/*! 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 58 A thin uniform rod is rotating a... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin uniform rod is rotating at an angular velocity of \(7.0 \mathrm{rad} / \mathrm{s}\) about an axis that is perpendicular to the rod at its center. As the drawing indicates, the rod is hinged at two places, one-quarter of the length from each end. Without the aid of external torques, the rod suddenly assumes a "u" shape, with the arms of the "u" parallel to the rotation axis. What is the angular velocity of the rotating "u"?

Short Answer

Expert verified
The angular velocity of the rotating 'u' is 28 rad/s.

Step by step solution

01

Understand the Problem

We have a thin rod rotating about its center and then it changes shape to form a 'u'. We need to find out how the change in shape affects its angular velocity.
02

Apply Conservation of Angular Momentum

Since there are no external torques acting on the system, angular momentum is conserved. Let the initial moment of inertia be \( I_1 \) and the angular velocity be \( \omega_1 = 7.0 \, \mathrm{rad/s} \). When the rod becomes a 'u', the new moment of inertia is \( I_2 \) and we need to find \( \omega_2 \). The conservation of angular momentum gives us: \( I_1 \omega_1 = I_2 \omega_2 \).
03

Calculate Initial Moment of Inertia

The rod's length is irrelevant as it cancels in the calculation. The initial moment of inertia of a rod about its center is \( I_1 = \frac{1}{12} M L^2 \), where \( M \) is the mass and \( L \) is the length of the rod.
04

Calculate New Moment of Inertia

For the 'u' shape, the arms are parallel to the rotational axis, and the center part allows for easy calculation as two individual rods joined together. Each arm has a moment of inertia \( \frac{1}{3} Ml^2 \) (where \( l = \frac{L}{4} \)), but the center remains the same with \( \frac{1}{12} M L^2 \). Total moment of inertia becomes \( I_2 = 2 \times \frac{1}{3}M(\frac{L}{4})^2 \).
05

Solve for New Angular Velocity

Substitute the values of moments of inertia back into the conservation equation: \( \frac{1}{12} M L^2 \times 7.0 = 2 \times \frac{1}{3} M (\frac{L}{4})^2 \times \omega_2. \) Simplifying, \( 7.0 \times 4 = \omega_2 \), hence \( \omega_2 = 28 \, \mathrm{rad/s} \).

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Key Concepts

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

Understanding Moment of Inertia
The **moment of inertia** is a measure of how much an object resists changes to its rotational motion. Think of it as the rotational equivalent of mass in linear motion. It depends on the mass of the object and how that mass is distributed in relation to the axis it rotates around.

- For simple shapes, there are standard formulas to find the moment of inertia. For example, a rod rotating about its center has a moment of inertia given by \( I = \frac{1}{12} ML^2 \). Here, \( M \) is the mass and \( L \) is the length of the rod.- When the shape of an object changes, its moment of inertia also changes, even if its mass remains constant.

In the exercise, the rod initially rotates as a full length object, requiring the initial formula above. When the rod bends to form a 'u', it becomes two separate elements—each "leg" and the middle. Each element contributes an individual term to the total moment of inertia, adjusting calculations to reflect this splitting of mass.
Angular Velocity in Rotational Motion
**Angular velocity** tells us how fast an object is rotating. It's the rate of change of angular displacement, generally expressed in radians per second (rad/s). In linear motion terms, it's similar to linear speed but for objects moving in circles.

- Imagine spinning a wheel: the faster you spin it, the higher its angular velocity.- It's key to note that angular velocity can change if the distribution of mass—or moment of inertia—changes in the object.

In the problem, the rod starts with an angular velocity of \(7.0\,\mathrm{rad/s}\). This is initial angular velocity before its shape changes. Because angular momentum is conserved (in this problem), when the shape of the rod changes, its moment of inertia changes, causing the new angular velocity \( \omega_2 \) to adjust so that the overall angular momentum remains constant.
Rotational Dynamics and Conservation Laws
Rotational dynamics deals with the motion of objects that rotate. These can be complex, as forces and torques need to be considered differently from straight-line motions. However, some concepts from linear motion, like conservation laws, also apply to rotation.

- **Conservation of Angular Momentum** is central in rotational dynamics. Similar to conservation of linear momentum, it states that the total angular momentum of a closed system remains constant if no external torques act on it.- In formula terms, this can be represented by \( I_1 \omega_1 = I_2 \omega_2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.

In the exercise, because no external torques act on the rod as it changes shape, we employ this conservation law. The change from a straight rod to a 'u' shape affects \( I \), and consequently, \( \omega \) must change to maintain the same angular momentum. This exercise demonstrates the interconnectedness of inertia, velocity, and conservation laws in rotational scenarios.

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Most popular questions from this chapter

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?

Concept Questions Two thin rods of length \(L\) are rotating with the same angular speed \(\omega\) (in \(\mathrm{rad} / \mathrm{s}\) ) about axes that pass perpendicularly through one end. \(\operatorname{Rod} \mathrm{A}\) is massless but has a particle of mass \(0.66 \mathrm{~kg}\) attached to its free end. Rod \(\mathrm{B}\) has a mass \(0.66 \mathrm{~kg}\), which is distributed uniformly along its length. (a) Which has the greater moment of inertia-rod A with its attached particle or rod B? (b) Which has the greater rotational kinetic energy? Account for your answers. Problem The length of each rod is \(0.75 \mathrm{~m}\), and the angular speed is \(4.2 \mathrm{rad} / \mathrm{s}\). Find the kinetic energies of rod A with its attached particle and of rod B. Make sure your answers are consistent with your answers to the Concept Questions.

A particle is located at each corner of an imaginary cube. Each edge of the cube is \(0.25 \mathrm{~m}\) long, and each particle has a mass of \(0.12 \mathrm{~kg}\). What is the moment of inertia of these particles with respect to an axis that lies along one edge of the cube?

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 300 -mile trip in a typical midsize car produces about \(1.2 \times 10^{9} \mathrm{~J}\) of energy. How fast would a \(13-\mathrm{kg}\) flywheel with a radius of \(0.30 \mathrm{~m}\) have to rotate to store this much energy? Give your answer in rev/min.

Three objects lie in the \(x, y\) plane. Each rotates about the \(z\) axis with an angular speed of \(6.00 \mathrm{rad} / \mathrm{s} .\) The mass \(m\) of each object and its perpendicular distance \(r\) from the \(z\) axis are as follows: \((1) m_{1}=6.00 k g\) and \(r_{1}=2.00 \mathrm{~m},(2)\) \(m_{2}=4.00 \mathrm{~kg}\) and \(r_{2}=1.50 \mathrm{~m},(3) \mathrm{m}_{3}=3.00 \mathrm{~kg}\) and \(r_{3}=3.00 \mathrm{~m} .\) (a) Find the tangential speed of each object. (b) Determine the total kinetic energy of this system using the expression \(\mathrm{KE}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+\frac{1}{2} m_{3} v_{3}^{2}\) (c) Obtain the moment of inertia of the system. (d) Find the rotational kinetic energy of the system using the relation \(\frac{1}{2} I \omega^{2}\) to verify that the answer is the same as that in (b).

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