/*! 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 48 A bucket of mass \(m\) is tied t... [FREE SOLUTION] | 91Ó°ÊÓ

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A bucket of mass \(m\) is tied to a massless cable that is wrapped around the outer rim of a uniform pulley of radius \(R,\) on a frictionless axle, similar to the system shown in Fig. E9.47. In terms of the stated variables, what must be the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket?

Short Answer

Expert verified
The moment of inertia of the pulley must be \( \frac{1}{2} mR^2 \) so that it always has half as much kinetic energy as the bucket.

Step by step solution

01

Relate the velocities

Because the bucket and the pulley are connected by the same cable, their translational and angular velocities are related by the equation \(v = R*\omega\). The velocity of the bucket, \(v\), is equal to the product of the radius of the pulley, \(R\), and its angular speed, \(\omega\).
02

Write down the expressions for kinetic energies

Next, we need to write down the expressions for the kinetic energies of the bucket and the pulley. The kinetic energy of the bucket (translational motion) is \( KE_{bucket} = \frac{1}{2} mv^2 \) and the kinetic energy of the pulley (rotational motion) is \( KE_{pulley} = \frac{1}{2} I \omega^2 \). We know from the problem that \(KE_{pulley} = \frac{1}{2} KE_{bucket}\).
03

Substitute for \(\omega\) and \(v\)

Substitute for \(\omega\) and \(v\) from Step 1 in the equation for kinetic energies from Step 2. This would give: \( \frac{1}{2} I (v/R)^2 = \frac{1}{2} (\frac{1}{2} mv^2) \). This simplifies to \( I = \frac{1}{2} mR^2 \).
04

Find the moment of inertia

The moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket should be \( \frac{1}{2} mR^2 \)

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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 a concept that describes the energy of an object due to its motion. It is a vital component in the study of dynamics in physics, enabling us to understand how objects behave as they move.
There are two primary forms of kinetic energy:
  • Translational Kinetic Energy: Associated with the linear movement of an object. Described by the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity.
  • Rotational Kinetic Energy: Related to the rotation of an object. This form of kinetic energy is described by \( KE = \frac{1}{2} I \omega^2 \), where \( I \) represents the moment of inertia and \( \omega \) the angular velocity.
In the given exercise, these two types of kinetic energy are examined. The kinetic energy of the bucket is translational while that of the pulley is rotational. Understanding how to calculate and compare these energies is crucial for solving motion-related problems.
Angular Velocity
Angular velocity describes how fast an object rotates or revolves relative to another point. It is usually represented by the symbol \( \omega \) and is measured in radians per second. Angular velocity indicates the rate of rotation, akin to how linear velocity indicates the rate of linear movement.
For instance, in the exercise, we know that the bucket and pulley system maintains the relationship \( v = R \omega \).
This formula helps us link linear and angular motion:
  • \( v \) is the linear velocity,
  • \( R \) is the radius of the pulley,
  • \( \omega \) is the angular velocity.
By connecting angular and linear velocities, it becomes possible to solve for quantities like the kinetic energies. This understanding is vital when working with rotational systems linked to translational motion.
Rotational Motion
Rotational motion involves objects turning around an axis. This type of motion adds complexity to problems due to additional factors like angular velocity and torque compared to simple linear motion.
Rotational motion’s core concept is the moment of inertia \( I \), which is akin to mass in linear motion. It indicates how much resistance an object has to changes in its rotational speed. Its calculation, as shown in the problem, affects how much rotational kinetic energy an object will have.
Through a well-defined relationship \( KE_{pulley} = \frac{1}{2} I \omega^2 \), the moment of inertia plays a crucial role in determining kinetic energy during rotational movements. In this exercise, ensuring the pulley's rotational kinetic energy equates to half that of the bucket emphasizes the importance of balance in energy comparisons.
Recognizing these complexities in rotational motion is essential in fields like engineering and physics, where mechanical designs often integrate rotary components.
Translational Motion
Translational motion refers to the straightforward linear movement of an object. It is the motion along a straight path and is governed by Newton's laws of motion.
In the context of the bucket and pulley system, the translational motion of the bucket is key to the problem. The bucket descends, moving linearly, while its speed is determined by the translational velocity \( v \).
This linear velocity directly connects to the rotational motion of the pulley through the relationship \( v = R \omega \), bridging two distinct types of motion.
Understanding translational motion helps predict and calculate positions, velocities, and accelerations of linear-moving objects. Solving these types of problems involves grasping Newtonian mechanics and their implications fully.

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Most popular questions from this chapter

The Crab Nebula is a cloud of glowing gas about 10 light-years across, located about 6500 light-years from the earth (Fig. P9.86). It is the remnant of a star that underwent a supernova \(e x\) plosion, seen on earth in 1054 A.D. Energy is released by the Crab Nebula at a rate of about \(5 \times 10^{31} \mathrm{~W},\) about \(10^{5}\) times the rate at which the sun radiates energy. The Crab Nebula obtains its energy from the rotational kinetic energy of a rapidly spinning neutron star at its center. This object rotates once every \(0.0331 \mathrm{~s},\) and this period is increasing by \(4.22 \times 10^{-13} \mathrm{~s}\) for each second of time that elapses. (a) If the rate at which energy is lost by the neutron star is equal to the rate at which energy is released by the nebula, find the moment of inertia of the neutron star. (b) Theories of supernovae predict that the neutron star in the Crab Nebula has a mass about 1.4 times that of the sun. Modeling the neutron star as a solid uniform sphere, calculate its radius in kilometers. (c) What is the linear speed of a point on the equator of the neutron star? Compare to the speed of light. (d) Assume that the neutron star is uniform and calculate its density. Compare to the density of ordinary rock \(\left(3000 \mathrm{~kg} / \mathrm{m}^{3}\right)\) and to the density of an atomic nucleus (about \(\left.10^{17} \mathrm{~kg} / \mathrm{m}^{3}\right) .\) Justify the statement that a neutron star is essentially a large atomic nucleus.

At \(t=0\) the current to a dc electric motor is reversed, resulting in an angular displacement of the motor shaft given by \(\theta(t)=(250 \mathrm{rad} / \mathrm{s}) t-\left(20.0 \mathrm{rad} / \mathrm{s}^{2}\right) t^{2}-\left(1.50 \mathrm{rad} / \mathrm{s}^{3}\right) t^{3} .\) (a) At what time is the angular velocity of the motor shaft zero? (b) Calculate the angular acceleration at the instant that the motor shaft has zero angular velocity. (c) How many revolutions does the motor shaft turn through between the time when the current is reversed and the instant when the angular velocity is zero? (d) How fast was the motor shaft rotating at \(t=0,\) when the current was reversed? (e) Calculate the average angular velocity for the time period from \(t=0\) to the time calculated in part (a).

Measuring \(I\). As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be \(0.640 \mathrm{~m}\). Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an \(8.20 \mathrm{~kg}\) block of wood from the free end of the rope, as in Fig. E9.49. You release the system from rest and find that the block descends \(12.0 \mathrm{~m}\) in \(4.00 \mathrm{~s}\). What is the moment of inertia of the wheel for this axis?

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of \(0.200 \mathrm{rev} / \mathrm{s}^{2}\). The design specifications call for it to have a rotational kinetic energy of \(240 \mathrm{~J}\) after it has turned through 30.0 revolutions. What should be the moment of inertia of the flywheel about its rotation axis?

The motor of a table saw is rotating at 3450 rev \(/\) min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter \(0.208 \mathrm{~m}\) is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.

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