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An impulsive force \(\vec{F}(t)=F_{x}(t) \hat{\mathrm{i}}\) acts for a short time \(\Delta t\) on a rotating rigid body constrained to rotate about the \(z\) axis with rotational inertia \(I\). Show that $$ \left(\int \tau_{z} d t\right) \hat{\mathrm{k}}=(\mid\langle\vec{F}\rangle R \Delta t) \hat{\mathrm{k}}=I\left(\omega_{2 z}-\omega_{1}\right) \hat{\mathrm{k}} $$ where \(\tau_{z} \hat{k}\) is the torque due to the force, \(R\) is the moment arm of the force, \(\langle\vec{F}\rangle\) is the average value of the force during the time it acts on the body, and \(\omega_{1}, \hat{k}\) and \(\omega_{2} z \hat{k}\) are the rotational velocities of the body just before and just after the force acts. (The quantity \(\left(\int \tau_{z} d t\right) \hat{\mathrm{k}}=(|\langle\vec{F}\rangle| R \Delta t) \hat{\mathrm{k}}\) is called the rotational impulse, in analogy with \(\langle\vec{F} / \Delta t\), the translational impulse.)

Short Answer

Expert verified
\[ \int \tau_z dt = (|\langle \vec{F} \rangle| R \Delta t) \hat{k} = I(\omega_2 - \omega_1) \hat{k} \].

Step by step solution

01

Identify the relationship between torque and angular momentum

First, recall the relationship between torque \( \tau \) and the change in angular momentum \( \Delta L \). By definition, torque is the rate of change of angular momentum. Hence, \( \tau = \frac{dL}{dt} \). Integrating both sides with respect to time, we get \[ \int \tau dt = \Delta L \].
02

Express the angular momentum in terms of rotational inertia and angular velocity

Angular momentum \( L \) for a rotating rigid body is given by \( L = I \omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. Therefore, the change in angular momentum \( \Delta L \) can be expressed as: \[ \Delta L = I (\omega_2 - \omega_1) \].
03

Relate torque to impulsive force

Torque \( \tau_z \) due to an impulsive force \( \vec{F} \) is given by \[ \tau_z = R \vec{F} \hat{k} \], where \( R \) is the moment arm of the force. Since the force is impulsive and acts for a short time \( \Delta t \), we can use the average force \( \langle \vec{F} \rangle \) over \( \Delta t \). Therefore, \[ \langle \tau_z \rangle = R \langle \vec{F} \rangle \], and the rotational impulse is \[ \int \tau_z dt = (|\langle \vec{F} \rangle| R \Delta t) \hat{k} \].
04

Combine the equations to show the final result

Comparing the expressions for the rotational impulse and the change in angular momentum, we get: \[ \int \tau_z dt = (|\langle \vec{F} \rangle| R \Delta t) \hat{k} = I(\omega_2 - \omega_1) \hat{k} \]. This completes the proof.

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Key Concepts

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

torque and angular momentum relationship
The relationship between torque and angular momentum is fundamental in rotational dynamics. Torque, usually denoted as \( \tau \), is essentially the rotational equivalent of force. It causes changes in the angular momentum of a system. Angular momentum (\( L \)) is a measure of the quantity of rotation of a body, which depends on the rotational inertia and the angular velocity.

To establish a clear relationship, we start with the definition of torque: \( \tau = \frac{dL}{dt} \). This means torque is the rate at which angular momentum changes over time. If you integrate this expression with respect to time, you get:

\[ \int \tau \, dt = \Delta L \]

This integral shows that the total torque applied over a period will be equal to the change in angular momentum, denoted as \( \Delta L \). It provides a direct link between how torque affects the rotational movement of an object.
rotational impulse
In rotational dynamics, impulse is an important concept. Just like translational impulse (momentum change due to a force acting over a time interval), rotational impulse refers to the change in angular momentum due to a torque acting over a time interval.

Consider an impulsive force \( \vec{F}(t) \) acting over a short time interval \( \Delta t \). The torque \( \tau_z \) produced by this force is related to the moment arm \( R \) and the force itself as:

\( \tau_z = R \vec{F} \)

Since the force acts impulsively over a short time, we often use the average force \( \langle \vec{F} \rangle \) over \( \Delta t \). The rotational impulse is given by:

\[ \int \tau_z \, dt = (| \langle \vec{F} \rangle | R \Delta t) \hat{k} \]

This equation shows how the rotational impulse (left side of the equation) is related to the applied torque and force (right side of the equation). Rotational impulse causes changes in the angular momentum of the body.
moment of inertia
Moment of inertia \( I \) is a measure of an object's resistance to change in its rotational motion. It is the rotational equivalent of mass in linear motion. For a solid object rotating around an axis, it depends not only on the object's mass but also on how that mass is distributed relative to the axis.

The angular momentum \( L \) of a rotating body can be expressed as the product of its moment of inertia \( I \) and its angular velocity \( \omega \):

\( L = I \omega \)

Therefore, the change in angular momentum \( \Delta L \) when the angular velocity changes from \( \omega_1 \) to \( \omega_2 \) is:

\[ \Delta L = I( \omega_2 - \omega_1 ) \]

This relationship shows how the inertia of the body influences the change in rotational motion. A larger moment of inertia means the body is more resistant to changes in its rotation. Understanding moment of inertia helps in analyzing and predicting rotational behavior in various physical systems.

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Most popular questions from this chapter

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

The rotational momentum of a flywheel having a rotational inertia of \(0.140 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis decreases from \(3.00\) to \(0.800 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) in \(1.50 \mathrm{~s}\). (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant rotational acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the average power of the flywheel?

The rotational inertia of a collapsing spinning star changes to \(\frac{1}{3}\) its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy?

Two particles, each of mass \(m\) and speed \(v\), travel in opposite directions along parallel lines separated by a distance \(d\). (a) In terms of \(m, v\), and \(d\), find an expression for the magnitude \(L\) of the rotational momentum of the two-particle system around a point midway between the two lines. (b) Does the expression change if the point about which \(L\) is calculated is not midway between the lines? (c) Now reverse the direction of travel for one of the particles and repeat (a) and (b).

A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g}\). Its axle radius is \(3.2 \mathrm{~mm}\), and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its translational acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) translational speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) rotational speed?

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