/*! 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 52 A solid disk rotates in the hori... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 52

A solid disk rotates in the horizontal plane at an angular velocity of \(0.067 \mathrm{rad} / \mathrm{s}\) with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is \(0.10 \mathrm{~kg} \cdot \mathrm{m}^{2}\). From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of \(0.40 \mathrm{~m}\) from the axis. The sand in the ring has a mass of \(0.50 \mathrm{~kg}\). After all the sand is in place, what is the angular velocity of the disk?

Short Answer

Expert verified
The final angular velocity is approximately \(0.0372\, \mathrm{rad/s}\).

Step by step solution

01

Understand the Problem

We have a rotating disk with a certain moment of inertia. Sand is added, forming a ring on the disk. We need to find the new angular velocity of the system.
02

Concept of Conservation of Angular Momentum

Since no external torques act on the system, the angular momentum before the sand is added will equal the angular momentum after the sand is added. Mathematically, this is expressed as \( L_i = L_f \), where \( L \) is the angular momentum.
03

Calculate Initial Angular Momentum

The initial angular momentum \( L_i \) is calculated by multiplying the initial angular velocity \( \omega_i = 0.067\, \mathrm{rad/s} \) by the moment of inertia of the disk \( I_d = 0.10\, \mathrm{kg \cdot m^2} \). Thus, \( L_i = I_d \cdot \omega_i = 0.10 \times 0.067 = 0.0067\, \mathrm{kg \cdot m^2/s} \).
04

Calculate Moment of Inertia of the Sand Ring

The sand forms a ring with mass \( m_s = 0.50\, \mathrm{kg} \) at a radius \( r = 0.40\, \mathrm{m} \) from the axis. Its moment of inertia is \( I_s = m_s \cdot r^2 = 0.50 \times (0.40)^2 = 0.08\, \mathrm{kg \cdot m^2} \).
05

Calculate Total Moment of Inertia After Sand is Added

The total moment of inertia \( I_f \) after the sand is added is the sum of the disk's moment of inertia and the sand's moment of inertia: \( I_f = I_d + I_s = 0.10 + 0.08 = 0.18\, \mathrm{kg \cdot m^2} \).
06

Apply Conservation of Angular Momentum

Since \( L_i = L_f \), we have \( I_d \cdot \omega_i = I_f \cdot \omega_f \). Substituting the known values: \( 0.0067 = 0.18 \cdot \omega_f \).
07

Solve for Final Angular Velocity

Solve the equation \( 0.0067 = 0.18 \cdot \omega_f \) for \( \omega_f \) to find the final angular velocity: \( \omega_f = \frac{0.0067}{0.18} \approx 0.0372\, \mathrm{rad/s} \).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in physics, particularly when analyzing rotational motion. Think of it as the rotational equivalent of mass in linear motion. It determines how difficult it is to change the rotational state of an object. For any rotating object, the moment of inertia (\(I\)) depends on how the mass is distributed about the axis of rotation.
  • If more mass is distributed further from the axis, the moment of inertia increases.
  • Conversely, mass closer to the axis decreases it.
In the original exercise, the moment of inertia computed for both the disk (\(0.10 \, \text{kg} \cdot \text{m}^2\)) and the sand ring (\(0.08 \, \text{kg} \cdot \text{m}^2\)) reveal how each contributes to the overall inertia of the system.This allows us to calculate the total inertia after changes have occurred, really highlighting its role in characterizing rotational motion.
Rotational Motion
Rotational motion is when an object spins around a specific axis. This motion can be described using principles similar to linear motion, but with a twist鈥攍iterally.
  • The angular position represents how far around the circle an object has rotated.
  • Angular displacement refers to the change between two angular positions.
  • Much like velocity in linear motion, angular velocity tells us how fast something is rotating.
Rotational dynamics focus on how various forces and torques affect objects in rotational motion. A critical aspect explained in the exercise is the conservation of angular momentum. This principle states: "If no external torques are acting on a system, the total angular momentum remains constant." This concept applies powerfully to the given disk as sand is added, forming a ring that modifies its overall rotational characteristics.
Angular Velocity
Angular velocity (\(\omega\)) is a measure of how quickly something spins around an axis, typically expressed in radians per second or degrees per second.In simpler terms:
  • It indicates the rate of rotation.
  • Angular velocity is positive for counterclockwise rotations and negative for clockwise rotations.
In the exercise, the initial angular velocity of the disk is set at \(0.067 \, \text{rad/s}\). When additional mass is added, as it does with the sand, the disk's angular velocity changes to maintain conservation of angular momentum.By applying the formula and solving, we find that the new angular velocity becomes approximately \(0.0372 \, \text{rad/s}\). This showcases interactions between mass distribution, angular momentum, and resulting changes in angular speed, underscoring the intimate link between these concepts in rotational dynamics.

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Most popular questions from this chapter

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of \(2.00 \mathrm{~N}\) and is applied perpendicular to the length of the stick at the free end. The other force has a magnitude of \(6.00 \mathrm{~N}\) and acts at a \(30.0^{\circ}\) angle with respect to the length of the stick. Where along the stick is the 6.00 -N force applied? Express this distance with respect to the end that is pinned.

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun. Assume that the earth is a uniform sphere and that its path around the sun is circular. For comparison, the total energy used in the United States in one year is about \(9.3 \times 10^{19} \mathrm{~J}\).

See Multiple-Concept Example 12 to review some of the concepts that come into play here. The crane shown in the drawing is lifting a 180 -kg crate upward with an acceleration of \(1.2 \mathrm{~m} / \mathrm{s}^{2} .\) The cable from the crate passes over a solid cylindrical pulley at the top of the boom. The pulley has a mass of \(130 \mathrm{~kg}\). The cable is then wound onto a hollow cylindrical drum that is mounted on the deck of the crane. The mass of the drum is \(150 \mathrm{~kg},\) and its radius is \(0.76 \mathrm{~m} .\) The engine applies a counterclockwise torque to the drum in order to wind up the cable. What is the magnitude of this torque? Ignore the mass of the cable.

Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of \(125 \mathrm{~kg} \cdot \mathrm{m}^{2} .\) When one person is standing at a distance of \(1.50 \mathrm{~m}\) from the center, the carousel has an angular velocity of \(0.600 \mathrm{rad} / \mathrm{s}\). However, as this person moves inward to a point located \(0.750 \mathrm{~m}\) from the center, the angular velocity increases to \(0.800 \mathrm{rad} / \mathrm{s}\). What is the person's mass?

ssm A stationary bicycle is raised off the ground, and its front wheel \((m=1.3 \mathrm{~kg})\) is rotating at an angular velocity of \(13.1 \mathrm{rad} / \mathrm{s}\) (see the drawing). The front brake is then applied for \(3.0 \mathrm{~s}\), and the wheel slows down to \(3.7 \mathrm{rad} / \mathrm{s}\). Assume that all the mass of the wheel is concentrated in the rim, the radius of which is \(0.33 \mathrm{~m}\). The coefficient of kinetic friction between each brake pad and the rim is \(\mu_{\mathrm{k}}=0.85 .\) What is the magnitude of the normal force that each brake pad applies to the rim?
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