potential energy is the stored energy of an object due to its position relative to other objects. In this exercise, the uniform hollow disk is initially supported at a certain height from the ground.

When we mention potential energy, we typically refer to gravitational potential energy, which can be calculated using the formula:
  • \( U = mgh \)
Here, \( m \) is the mass of the disk, \( g \) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\) on Earth), and \( h \) is the height above the reference point (usually the ground). When the disk is suspended, its potential energy is maximum because all its energy is stored as height-based energy in this scenario.

When one of the wires breaks and the disk starts to roll down, the potential energy begins to change into kinetic energy. This transformation is crucial as it manifests the principle of energy conservation, where the total mechanical energy (potential plus kinetic) in an isolated system remains constant.
Kinetic Energy
Kinetic energy is the energy of motion. In our scenario, it forms when the disk begins to fall and roll. In mechanics, kinetic energy has two components when considering a rolling object: translational kinetic energy and rotational kinetic energy.

  • Translational Kinetic Energy: This is associated with the movement of the object as a whole moving forward. It is given by the expression \( K_{trans} = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the disk's center of mass.
  • Rotational Kinetic Energy: This accounts for the spinning motion of the disk and is defined as \( K_{rot} = \frac{1}{2} I \omega^2 \), with \( I \) representing the moment of inertia and \( \omega \) the angular velocity. For a hollow disk, \( I = \frac{1}{2} mR^2 \), where \( R \) is the radius.
Combining these forms gives us the total kinetic energy as the disk moves and spins. Energy conservation means that as potential energy decreases, kinetic energy increases by the same amount, ensuring net energy remains constant. Understanding how these two forms of kinetic energy interact aids in solving the problem of finding the disk's speed after rolling down a certain height.
Rolling Motion
Rolling motion involves both translation and rotation. When the wire breaks and the disk rolls down without slipping, it means the point of contact with the surface does not slide.

A key relationship in rolling motion is between the angular velocity \( \omega \) and the linear velocity \( v \) of the object's center of mass. This relationship is:
  • \( v = R \omega \)
Where \( R \) is the radius of the disk, reflecting that as the disk rotates, each portion of its circumference moves linearly.

For rolling without slipping:
  • The disk's linear velocity at its edge matches its rotation, ensuring no slip.
  • Translation and rotation energies interlink, making energy conservation easy to apply.
Hence, formulating the energy conservation equation considers both translation and rotational components. Understanding this conjunction aids in determining the evolution of speed and energy as the disk moves downward.

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Most popular questions from this chapter

A thin, uniform metal bar, 2.00 \(\mathrm{m}\) long and weighing \(90.0 \mathrm{N},\) is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 \(\mathrm{m}\) below the ceiling by a small 3.00 -kg ball, initially traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s} .\) The ball rebounds in the opposite direction with a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) . (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

The Spinning Figure Skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center (Fig. Elo.43). When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled, hollow cylinder. His hands and arms have a combined mass of 8.0 \(\mathrm{kg} .\) When outstretched, they span 1.8 \(\mathrm{m}\) ; when wrapped, they form a cylinder of radius 25 \(\mathrm{cm} .\) The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . If his original angular speed is \(0.40 \mathrm{rev} / \mathrm{s},\) what is his final angular speed?

A solid wood door 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high is hinged along one side and has a total mass of 40.0 \(\mathrm{kg}\) . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 \(\mathrm{kg}\) , traveling perpendicular to the door at 12.0 \(\mathrm{m} / \mathrm{s}\) just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

A cord is wrapped around the rim of a solid uniform wheel 0.250 \(\mathrm{m}\) in radius and of mass 9.20 \(\mathrm{kg} .\) A steady horizontal pull of 40.0 \(\mathrm{N}\) to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

When an object is rolling without slipping, the rolling friction force is much less the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3\() .\) When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\) . If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\) , rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal speed \(\omega_{0}\) about the kinetic friction coefficient is \(\mu_{\mathrm{k}}\) (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force. on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0 .\) Rolling without slipping sets in when \(v_{x}=R \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

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