/*! 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 74 A sphere consists of a solid woo... [FREE SOLUTION] | 91Ó°ÊÓ

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A sphere consists of a solid wooden ball of uniform density 800 \(\mathrm{kg} / \mathrm{m}^{3}\) and radius 0.30 \(\mathrm{m}\) and is covered with a thin coating of lead foil with area density 20 \(\mathrm{kg} / \mathrm{m}^{2}\) . Calculate the moment of inertia of this sphere about an axis passing through its center.

Short Answer

Expert verified
The total moment of inertia of the sphere is 4.616 kg·m².

Step by step solution

01

Calculate the Mass of the Wooden Sphere

First, find the volume of the wooden sphere using the formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \). Given that the radius \( r \) is 0.30 m, calculate: \( V = \frac{4}{3} \pi (0.30)^3 = 0.1131 \text{ m}^3 \). With a density of 800 kg/m³, the mass of the wooden sphere \( m_{wood} \) is the product of volume and density: \( m_{wood} = 0.1131 \times 800 = 90.48 \text{ kg} \).
02

Calculate the Moment of Inertia of the Wooden Sphere

The moment of inertia for a uniform solid sphere is given by the formula \( I_{wood} = \frac{2}{5} m_{wood} r^2 \). Substituting in the values we have: \( I_{wood} = \frac{2}{5} \times 90.48 \times (0.30)^2 = 3.26 \text{ kg} \cdot \text{m}^2 \).
03

Calculate the Mass of the Lead Foil

The area of the sphere's surface is calculated using \( A = 4 \pi r^2 \). Plug in the radius \( r = 0.30 \): \( A = 4 \pi (0.30)^2 = 1.131 \text{ m}^2 \). With given area density of 20 kg/m², the mass of the lead foil \( m_{foil} \) is: \( m_{foil} = 1.131 \times 20 = 22.62 \text{ kg} \).
04

Calculate the Moment of Inertia of the Lead Foil

The moment of inertia for a hollow spherical shell is given by \( I_{foil} = \frac{2}{3} m_{foil} r^2 \). Using the mass calculated previously: \( I_{foil} = \frac{2}{3} \times 22.62 \times (0.30)^2 = 1.356 \text{ kg} \cdot \text{m}^2 \).
05

Calculate the Total Moment of Inertia

The total moment of inertia of the sphere, which is the sum of the moments of inertia of the wooden sphere and the lead foil, is given by \( I_{total} = I_{wood} + I_{foil} \). Substituting in our results: \( I_{total} = 3.26 + 1.356 = 4.616 \text{ kg} \cdot \text{m}^2 \).

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Key Concepts

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

Solid Sphere
A solid sphere, in physics, is an object that is completely filled with mass. Unlike hollow spheres or shells, every point from the center to the outermost surface is solid. The moment of inertia of a solid sphere can tell us a lot about its rotational behavior, giving insights into how an object's mass is distributed in relation to its axis of rotation.
The equation for calculating the moment of inertia for a uniform solid sphere rotating about an axis through its center is:
  • \( I = \frac{2}{5} m r^2 \)
Here, \(m\) is the mass of the sphere and \(r\) is its radius. The formula reflects the distribution of the sphere's mass from its core outwards.
Solid spheres are often modeled in physics problems because they provide a uniform distribution of mass, simplifying calculations. This characteristic makes them an ideal subject when studying moments of inertia in rotational dynamics.
Hollow Spherical Shell
The hollow spherical shell is quite different from a solid sphere, as it only has its mass distributed along its outer surface with nothing inside. Calculating the moment of inertia for a hollow sphere involves its mass concentrated along a very thin shell encircling the core.
Because its mass is concentrated further from the center as compared to a solid sphere, the formula for the moment of inertia is also different:
  • \( I = \frac{2}{3} m r^2 \)
This equation implies that the rotational inertia is affected by the radius more significantly, given that the mass is at a distance from the axis of the motion.
It's crucial to distinguish between these two types of spherical objects because despite their geometric similarities, their physical behaviors under rotational circumstances can greatly differ.
Uniform Density
Uniform density refers to the consistent distribution of mass throughout an object. For a sphere with uniform density, every portion or section of the sphere has the same density, composed equally throughout its volume. This uniformity allows for straightforward calculations of its physical characteristics such as mass and volume.
The density equation is given by:
  • \( \rho = \frac{m}{V} \)
where \( \rho \) is density, \( m \) is mass, and \( V \) is volume. In uniform density objects, once the volume is known, the mass can be easily determined through this relationship. This is particularly useful for solid spheres, ensuring calculations for moment of inertia reflect the real distribution of mass inside the sphere.
Assuring uniform density simplifies understanding and describing how the object will react to various forces and moments, like rotation.
Area Density
Area density is a measure of how much mass exists per unit area on a surface, especially applied in objects like thin films or coatings. This is important for objects that are not completely solid, like the lead foil coating on our hollow spherical shell.
The formula for area density \( \sigma \) is:
  • \( \sigma = \frac{m}{A} \)
where \( m \) is mass and \( A \) is the area. This relationship is essential when considering coatings and their contribution to the overall mass and moment of inertia of an object.
When you have a thin layer over an object, the area density helps calculate its mass contribution without having to dive into volumetric properties. For a sphere coated with lead foil, understanding the area density is key to comprehending how this additional mass affects the sphere’s total moment of inertia.

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Most popular questions from this chapter

A cylinder with radius \(R\) and mass \(M\) has density that increases linearly with distance \(r\) from the cylinder axis, \(\rho=\alpha r\) where \(\alpha\) is a positive constant. (a) Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of \(M\) and \(R .\) (b) Is your answer greater or smaller than the moment of inertia of a cylinder of the same mass and radius but of uniform density? Explain why this result makes qualitative sense.

A flywheel has angular acceleration \(\alpha_{z}(t)=$$8.60 \mathrm{rad} / \mathrm{s}^{2}-\left(2.30 \mathrm{rad} / \mathrm{s}^{3}\right) t,\) where counterclockwise rotation is positive. (a) If the flywheel is at rest at \(t=0,\) what is its angular velocity at 5.00 s? (b) Through what angle (in radians) does the flywheel turn in the time interval from \(t=0\) to \(t=5.00\) s?

The flywheel of a gasoline engine is required to give up 500 \(\mathrm{J}\) of kinetic energy while its angular velocity decreases from 650 \(\mathrm{rev} / \mathrm{min}\) to 520 \(\mathrm{rev} / \mathrm{min}\) . What moment of inertia is required?

(a) Derive an equation for the radial acceleration that includes \(v\) and \(\omega,\) but not \(r .\) (b) You are designing a merry-go-roundfor which a point on the rim will have a radial acceleration of 0.500 \(\mathrm{m} / \mathrm{s}^{2}\) when the tangential velocity of that point has magnitude 2.00 \(\mathrm{m} / \mathrm{s} .\) What angular velocity is required to achieve these values?

A thin, uniform rod is bent into a square of side length \(a\) . If the total mass is \(M\) , find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (Hint: Use the parallel-axis theorem.)

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