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Chapter 9: Problem 47

A pulley on a frictionless axle has the shape of a uniform solid disk of mass \(2.50 \mathrm{~kg}\) and radius \(20.0 \mathrm{~cm}\). A \(1.50 \mathrm{~kg}\) stone is attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has \(4.50 \mathrm{~J}\) of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

Short Answer

Expert verified
The stone must fall \(24.5 \) cm so the pulley acquires kinetic energy of \(4.50 \mathrm{~J}\). The pulley has 14.29% of the total kinetic energy.

Step by step solution

01

Determine the moment of inertia of the pulley

The moment of inertia, I, of a uniform disk is given by \( I = 0.5MR^2 \) where M is the mass and R is the radius. Here, Mass M = 2.50 kg and Radius R = 20.0 cm or 0.2 m. Substituting these values into the formula gives us \( I = 0.5 * 2.50kg * (0.2m)^2 = 0.05 kg.m^2 \)
02

Determine the rotational kinetic energy of the pulley

The kinetic energy (K) of an object rotating about an axis is given by \( K = 0.5I蠅^2 \), where 蠅 is the angular velocity. We are told that the pulley has 4.50 J of kinetic energy. We equate this value to the kinetic energy formula, thus \( 4.5 J = 0.5 * 0.05 kg.m^2 * 蠅^2 \). Solving for 蠅, we find that \( 蠅 = sqrt((4.5 J) / (0.5 * 0.05 kg.m^2)) = 30 rad/s \).
03

Apply conservation of energy principles

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed. When the stone falls, potential energy is converted into kinetic energy. The potential energy of the stone is given by \( mgh \), where m is mass, g is acceleration due to gravity, and h is height. Setting this equal to the kinetic energy gained by the stone and pulley, we have \( (1.5 kg * 9.8 m/s^2 * h) + 4.5 J = 0.5 * 1.5kg * v^2 \), where v is the velocity of the stone. We know the pulley and the stone move together so their linear velocities are the same, \( v = R蠅 \). Substituting for v, the equation becomes \( 1.5kg * 9.8 m/s^2 * h + 4.5J = 0.5 * 1.5kg * (0.2m * 30 rad/s)^2 \). Solving for h, we get h = 0.245 m or 24.5 cm.
04

Determine the percent of the total kinetic energy the pulley has

The total kinetic energy is the sum of the kinetic energy of both the pulley and the stone. The kinetic energy of the stone is \( Ks = 0.5 * 1.5kg * (0.2m * 30 rad/s)^2 = 27J \). So, the total kinetic energy is \( 27J + 4.5J = 31.5J \). The percentage of the total kinetic energy that the pulley has is thus \( (4.5J / 31.5J) * 100 = 14.29% \).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in the study of rotational motion, much like mass in linear motion. It's a measure of an object's resistance to changes in its rotational motion.

Moment of inertia depends not only on the mass of an object but also on the distribution of that mass relative to the axis of rotation. For a uniform solid disk like a pulley, the formula to calculate moment of inertia is \( I = 0.5MR^2 \), where \( M \) is the mass and \( R \) is the radius of the disk. In our exercise, substituting the given mass of \( 2.50 \text{kg} \) and radius of \( 20.0 \text{cm} \), or \( 0.2 \text{m} \), we find a moment of inertia of \( 0.05 \text{kg}\cdot\text{m}^2 \). This value is pivotal as it sets the stage for understanding how the pulley reacts to rotational forces and will influence the pulley's kinetic energy as well as how far the attached stone must fall to reach a certain rotational speed.
Conservation of Energy
Conservation of energy is a principle stating that the total energy in an isolated system remains constant over time. This means that energy can neither be created nor destroyed, but it can be transformed from one form to another.

In the context of our problem, when the stone falls, its potential energy - due to its height above the ground - is converted into two types of kinetic energy: the stone's linear kinetic energy and the pulley's rotational kinetic energy. Mathematically, we express this conservation by equating the potential energy lost by the stone, \( mgh \), with the total kinetic energy it gained along with the pulley. By wisely using this principle, we were able to calculate the distance the stone must fall, which is \( 24.5 \text{cm} \), to impart \( 4.50 \text{J} \) of kinetic energy to the pulley.
Angular Velocity
Angular velocity represents the rate at which an object rotates or revolves. It's analogous to linear velocity but in the context of circular motion. The faster the rotation, the higher the angular velocity.

To find the angular velocity, we utilize the formula for rotational kinetic energy: \( K = 0.5I\omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. From our given data, we set the pulley's kinetic energy to \( 4.50 \text{J} \) and solve for \( \omega \) to get \( 30 \text{rad/s} \). This solution tells us how quickly the pulley is spinning once the stone has fallen the required distance to reach the specified kinetic energy.

Understanding angular velocity is essential not only for solving physics problems but also for real-world applications, like calculating the speed of wheels and gears in machinery.

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Most popular questions from this chapter

A flywheel with radius \(0.300 \mathrm{~m}\) starts from rest and accelerates with a constant angular acceleration of \(0.600 \mathrm{rad} / \mathrm{s}^{2} .\) For a point on the rim of the flywheel, what are the magnitudes of the tangential, radial, and resultant accelerations after \(2.00 \mathrm{~s}\) of acceleration?

A safety device brings the blade of a power mower from an initial angular speed of \(\omega_{1}\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_{3}\) that was three times as great, \(\omega_{3}=3 \omega_{1} ?\)

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(v=1.25 \mathrm{~m} / \mathrm{s} .\) Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the \(\mathrm{CD}\) is played. (See Exercise \(9.20 .\) ) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta)=r_{0}+\beta \theta,\) where \(r_{0}\) is the radius of the spiral at \(\theta=0\) and \(\beta\) is a constant. On a \(\mathrm{CD}, r_{0}\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d \theta,\) the distance scanned along the track is \(d s=r d \theta .\) Using the above expression for \(r(\theta),\) integrate \(d s\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) since the track is scanned at a constant linear speed \(v,\) the distance \(s\) found in part (a) is equal to vi. Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta ;\) choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_{z}\) and the angular acceleration \(\alpha_{z}\) as functions of time. Is \(\alpha_{z}\) constant? (d) On a CD, the inner radius of the track is \(25.0 \mathrm{~mm}\), the track radius increases by \(1.55 \mu \mathrm{m}\) per revolution, and the playing time is \(74.0 \mathrm{~min} .\) Find \(r_{0}, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_{z}\) (in rad/s) versus \(t\) and \(\alpha_{z}\) (in rad/s \(^{2}\) ) versus \(t\) between \(t=0\) and \(t=74.0 \mathrm{~min}\)

A uniform, solid disk with mass \(m\) and radius \(R\) is pivoted about a horizontal axis through its center. A small object of the same mass \(m\) is glued to the rim of the disk. If the disk is released from rest with the small object at the end of a horizontal radius, find the angular speed when the small object is directly below the axis.

A disk of radius \(25.0 \mathrm{~cm}\) is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. \(\mathbf{P 9 . 6 1}\) ). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is \(1.80 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find \(A\). (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of \(15.0 \mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches \(15.0 \mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .)\)
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