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Chapter 9: Problem 74

A sphere consists of a solid wooden ball of uniform density 800 \(\mathrm{kg} / \mathrm{m}^{3}\) and radius 0.20 \(\mathrm{m}\) 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 is approximately 1.612 kg m虏.

Step by step solution

01

Find the Volume of the Sphere

Calculate the volume of the solid wooden ball using the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \). Given that \( r = 0.2 \) m, we have \[ V = \frac{4}{3} \pi (0.2)^3 = \frac{4}{3} \pi \times 0.008 = 0.0335 \pi \approx 0.0335 \times 3.1416 = 0.105 \text{ cubic meters.} \]
02

Calculate the Mass of the Wooden Ball

The mass of the wooden ball can be found using its volume and density: \( m = \rho V \). With \( \rho = 800 \) kg/m\(^3\) and \( V = 0.105 \) m\(^3\), the mass is \[ m = 800 \times 0.105 = 84 \text{ kg.} \]
03

Calculate the Mass of the Lead Coating

Find the surface area of the sphere: \( A = 4 \pi r^2 \). With \( r = 0.2 \) meters, we have \( A = 4 \pi (0.2)^2 = 0.16 \pi \approx 0.502 \text{ square meters.} \) Multiply this by the area density to find the mass of the coating, \( m_{lead} = 20 \times 0.502 \approx 10.04 \) kg.
04

Calculate the Moment of Inertia of Wooden Ball

Use the formula for the moment of inertia of a solid sphere: \( I = \frac{2}{5} m r^2 \). Substitute the mass of the wooden ball (84 kg) and its radius (0.2 m): \[ I = \frac{2}{5} \times 84 \times (0.2)^2 = \frac{2}{5} \times 84 \times 0.04 = 1.344 \text{ kg m}^2. \]
05

Calculate the Moment of Inertia of Lead Coating

The lead coating is considered as a thin spherical shell. Use its moment of inertia formula: \( I = \frac{2}{3} m r^2 \). With \( m = 10.04 \) kg, we have \[ I = \frac{2}{3} \times 10.04 \times (0.2)^2 = \frac{2}{3} \times 10.04 \times 0.04 \approx 0.2677 \text{ kg m}^2. \]
06

Total Moment of Inertia

Add the moments of inertia from the wooden ball and lead coating: \[ I_{total} = I_{wood} + I_{lead} = 1.344 + 0.2677 = 1.6117 \text{ kg m}^2. \]

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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 is a three-dimensional object with every point on its surface equidistant from its center. Imagine a perfectly round ball where not only the outer surface is solid, but everything within it is also uniformly filled with material. This makes it different from a shell, which is hollow inside. Plenty of examples in real life can be categorized as solid spheres, like a marble or a basketball filled with material.
  • The key property of a solid sphere is uniformity. This means that the material's distribution inside is the same throughout.
  • Such uniformity leads to predictable behaviors, such as the calculation of moment of inertia, which we'll touch upon later.
Understanding the solid sphere concept is crucial when dealing with the physics of rotations and moments of inertia.
Spherical Shell
A spherical shell, unlike a solid sphere, is essentially hollow. It is like the outer layer of an onion with nothing but emptiness inside. Despite this difference, both the solid sphere and spherical shell share similar basic properties, like their round shape.
  • The main difference lies in mass distribution. A spherical shell's mass is concentrated on its surface, resulting in different calculations for physical properties.
  • Its applications include things like domes or thin-walled containers. Understanding this concept is essential when working with rotational dynamics and inertia calculations.
The importance of differentiating between a solid sphere and a spherical shell becomes evident in the calculation of their moments of inertia, which vary due to their mass distributions.
Volume Calculation
Volume calculation is crucial when determining the physical properties of three-dimensional objects like spheres. For spheres, the formula is quite standard: \[ V = \frac{4}{3}\pi r^3 \]This formula calculates how much space a sphere occupies based on its radius, a key measurement from its center to any point on its surface.
  • A larger radius means a larger volume, which affects an object's mass and other properties.
  • The formula emphasizes the cubic relationship of volume to the radius, showing that small changes in radius can create large changes in volume.
Correct volume calculation is vital, especially in physics and engineering, where the volume is foundational to further calculations involving mass or density.
Density
Density is a measure of how much mass is packed into a unit volume of a substance. It's like figuring out how tightly stuffed a suitcase is.
  • Mathematically, it is expressed as \( \rho = \frac{m}{V} \), with \( m \) being mass and \( V \) being volume.
  • A consistent density implies uniform distribution of material, important in predicting an object鈥檚 mechanical properties.
In our context, knowing the density allows us to compute the mass from the given volume, which is essential in evaluating the moment of inertia. The concept of density pops up in various scientific fields, signifying the importance of material consistency.
Mass Calculation
Calculating mass involves understanding both the object's volume and its density鈥攖wo sides of the same coin. Essentially, the mass is determined by multiplying an object's density by its volume:\[ m = \rho V \]Each sphere, regardless of it being solid or a shell, will have mass depending on its specific attributes.
  • For a solid sphere, this means combining its entire volume with its uniform density.
  • For a spherical shell, focus only on the mass distributed across its surface, like a coat of paint.
Effective mass calculation is crucial in physics, influencing the calculation of other physical properties such as inertia and gravitational force.

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Most popular questions from this chapter

A bucket of mass \(m\) is tied to a massless cable that is wrapped around the outer rim of a frictionless uniform pulley of radius \(R\) , similar to the system shown in Fig. \(9.32 .\) In terms of the stated variables, what must be the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

In. A vacuum cleaner belt is looped over a shaft of radius 0.45 \(\mathrm{cm}\) and a wheel of radius 2.00 \(\mathrm{cm} .\) The arrangement of the bell, shaft, and wheel is similar to that of the chain and sprockets in Fig. 9.14 . The motor turns the shaft at 60.0 rev/s and the moving belt turns the wheel, which in turn is connected by another shaft to the roller that beats the dirt out of the rug being vacuumed. Assume that the belt doesn't ship on either the shaft or the wheel. (a) What is the speed of a point on the belt? (b) What is the angular velocity of the wheel, in rad/s?

Engineers are designing a system by which a falling mass \(m\) imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum (Fig. 9.34\()\) . There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 \(\mathrm{m} / \mathrm{s}^{2} .\) In the earth tests, when \(m\) is set to 15.0 \(\mathrm{kg}\) and allowed to fall through \(5.00 \mathrm{m},\) it gives 250.0 \(\mathrm{J}\) of kinetic energy to the drum. (a) If the system is operated on Mars, through what distance would the \(15.0-0\) mass have to fall to give the same amount of kinetic energy to the drum? (b) How fast would the 15.0 \(\mathrm{kg}\) mass be moving on Mars just as the drum gained 250.0 \(\mathrm{J}\) of kinetic energy?

A flywheel with a radius of 0.300 \(\mathrm{m}\) starts from rest and accelerates with a constant angular acceleration of 0.600 \(\mathrm{rad} / \mathrm{s}^{2}\) . Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim (a) at the start; (b) after it has turned through \(60.0^{\circ} ;\) (c) after it has turned through \(120.0^{\circ} .\)
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