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Chapter 7: Problem 26

Culculate moment of inertia of a uniform solid cone of mass \(m\), base radius \(R\) and height \(H\) about its own axis.

Short Answer

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

Step by step solution

01

Understand the Moment of Inertia

The moment of inertia of an object is a measure of how resistant it is to rotational motion about an axis. For a solid object, it is often calculated as an integral that adds up contributions from each small mass element of the object.
02

Set Up the Problem

Consider a uniform solid cone of mass \(m\), base radius \(R\), and height \(H\). We want to find the moment of inertia \(I\) about its own symmetry axis, which runs from the apex to the center of the base.
03

Express Mass Element in Differential Form

For a cone, use cylindrical coordinates. Consider a thin disk of thickness \(dz\) at a height \(z\) from the apex. The volume of this disk is \(\pi r^2 dz\), where \(r\) is the radius of the disk at height \(z\).
04

Relate Disk Radius to Height

The radius of the disk \(r\) at any height \(z\) is given by the ratio \(\frac{R}{H}z\), because the cone narrows linearly from base to apex.
05

Elemental Moment of Inertia

The moment of inertia of a thin disk about its central axis is \(\frac{1}{2}dm \, r^2\). Substitute \(dm = \rho \pi r^2 dz\), where \(\rho = \frac{m}{\frac{1}{3}\pi R^2 H}\) is the mass density of the cone.
06

Substitute and Simplify

Substitute \(r = \frac{R}{H}z\) and \(dm\) into the expression for the elemental moment of inertia, \(dI = \frac{1}{2} \rho \pi \left(\frac{R}{H}z\right)^4 dz\). Integrate this expression from \(z = 0\) to \(z = H\).
07

Integration

Recalculate the integral:\[ I = \int_0^H \frac{1}{2} \rho \pi \left(\frac{R}{H}z\right)^4 dz = \frac{1}{2} \rho \pi \left(\frac{R}{H}\right)^4 \int_0^H z^4 \, dz. \]Calculate the definite integral of \(z^4\) from 0 to \(H\).
08

Calculate the Definite Integral

Perform the integration:\[ \int_0^H z^4 \, dz = \left[ \frac{z^5}{5} \right]_0^H = \frac{H^5}{5}. \]
09

Finalize the Expression for \(I\)

Substitute the integral result back into the expression for \(I\):\[ I = \frac{1}{2} \frac{m}{\frac{1}{3} \pi R^2 H} \pi \left(\frac{R}{H}\right)^4 \frac{H^5}{5}. \] Simplify to get:\[ I = \frac{3}{10}mR^2. \]

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Key Concepts

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

Rotational Motion
When we think about objects in motion, often we consider linear movement 鈥� objects moving straight in one direction. However, when it comes to objects spinning around an axis, we delve into the world of rotational motion. This concept is crucial for understanding how objects like wheels, planets, and even a solid cone spin.
  • Rotational motion involves an object rotating around its own axis, similar to how Earth rotates daily.
  • The resistance an object offers to change in its rotational motion is known as the moment of inertia.
In this exercise, we consider the rotational motion of a solid cone around its central axis, from the apex (top point) to the base's center. The moment of inertia gives us an idea of how much force is needed to change its rotational speed.
Integral Calculus
Integral calculus is a powerful mathematical tool that helps us calculate quantities when simple multiplication doesn't suffice. It allows us to add up small parts to find a whole. In the context of rotational motion, we use integrals to calculate the moment of inertia for complex shapes like a cone.
  • Calculating moment of inertia involves summing up contributions from tiny mass elements across the entire object.
  • Mathematically, this is expressed through an integral, making it possible to account for objects of non-uniform shape.
In the solution given, integral calculus enables us to integrate over the height of the cone to find its overall moment of inertia based on the contributions from each thin disk slice of the cone.
Mass Density
Mass density is a measure that describes how much mass is packed into a given volume of an object. For a solid object like a cone, uniform mass density means the mass is evenly distributed throughout the object.
  • In this problem, it鈥檚 crucial to know the mass density to express the mass of an infinitely small part of the cone, dm, during the integration.
  • The mass density, denoted \( \rho \), can be calculated using the total mass and volume of the cone: \( \rho = \frac{m}{\frac{1}{3} \pi R^2 H} \).
By expressing the mass density in terms of the cone's dimensions, we can determine how each slice of the cone contributes to the moment of inertia.
Cylindrical Coordinates
Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height component. They are invaluable when dealing with symmetrical problems involving cylinders or cones.
  • This coordinate system simplifies the problem of calculating moments of inertia for objects like cones by focusing on radius, angle, and height.
  • We express the position and size of small mass elements in terms of these coordinates, leading to more straightforward calculations.
Using cylindrical coordinates, we describe a disk at height \( z \) within the cone with a radius determined as \( r = \frac{R}{H}z \), aiding in setting up the integral for each disk's contributions to moment of inertia.

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Most popular questions from this chapter

A particle of mass \(m\) is projected from the ground with an initial speed \(u\) at an angle \(\alpha\). Find the magnitude of its angular momentum at the highest point of its trajectory about the point of projection.

A rod of uniform cross-scction of mass \(M\) and lenghth \(L\) is hinged about an end to swing frecly in a vertical planc. However, its density is non uniform and varies lincarly from hinged end to the frec cnd doubling its value. Find the moment of incrtia of the rod, about the rotation axis passing through the hinge point.

A rigid body of mass \(M\) and radius \(R\) rolling without slipping on an inclined planc. The magnitud of Corce of friction Column-I Column-II (a) For ring (p) \(\frac{M g \sin \theta}{2.5}\) (b) For solid sphere (q) \(\frac{M g \sin \theta}{3}\) (c) For solid eylinder (r) \(\frac{M g \sin \theta}{3.5}\) (d) For hollow sphere (s) \(\frac{M g \sin \theta}{2}\)

A particle of mass \(m\) is projected from origin \(O\) with spced \(u\) at an angle \(\theta\) with positive \(x\) -axis. Positive \(y\) -axis is in vertically upward dircction. Find the angular momentum of particle at any time \(t\) about \(O\) before the particle strikes the ground again.

A uniform thin rod of mass \(m\) and lenght \(l\) is standing on a smooth horizontal surface. \(\Lambda\) slight disturbance causes the lower end to slip on the smooth surface and the rod starts falling. Find the velocity of centre of mass of the rod at the instant when it makes an angle \(\theta\) with horizontal.
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