/*! 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 70 A solid disk rotates in the hori... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 of the disk is approximately \(0.0372\, \mathrm{rad/s}\).

Step by step solution

01

Identify the Principle

The principle here is conservation of angular momentum. The total angular momentum before and after the sand falls must be equal, as no external torques are acting on the system. Let the initial angular momentum be \( L_i \) and the final angular momentum be \( L_f \). Therefore, \( L_i = L_f \).
02

Establish Initial Angular Momentum

Initially, the angular momentum \( L_i \) can be calculated using the formula \( L = I \cdot \omega \), where \( I \) is moment of inertia and \( \omega \) is angular velocity. Here, the initial moment of inertia \( I_i = 0.10 \: \mathrm{kg} \cdot \mathrm{m}^2 \) and the initial angular velocity \( \omega_i = 0.067 \: \mathrm{rad/s} \). Hence, \( L_i = 0.10 \: \mathrm{kg} \cdot \mathrm{m}^2 \cdot 0.067 \: \mathrm{rad/s} \).
03

Calculate the Initial Angular Momentum

Perform the multiplication: \( L_i = 0.10 \times 0.067 = 0.0067 \: \mathrm{kg} \cdot \mathrm{m}^2/s \).
04

Establish Final Angular Momentum Terms

After the sand is dropped, the total moment of inertia becomes \( I_f = I_d + I_s \), where \( I_d \) is moment of inertia of the disk and \( I_s \) is moment of inertia of the sand ring. The sand forms a ring at distance \( r = 0.40 \: m \) from the axis: \( I_s = m_{sand} \cdot r^2 = 0.50 \cdot 0.40^2 \).
05

Calculate Moment of Inertia for Sand

Calculate \( I_s \): \( I_s = 0.50 \cdot (0.40)^2 = 0.50 \cdot 0.16 = 0.08 \: \mathrm{kg} \cdot \mathrm{m}^2 \).
06

Find the Total Final Moment of Inertia

Now, find \( I_f \) by summing the disk's and the sand's moments of inertia: \( I_f = 0.10 + 0.08 = 0.18 \: \mathrm{kg} \cdot \mathrm{m}^2 \).
07

Determine Final Angular Velocity

Use the conservation of angular momentum \( L_i = L_f \). Since \( L_f = I_f \cdot \omega_f \), substitute the values to solve for \( \omega_f \). Thus, \( 0.0067 = 0.18 \cdot \omega_f \), then solve for \( \omega_f \).
08

Solve for Final Angular Velocity

Divide both sides by the final moment of inertia to get \( \omega_f = \frac{0.0067}{0.18} \approx 0.0372 \: \mathrm{rad/s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Velocity
Angular velocity is a measure of how rapidly something spins around a central point or axis. It's like the rotational counterpart of linear velocity, describing how fast an object goes around in circles. In the problem we're examining, we deal with the angular velocity of a solid disk before and after sand is added to it. The initial angular velocity given is \( 0.067 \, \mathrm{rad/s} \). This refers to how fast the disk spins initially.

When we talk about angular velocity, we usually use the symbol \( \omega \). Angular velocity is essential in determining how the time rate of change in the angle is occurring, usually measured in radians per second (rad/s). This standard unit helps us avoid confusion with linear measurements like meters per second.
  • Before the sand is added, \( \omega_i = 0.067 \, \mathrm{rad/s} \).
  • After the sand is added, due to conservation of angular momentum, the angular velocity decreases, as seen with \( \omega_f \).
Understanding angular velocity's role in rotational dynamics helps us predict how changes in a system, like adding mass, impact overall rotational speed.
Moment of Inertia
Moment of inertia is a key concept in rotations. Think of it like rotational mass. It quantifies how difficult it is to change an object's rotational speed. The heavier and the more spread out an object is from its rotation point, the higher the moment of inertia.

In the exercise, you've been given the moment of inertia of the disk, \( I_d = 0.10 \, \mathrm{kg} \cdot \mathrm{m}^2 \). This tells you how resistant the disk is to changes in its rotation.

Moment of Inertia for Added Sand

When the sand drops onto the disk forming a ring, it also has a moment of inertia. This part of the exercise involved calculating the sand's effect.
  • Sand's moment of inertia \( I_s = 0.08 \, \mathrm{kg} \cdot \mathrm{m}^2 \).
  • Total moment of inertia \( I_f = I_d + I_s = 0.18 \, \mathrm{kg} \cdot \mathrm{m}^2 \).
The combined moments explain how more mass distributed further out changes the system's dynamics and, notably, its angular velocity.
Rotational Dynamics
Rotational dynamics is the branch of physics that deals with objects rotating. It's akin to linear dynamics but for spinning objects. When an object spins, factors like angular velocity, moment of inertia, and torque tell us a lot about its motion and stability. The central law of rotational dynamics we applied here is conservation of angular momentum.

Conservation of angular momentum states that in an isolated system, with no external torques, the angular momentum remains constant. Hence, when sand falls on our disk, even though the mass changes, the total angular momentum does not.
  • Initial angular momentum: \( L_i = I_i \cdot \omega_i = 0.0067 \, \mathrm{kg} \cdot \mathrm{m}^2/s \).
  • Final angular momentum: remains \( L_f = L_i \), now expressed as \( I_f \cdot \omega_f \).
This exercise beautifully illustrates rotational dynamics by showing how added mass affects speed. After the sand is added, the decreased angular velocity demonstrates the inverse relationship between mass distribution (moment of inertia) and spin speed (angular velocity). This is a classic application of rotational dynamics in a real-world scenario. By understanding these concepts, one can better predict and analyze the behavior of rotating systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A block (mass \(=2.0 \mathrm{kg}\) ) is hanging from a massless cord that is wrapped around a pulley (moment of inertia \(=1.1 \times 10^{-3} \mathrm{kg} \cdot \mathrm{m}^{2}\) ), as the drawing shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of \(0.040 \mathrm{m}\) during the block's descent. Find the angular acceleration of the pulley and the tension in the cord.

Two thin rectangular sheets \((0.20 \mathrm{m} \times 0.40 \mathrm{m})\) are identical. In the first sheet the axis of rotation lies along the \(0.20-\mathrm{m}\) side, and in the second it lies along the \(0.40-\mathrm{m}\) side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in 8.0 s. How long does it take for the second sheet, starting from rest, to reach the same angular velocity?

A ceiling fan is turned on and a net torque of \(1.8 \mathrm{N} \cdot \mathrm{m}\) is applied to the blades. The blades have a total moment of inertia of \(0.22 \mathrm{kg} \cdot \mathrm{m}^{2}\) What is the angular acceleration of the blades?

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of the star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius \(R\) that is initially rotating at 2.0 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is \(4.0 R\). Assume that all of the star's original mass is contained in the shell.

A 15.0-m length of hose is wound around a reel, which is initially at rest. The moment of inertia of the reel is \(0.44 \mathrm{kg} \cdot \mathrm{m}^{2}\), and its radius is \(0.160 \mathrm{m} .\) When the reel is turning. friction at the axle exerts a torque of magnitude \(3.40 \mathrm{N} \cdot \mathrm{m}\) on the reel. If the hose is pulled so that the tension in it remains a constant \(25.0 \mathrm{N}\), how long does it take to completely unwind the hose from the reel? Neglect the mass and thickness of the hose on the reel, and assume that the hose unwinds without slipping.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.