/*! 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 100 Calculate the moment of inertia ... [FREE SOLUTION] | 91Ó°ÊÓ

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Calculate the moment of inertia of a uniform solid cone about an axis through its center (Fig. P9.100). The cone has mass \(M\) and altitude \(h .\) The radius of its circular base is \(R .\)

Short Answer

Expert verified
The moment of inertia of the cone about its central axis is \( \frac{3MR^2}{10} \).

Step by step solution

01

Define the Problem

We need to calculate the moment of inertia of a uniform solid cone. The axis of rotation is through the center of the cone and perpendicular to its base. The cone has mass \( M \), height \( h \), and base radius \( R \).
02

Express Volume Element in Cylindrical Coordinates

Consider a thin disk of thickness \( dz \) at a height \( z \) from the base of the cone. The radius of the disk, \( r(z) \), is a function of \( z \) given by \( r(z) = \frac{R}{h} z \) due to the linear taper of the cone.
03

Calculate Volume Element

The volume of the thin disk \( dV \) is the area of the disk times its thickness: \[ dV = \pi \left( \frac{R}{h} z \right)^2 dz. \]
04

Express Mass Element

The mass element \( dm \) associated with the disk is: \[ dm = \rho dV = \rho \pi \left( \frac{R}{h} z \right)^2 dz, \] where \( \rho \) is the density of the cone. \( \rho \) can be calculated as \( \frac{M}{V} \), with \( V \) the total volume \( V = \frac{1}{3} \pi R^2 h \).
05

Set Up Moment of Inertia Integral

The moment of inertia \( dI \) of each disk about the axis is given by \( dI = \frac{1}{2} r(z)^2 dm \). Therefore, the total moment of inertia \( I \) is the integral over the height of the cone: \[ I = \int_0^h \frac{1}{2} \left( \frac{R}{h} z \right)^2 dm. \]
06

Integrate to Find Total Moment of Inertia

Substitute \( dm \) from Step 4 and simplify: \[ I = \frac{3M}{2h^3} \int_0^h z^2 \left( \frac{R}{h} z \right)^4 dz = \frac{3M R^2}{2h^5} \int_0^h z^4 dz. \] Evaluate the integral \( \int_0^h z^4 dz = \frac{h^5}{5} \): \[ I = \frac{3M R^2}{10}. \]

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Key Concepts

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

Solid Cone
A solid cone is a three-dimensional geometric shape that tapers smoothly from a flat, circular base to a single point called the apex. Think of an ice cream cone or a funnel for a real-life example.

In geometry, a solid cone is defined by its height, radius, and the slant height. The total volume of a cone can be calculated using the formula: \[ V = \frac{1}{3} \pi R^2 h \] where \( R \) is the base radius, and \( h \) is the height from base to apex.

When studying the moment of inertia, knowing the shape and dimensions helps us understand how mass is distributed in the cone, crucial for calculations in rotational dynamics.
Cylindrical Coordinates
Cylindrical coordinates are a three-dimensional extension of two-dimensional polar coordinates. They are especially useful in dealing with objects having cylindrical symmetry, like our cone.

In cylindrical coordinates, the position of a point is determined by:
  • \( r \): the radial distance from the origin.
  • \( \theta \): the angular position around the z-axis.
  • \( z \): the height along the z-axis.
To express the cone, at a given height \( z \), a small disk can be sliced with radius \( r(z) = \frac{R}{h} z \). This changes linearly with \( z \), tapering as it reaches the apex.

Using cylindrical coordinates helps simplify integration over curved surfaces and complex shapes, making it handy in tasks such as finding the moment of inertia.
Rotational Dynamics
Rotational dynamics is the branch of physics that deals with the motion of objects when they rotate. This field examines how forces cause rotational motion and how mass distribution affects this motion.

The moment of inertia plays a key role because it measures how difficult it is to change an object’s rotational motion. Imagine trying to spin an object around a central axis. The moment of inertia tells us how much torque (rotational force) is needed.

For our cone, we calculate the moment of inertia to understand how it would behave under rotational motion. It's like understanding how a topspin or spin of a wheel works. The formula for the moment of inertia for an element is: \[ dI = \frac{1}{2} r(z)^2 dm \] where \( dm \) is the mass of the element being rotated.
Mass Distribution
Mass distribution refers to how mass is spread across a body and affects the object's structural and mechanical behavior. In a solid cone, mass distribution is not uniform across all dimensions due to its shape.

To calculate quantities like the moment of inertia, we consider small elements of mass (like slices or disks) within the cone, calculate their contribution, and integrate across the entire shape.

In the case of this exercise, we utilized the disk method where each small disk has a mass element: \[ dm = \rho \pi \left( \frac{R}{h} z \right)^2 dz \] with \( \rho \) representing the density. This approach ensures we accurately account for the tapered shape, vital for realistic modeling of rotational behaviors in physical systems. Understanding mass distribution allows engineers to design stable structures and machines efficiently.

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Most popular questions from this chapter

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.

The Kinetic Energy of Walking. If a person of mass \(M\) simply moved forward with speed \(V\), his kinetic energy would be \(\frac{1}{2} M V^{2}\).However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 13\(\%\) of a person's mass, while the legs and feet together account for 37\(\%\) . For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about \(\pm 30^{\circ}\) (a total of \(60^{\circ} )\) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a 75 -kg person walking at 5.0 \(\mathrm{km} / \mathrm{h}\) , having arms 70 \(\mathrm{cm}\) long and legs 90 \(\mathrm{cm}\) long. (a) What is the average angular velocity of his arms and legs? (b) Using the average angular velocity from part (a), calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. (c) What is the total kinetic energy due to both his forward motion and his rotation? (d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

A passenger bus in Zurich, Switzerland, derived its motive power from the energy stored in a large flywheel. The wheel was brought up to speed periodically, when the bus stopped at a station, by an electric motor, which could then be attached to the electric power lines. The flywheel was a solid cylinder with mass 1000 \(\mathrm{kg}\) and diameter 1.80 \(\mathrm{m}\) ; its top angular speed was 3000 rev/min. (a) At this angular speed, what is the kinetic energy of the flywheel? (b) If the average power required to operate the bus is \(1.86 \times 10^{4} \mathrm{W},\) how long could it operate between stops?

A hollow spherical shell has mass 8.20 \(\mathrm{kg}\) and radius 0.220 \(\mathrm{m} .\) It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.890 \(\mathrm{rad} / \mathrm{s}^{2} .\) What is the kinetic energy of the shell after it has turned through 6.00 \(\mathrm{rev} ?\)

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