/*! 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 54 A uniform solid disk and a unifo... [FREE SOLUTION] | 91Ó°ÊÓ

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A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height \(h\). If they are released from rest and roll without slipping, which object reaches the bottom first? Verify your answer by calculating their speeds when they reach the bottom in terms of \(h\)

Short Answer

Expert verified
The disk reaches the bottom first. The final velocities of the disk and hoop, when they reach the bottom, are \(\sqrt{{4gh}/{3}}\) and \(\sqrt{{gh}}\) respectively.

Step by step solution

01

Define the Basic Concepts

Let's gather all the concepts used in the problem: \n - Kinetic energy: \(0.5*m*v^2\) \n - Rotational kinetic energy: \(0.5*I*ω^2\) \n - Potential energy at height \(h\): \(m*g*h\) \n - Moment of inertia for disc: \(0.5*m*r^2\) \n - Moment of inertia for hoop: \(m*r^2\) \n - Relationship between angular & linear velocity: \(v=ω*r\) \n Then, as there is no external work done, by conservation of energy, potential energy will convert to kinetic energy and rotational kinetic energy.
02

Find the final velocity of the disk

First let's find the final velocity of the disc using energy conservation \((m * g * h = 0.5 * m * v^2 + 0.5 * I_{disc} * ω^2)\). Substituting the moment of inertia of the disk, the relationship between angular and linear velocity, and simplifying, we get the final velocity of the disk as \(v_{disk} = \sqrt{{4gh}/{3}}\).
03

Find the final velocity of the hoop

Next, let's find the final velocity of the hoop using energy conservation \((m * g * h = 0.5 * m * v^2 + 0.5 * I_{hoop} * ω^2)\). Similarly substituting the moment of inertia of the hoop, the relationship between angular and linear velocity, and simplifying, we get the final velocity of the hoop as \(v_{hoop} = \sqrt{{gh}}\).
04

Comparing the final velocities

By comparing \(\sqrt{{4gh}/{3}} > \sqrt{{gh}}\), it's evident that the disk reaches bottom first because it has higher speed when reaching the bottom.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It can be calculated using the formula \( 0.5 \cdot m \cdot v^2 \), where \(m\) is the mass of the object and \(v\) is its linear velocity. For objects rolling down an incline, such as a disk or a hoop, kinetic energy plays a crucial role as the object converts potential energy into kinetic energy as it descends.

Understanding how kinetic energy relates to other forms of energy is pivotal when analyzing the motion of objects on an incline. In the case of the uniform solid disk and the hoop, as they roll without slipping, their total kinetic energy is the sum of their linear kinetic energy and rotational kinetic energy at the bottom of the incline.
Rotational Kinetic Energy
Rotational kinetic energy is the energy an object has due to its rotation. It differs from linear kinetic energy in that it specifically relates to the motion around an axis. The formula to calculate rotational kinetic energy is \( 0.5 \cdot I \cdot \omega^2 \), where \(I\) denotes the moment of inertia and \(\omega\) is the angular velocity.

For the uniform solid disk and hoop scenario, each of these objects will have a portion of their total kinetic energy in the form of rotational kinetic energy. This component of kinetic energy is integral to understanding which object reaches the bottom of the incline first because it relates directly to how the object's mass is distributed around its rotation axis.
Moment of Inertia
The moment of inertia is essentially a measure of an object's resistance to changes in its rotational motion. It depends on the distribution of the object's mass relative to the axis of rotation. In mathematical terms, it's defined with an integral over the object's mass. However, for common shapes, there are standard equations. For instance, the moment of inertia for a uniform solid disk is \( 0.5 \cdot m \cdot r^2 \) and for a hoop, it's \( m \cdot r^2 \).

The higher the moment of inertia, the harder it is to change the rotational motion. When comparing the disk and the hoop on an incline, their different moments of inertia mean they will not accelerate to the bottom at the same rate, even if their masses are the same.
Conservation of Energy
The conservation of energy principle tells us that in a closed system (one without external forces doing work), energy cannot be created or destroyed, only transformed from one form to another. For the objects rolling down the incline, the gravitational potential energy they possess at the top (calculated as \(m \cdot g \cdot h\)) will convert entirely into various forms of kinetic energy (both linear and rotational) by the time they reach the bottom, assuming no energy is lost to friction or air resistance.

Applying this principle to the disk and hoop rolling down the incline allows us to calculate their speeds at the bottom. By equating the potential energy at the top to the total kinetic energy at the bottom, we can solve for their final velocities.
Angular Velocity
Angular velocity, represented by the Greek letter \(\omega\), is a vector quantity that describes the rate of rotation of an object around an axis. The magnitude of angular velocity is measured in radians per second and signifies how quickly an object rotates. For rotating objects, angular velocity is linked to linear velocity through the relationship \(v = \omega \cdot r\), where \(v\) is linear velocity, \(\omega\) is angular velocity, and \(r\) is the radius of the rotating object.

This relationship helps us understand how the speed of an object's point on its outer edge is determined by how fast it's spinning – a crucial concept when comparing how quickly different objects roll down an incline.
Linear Velocity
Linear velocity is the rate at which an object changes its position. It is a vector quantity having both magnitude and direction. In the context of an object rolling down an incline, linear velocity represents how fast the center of mass of the object is moving along the incline path. It's calculated as the distance covered per unit of time and, for circular motion, is directly related to angular velocity.

Understanding linear velocity is fundamental in determining which of our two rolling objects, the disk or the hoop, would reach the bottom of the incline first. It is this linear velocity at the bottom of the incline that we solve for in our exercise using the conservation of energy and the objects' characteristics.

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Most popular questions from this chapter

(a) Find the angular speed of the Earth's rotation on its axis. As the Earth turns toward the east, we see the sky turning toward the west at this same rate. (b) The rainy Pleiads wester And seek beyond the sea The head that I shall dream of That shall not dream of me. Cambridge, England, is at longitude \(0^{\circ},\) and Saskatoon, Saskatchewan, is at longitude \(107^{\circ}\) west. How much time elapses after the Pleiades set in Cambridge until these stars fall below the western horizon in Saskatoon?

The combination of an applied force and a friction force produces a constant total torque of \(36.0 \mathrm{N} \cdot \mathrm{m}\) on a wheel rotating about a fixed axis. The applied force acts for \(6.00 \mathrm{s} .\) During this time the angular speed of the wheel increases from 0 to 10.0 rad/s. The applied force is then removed, and the wheel comes to rest in 60.0 s. Find (a) the moment of inertia of the wheel, (b) the magnitude of the frictional torque, and (c) the total number of revolutions of the wheel.

A merry-go-round is stationary. A dog is running on the ground just outside its circumference, moving with a constant angular speed of \(0.750 \mathrm{rad} / \mathrm{s} .\) The dog does not change his pace when he sees what he has been looking for: a bone resting on the edge of the merry-go-round one third of a revolution in front of him. At the instant the dog sees the bone \((t=0),\) the merry-go-round begins to move in the direction the dog is running, with a constant angular acceleration of \(0.0150 \mathrm{rad} / \mathrm{s}^{2},\) (a) At what time will the dog reach the bone? (b) The confused dog keeps running and passes the bone. How long after the merry-go-round starts to turn do the dog and the bone draw even with each other for the second time?

Consider a tall building located on the Earth's equator. As the Earth rotates, a person on the top floor of the building moves faster than someone on the ground with respect to an inertial reference frame, because the latter person is closer to the Earth's axis. Consequently, if an object is dropped from the top floor to the ground a distance \(h\) below, it lands east of the point vertically below where it was dropped. (a) How far to the east will the object land? Express your answer in terms of \(h, g,\) and the angular speed \(\omega\) of the Earth. Neglect air resistance, and assume that the free-fall acceleration is constant over this range of heights. (b) Evaluate the eastward displacement for \(h=50.0 \mathrm{m}\) (c) In your judgment, were we justified in ignoring this aspect of the Coriolis effect in our previous study of free fall?

An object with a weight of \(50.0 \mathrm{N}\) is attached to the free end of a light string wrapped around a reel of radius \(0.250 \mathrm{m}\) and mass \(3.00 \mathrm{kg} .\) The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center. The suspended object is released \(6.00 \mathrm{m}\) above the floor. (a) Determine the tension in the string, the acceleration of the object, and the speed with which the object hits the floor. (b) Verify your last answer by using the principle of conservation of energy to find the speed with which the object hits the floor.

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