/*! 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 15 A nonrotating cylindrical disk o... [FREE SOLUTION] | 91Ó°ÊÓ

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A nonrotating cylindrical disk of moment of inertia \(I\) is dropped onto an identical disk rotating at angular speed \(\omega\). Assuming no external torques, what is the final common angular speed of the two disks?

Short Answer

Expert verified
The final common angular speed is \( \frac{1}{2} \omega \).

Step by step solution

01

Identify initial conditions

We have two identical disks. The first disk is rotating with an angular speed \( \omega \) and a moment of inertia \( I \), while the second disk is not rotating, so its initial angular speed is \( 0 \).
02

Use conservation of angular momentum

Since no external torques are acting on the system, the angular momentum is conserved. This means the initial total angular momentum equals the final total angular momentum.
03

Calculate initial total angular momentum

The initial angular momentum \( L_i \) is given by the rotating disk: \( L_i = I \times \omega \). The nonrotating disk has an initial angular momentum of 0.
04

Set up equation for final angular momentum

After the disks are in contact, they will rotate together at a common angular speed \( \omega_f \). The final total angular momentum \( L_f \) is given by \( (I + I) \times \omega_f = 2I \times \omega_f \).
05

Equate initial and final angular momentum

Using angular momentum conservation:\[ I \times \omega = 2I \times \omega_f \]
06

Solve for the final angular speed

Divide both sides by \( 2I \) to solve for \( \omega_f \):\[ \omega_f = \frac{1}{2} \omega \]

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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 rotational dynamics. It measures an object's resistance to changes in its rotation. Think of it like mass in linear motion but for rotating objects.

For a cylindrical disk, the moment of inertia depends on its mass and how that mass is distributed relative to the axis of rotation. Mathematically, it is denoted as \( I \). If you have a thicker or denser disk, it means a higher moment of inertia, making it harder to start or stop spinning.

In our exercise, each disk has the same moment of inertia \( I \), meaning they both have the same resistance to rotational acceleration or deceleration.
Angular Speed
Angular speed, represented by \( \omega \), describes how fast an object is rotating. It's similar to linear speed but instead of moving through space, it's about how fast an object completes rotations in a circle.

Initial conditions often involve angular speed, as seen where one disk rotates with angular speed \( \omega \) while the other is at rest. The final result we are looking for is the new common angular speed \( \omega_f \) once they rotate together. This change is calculated by using conservation laws in rotational dynamics.

It's useful to remember that even when objects combine and move as one, their individual speeds initially can significantly influence the resulting angular speed, as we solve for \( \omega_f = \frac{1}{2} \omega \).
Rotational Dynamics
Rotational dynamics is the branch of physics that deals with the motion of rotating systems and how forces affect this motion. It's analogous to linear dynamics but focuses on rotations, moments, and torques.

In our exercise, the falling disk interacts with the rotating disk, and thanks to rotational dynamics, they come to rotate at a shared speed. Even without directly applying forces, the fusion of their initial states is governed by rotational principles. This means considering their combined moments of inertia and changes in rotational speed.

Understanding these principles helps you predict how objects behave in real scenarios involving spinning or rotating, such as wheels, disks, or planets.
Angular Momentum Conservation
Angular momentum conservation is a powerful principle in physics. It states that if no external torques act on a system, the total angular momentum remains constant over time.

In our scenario, we see this principle in action. Initially, only one disk contributes to the angular momentum, but as they combine, the total angular momentum must remain as it was due to the lack of external forces or torques.

This is why the initial momentum \( I \times \omega \) equates to the final momentum \( 2I \times \omega_f \). Solving this gives us the final angular speed, demonstrating how initial conditions influence the combined system's results through momentum conservation.

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Most popular questions from this chapter

(II) A person stands on a platform, initially at rest, that can rotate frecly without friction. The moment of inertia of the person plus the platform is \(I_{\mathrm{P}}\) . The person holds a spinning bicycle whecl with its axis horizontal. The whecl has moment of inertia $$I_{\mathrm{w}} \text { and angular velocity } \omega_{\mathrm{W}} $$ What will be the angular velocity $$ \omega_{\mathbf{Y}} $$ of the platform if bthe person moves the axis of the wheel so that it points (a) vertically upward, (b) at a \(60^{\circ}\) angle to the vertical, \((c)\) vertically downward? (d) What will \(\omega_{\mathrm{P}}\) be if the person reaches up and stops the whecl in part \((a) ?\)

A potter's wheel is rotating around a vertical axis through its center at a frequency of \(1.5 \mathrm{rev} / \mathrm{s} .\) The wheel can be considered a uniform disk of mass \(5.0 \mathrm{~kg}\) and diameter \(0.40 \mathrm{~m} .\) The potter then throws a \(2.6-\mathrm{kg}\) chunk of clay, approximately shaped as a flat disk of radius \(8.0 \mathrm{~cm},\) onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it?

Determine the angular momentum of the Earth (a) about its rotation axis (assume the Earth is a uniform sphere), and \((b)\) in its orbit around the Sun (treat the Earth as a particle orbiting the Sun). The Earth has mass \(=6.0 \times 10^{24} \mathrm{~kg}\) and radius \(=6.4 \times 10^{6} \mathrm{~m},\) and is \(1.5 \times 10^{8} \mathrm{~km}\) from the Sun.

Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process, and winding up with a radius \(1.0 \%\) of its existing radius. Assuming the lost mass carries away no angular momentum, what would the Sun's new rotation rate be? (Take the Sun's current period to be about 30 days.) What would be its final kinetic energy in terms of its initial kinetic energy of today?

(1) What are the \(x, y,\) and \(z\) components of the angular momentum of a particle located at \(\vec{\mathbf{r}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}\) which has momentum \(\mathbf{p}=p_{x} \mathbf{i}+p_{y} \mathbf{j}+p_{z} \mathbf{k} ?\)

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