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\- 52 (co A cockroach of mass \(m\) lies on the rim of a uniform disk of mass \(4.00 m\) that can rotate freely about its center like a merrygo-round. Initially the cockroach and disk rotate together with an angular velocity of \(0.260 \mathrm{rad} / \mathrm{s}\). Then the cockroach walks halfway to the center of the disk. (a) What then is the angular velocity of the cockroach- disk system? (b) What is the ratio \(K / K_{0}\) of the new kinetic energy of the system to its initial kinetic energy? (c) What accounts for the change in the ki- \(\quad\) netic energy?

Short Answer

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We have a system consisting of a disk and a cockroach, which initially rotate together with a specific angular velocity. As the cockroach walks towards the center of the disk, its position changes, affecting the system's rotation and kinetic energy. We need to determine the new angular velocity and compare the kinetic energies before and after the cockroach moves.

Step by step solution

01

Understanding the problem

We have a system consisting of a disk and a cockroach, which initially rotate together with a specific angular velocity. As the cockroach walks towards the center of the disk, its position changes, affecting the system's rotation and kinetic energy. We need to determine the new angular velocity and compare the kinetic energies before and after the cockroach moves.
02

Apply conservation of angular momentum

Since no external torques act on the system, angular momentum is conserved. The initial angular momentum is given by the sum of the angular momentum of the disk and the cockroach: \[ L_i = I_d \omega_i + I_c \omega_i \]where \( I_d \) is the moment of inertia of the disk \( \frac{1}{2}(4m)R^2 \) and \( I_c \) is the moment of inertia of the cockroach \( mR^2 \). After the cockroach moves, the final angular momentum \( L_f \) is:\[ L_f = I_d \omega_f + m(\frac{R}{2})^2 \omega_f \]

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Key Concepts

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

Conservation of Angular Momentum
Conservation of Angular Momentum is a fundamental principle in physics that states that within a closed system, without external influences like torque, the total angular momentum remains constant. This idea relates directly to the exercise where the cockroach moves closer to the center of the rotating disk. Initially, both the cockroach and the disk are rotating as a single system with a certain angular velocity. Angular momentum, denoted by \( L \), is the product of the moment of inertia \( I \) and angular velocity \( \omega \).
The principle tells us that
  • Initial Angular Momentum \( = I_d \omega_i + I_c \omega_i \)
  • Final Angular Momentum \( = I_d \omega_f + I_c' \omega_f \)
where \( I_c \) changes when the cockroach moves, since moment of inertia depends on mass distribution. Thus, because there's no net torque: \[ I_d \omega_i + I_c \omega_i = I_d \omega_f + m\left(\frac{R}{2}\right)^2 \omega_f \]This equation allows us to solve for \( \omega_f \), showing conservation of angular momentum.
Kinetic Energy
Kinetic Energy in a rotating system is associated with the motion of its parts, and in this scenario, it's important to consider how it changes when the cockroach shifts its position. The kinetic energy of a rotating object is given by:
\[ K = \frac{1}{2} I \omega^2 \]
Before the cockroach moves, the total initial kinetic energy \( K_0 \) is:
  • \( K_0 = \frac{1}{2} I_d \omega_i^2 + \frac{1}{2} I_c \omega_i^2 \)
After the cockroach has moved, the new kinetic energy \( K \) is:
  • \( K = \frac{1}{2} I_d \omega_f^2 + \frac{1}{2} I_c' \omega_f^2 \)
The ratio \( K / K_0 \) shows how kinetic energy changes. Notably, this shift occurs because while angular momentum is conserved, kinetic energy can change due an internal redistribution of mass. Such kinetic energy changes are common in systems where mass repositions itself internally.
Rotational Motion
Rotational Motion is an essential aspect of systems like the cockroach on the disk. Here, the entire setup can rotate about a fixed axis, much like a merry-go-round. The fundamental quantities characterizing rotational motion include angular velocity \( \omega \), moment of inertia \( I \), and angular momentum \( L \).
For the disk and cockroach:
  • Angular velocity \( \omega \) measures how fast the system rotates.
  • Moment of inertia \( I \) is a measure of how the mass is distributed relative to the axis.
  • As mass distribution changes, so does \( I \), thereby affecting \( \omega \).
The cockroach walking towards the center changes \( I \), and therefore impacts \( \omega \). This adjustment reflects the inherent link between linear and rotational dynamics, with rotational motion being integral in our understanding of how force and movement are related in rotating systems.
Moment of Inertia
Moment of Inertia, often symbolized by \( I \), measures an object's resistance to changes in its rotational state. It depends significantly on how an object's mass is distributed in relation to the axis of rotation. The farther the mass is from the axis, the greater the moment of inertia.
For the cockroach-disk system:
  • Initially, the disk has a moment of inertia \( I_d = \frac{1}{2}(4m)R^2 \).
  • The cockroach has \( I_c = mR^2 \).
When the cockroach moves to half the radius, its moment of inertia changes to \( m\left(\frac{R}{2}\right)^2 \). This reduction in \( I_c \) leads to a change in angular velocity of the system in order to keep angular momentum constant. Understanding how moment of inertia works is key to mastering rotational dynamics, especially when mass repositions within a system, as it tells us how hard it is to rotate.

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Most popular questions from this chapter

Shows three rotating, uniform disks that are coupled by belts. One belt runs around the rims of disks \(A\) and \(C\). Another belt runs around a central hub on disk \(A\) and the rim of disk \(B .\) The belts move smoothly without slippage on the rims and hub. Disk \(A\) has radius \(R ;\) its hub has radius \(0.5000 R ;\) disk \(B\) has radius \(0.2500 R ;\) and disk \(C\) has radius \(2.000 R .\) Disks \(B\) and \(C\) have the same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of disk \(C\) to that of disk \(B\) ?

At time \(t=0\), a \(2.0 \mathrm{~kg}\) particle has the position vector \(\vec{r}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}-(2.0 \mathrm{~m}) \hat{\mathrm{j}}\) relative to the origin. Its velocity is given by \(\vec{v}=\left(-6.0 t^{2} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}\) for \(t \geq 0\) in seconds. About the origin, what are (a) the particle's angular momentum \(\vec{L}\) and \((\mathrm{b})\) the torque \(\vec{\tau}\) acting on the particle, both in unit-vector notation and for \(t>0 ?\) About the point \((-2.0 \mathrm{~m},-3.0 \mathrm{~m}, 0)\), what are (c) \(\vec{L}\) and (d) \(\vec{\tau}\) for \(t>0 ?\)

A \(0.400 \mathrm{~kg}\) ball is shot directly upward at initial speed \(40.0 \mathrm{~m} / \mathrm{s}\). What is its angular momentum about \(P, 2.00 \mathrm{~m}\) horizontally from he launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about \(P\) due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground?

An automobile has a total mass of \(1700 \mathrm{~kg}\). It accelerates from rest to \(40 \mathrm{~km} / \mathrm{h}\) in \(10 \mathrm{~s}\). Assume each wheel is a uniform \(32 \mathrm{~kg}\) disk. Find, for the end of the \(10 \mathrm{~s}\) interval, (a) the rotational kinetic energy of each wheel about its axle, (b) the total kinetic energy of each wheel, and (c) the total kinetic energy of the automobile.

Force \(\vec{F}=(-8.0 \mathrm{~N}) \hat{\mathrm{i}}+(6.0 \mathrm{~N}) \hat{\mathrm{j}}\) acts on a particle with position vector \(\vec{r}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}+(4.0 \mathrm{~m}) \hat{\mathrm{j}}\). What are (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of \(\vec{r}\) and \(\vec{F}\) ?

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