/*! 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 76 While redesigning a rocket engin... [FREE SOLUTION] | 91Ó°ÊÓ

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While redesigning a rocket engine, you want to reduce its weight by replacing a solid spherical part with a hollow spherical shell of the same size. The parts rotate about an axis through their center. You need to make sure that the new part always has the same rotational kinetic energy as the original part had at any given rate of rotation. If the original part had mass \(M,\) what must be the mass of the new part?

Short Answer

Expert verified
The new part must have a mass of \( \frac{3}{5}M \).

Step by step solution

01

Define Initial Conditions

We start by considering the solid sphere with mass \( M \) and radius \( R \). Recall that for a solid sphere, the moment of inertia \( I_s \) about its center is given by \( \frac{2}{5}MR^2 \).
02

Calculate Kinetic Energy of Original Part

The rotational kinetic energy \( KE \) for a solid object is given by \( \frac{1}{2} I \omega^2 \), where \( \omega \) is the angular velocity. So for the solid sphere, \( KE = \frac{1}{2} \cdot \frac{2}{5}MR^2 \cdot \omega^2 = \frac{1}{5}MR^2 \omega^2 \).
03

Define Inertia of Hollow Shell

For a hollow spherical shell of mass \( m \) and radius \( R \), the moment of inertia \( I_h \) is \( \frac{2}{3}mR^2 \).
04

Set Equal Kinetic Energies

To maintain the same rotational kinetic energy, we require \( \frac{1}{5}MR^2 \omega^2 = \frac{1}{2} \cdot \frac{2}{3}mR^2 \omega^2 \).
05

Solve for New Mass

By cancelling \( R^2 \omega^2 \) from both sides, the equation becomes \( \frac{1}{5}M = \frac{1}{3}m \). Solving for \( m \), you get: \[ m = \frac{3}{5}M \].

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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 as the rotational equivalent of mass in linear motion. The moment of inertia depends on the object's shape and mass distribution. For instance, a solid object has its mass spread throughout its volume, affecting how it rotates.

In mathematical terms, the moment of inertia, often symbolized as \(I\), is calculated using integrals over the mass distribution of the object. For simpler shapes, there are standard formulas:
  • For a solid sphere, \(I = \frac{2}{5}MR^2\).
  • For a hollow spherical shell, \(I = \frac{2}{3}mR^2\).
These formulas reveal how different geometries affect rotation, influencing the moment of inertia. The larger the moment of inertia, the more torque needed to accelerate or decelerate its rotation.
Solid Sphere
A solid sphere is a three-dimensional object where all the mass is distributed uniformly throughout its volume. This uniform distribution causes the properties of rotation to be distinct compared to other shapes.

In the context of our problem, the solid sphere's mass is concentrated from the center to the outer surface, contributing to its moment of inertia: \(I = \frac{2}{5}MR^2\). This equation highlights that the mass \(M\) and the radius \(R\) both play crucial roles in how the sphere resists changes to its rotation. When substituting a hollow shell for a solid sphere, its density critically changes, affecting the moment of inertia and consequently the rotational energy.
Hollow Spherical Shell
Unlike a solid sphere, a hollow spherical shell has all its mass located at a fixed distance from the center. Its mass is concentrated along the shell's inner surface, not throughout the entire volume.

The greater emphasis on mass location results in a different moment of inertia formula: \(I = \frac{2}{3}mR^2\). Given the same radius, a hollow shell has a higher moment of inertia compared to a solid sphere if they both have the same mass. This means you'll need less mass \(m = \frac{3}{5}M\) in a hollow shell to achieve the same rotational kinetic energy as a solid sphere of mass \(M\). Using this design principle helps engineers optimize the weight of rotating components while maintaining functionality.
Angular Velocity
Angular velocity is the rate at which an object rotates around an axis. It tells us how quickly the angle changes in a given amount of time. Unlike linear velocity, which uses units like meters per second, angular velocity is often measured in radians per second.

Understanding angular velocity is crucial because it directly impacts rotational kinetic energy, defined in the formula \(KE = \frac{1}{2} I \omega^2\). When you adjust the shape or mass of rotating objects, such as swapping a solid sphere for a hollow shell, it’s crucial to keep angular velocity changes in mind to ensure the energy remains constant. The importance of angular velocity becomes evident in fields like mechanical engineering or physics, where rotation is critical, from engines to planetary motion.

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Most popular questions from this chapter

A frictionless pulley has the shape of a uniform solid disk of mass 2.50 \(\mathrm{kg}\) and radius 20.0 \(\mathrm{cm} . \mathrm{A}\) 1.50-\textrm{kg} \text { stone is } attached to a very light wire that is wrapped around the rim of the pulley (Fig. E9.47), and the system is released from rest. (a) How far must the stone fall so that the pulley has 4.50 J of kinetic energy? (b) What percent of the total kinetic energy does the pulley have?

A turntable rotates with a constant 2.25 \(\mathrm{rad} / \mathrm{s}^{2}\) angular acceleration. After 4.00 s it has rotated through an angle of 60.0 rad. What was the angular velocity of the wheel at the beginning of the 4.00 -s interval?

A slender rod with length \(L\) has a mass per unit length that varies with distance from the left end, where \(x=0\) , according to \(d m / d x=\gamma x,\) where \(\gamma\) has units of \(\mathrm{kg} / \mathrm{m}^{2}\) . (a) Calculate the total mass of the rod in terms of \(\gamma\) and \(L .\) (b) Use Eq. \((9.20)\) to calculate the moment of inertia of the rod for an axis at the left end, perpendicular to the rod. Use the expression you derived in part (a) to express \(I\) in terms of \(M\) and \(L .\) How does your result compare to that for a uniform rod? Explain this comparison. (c) Repeat part (b) for an axis at the right end of the rod. How do the results for parts (b) and (c) compare? Explain this result.

Centrifuge. An advertisement claims that a centrifuge takes up only 0.127 m of bench space but can produce a radial acceleration of 3000\(g\) at 5000 rev/min. Calculate the required radius of the centrifuge. Is the claim realistic?

A uniform 2.00 -m ladder of mass 9.00 kg is leaning against a vertical wall while making an angle of \(53.0^{\circ}\) with the floor. A worker pushes the ladder up against the wall until it is vertical. What is the increase in the gravitational potential energy of the ladder?

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