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Calculate the moment of inertia of a uniform solid cone about an axis through its center (Fig. \(\mathrm{P} 9.90\) ). 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 center axis is \(I = \frac{3MR^2}{10}\).

Step by step solution

01

Understand the problem

The goal is to find the moment of inertia of a uniform cone with mass M, height h, and base radius R. The moment of inertia I about the center axis of the cone is to be calculated. In terms of moments of inertia, we need to use the formula \(I = \int r^2 dm\), where r is the distance to the axis and dm is the mass of an infinitesimal portion of the body.
02

Express dm in terms of radial distance

To perform integration, express the infinitesimal mass element dm in terms of radial elements. Based on the given information, \(dm = 蟻dV\), where 蟻 is the mass density of the cone and dV is the infinitesimal volume. Mass density 蟻 can be related to the total mass M of the cone as \(\frac{M}{\frac{1}{3}蟺R^2h}\). The volume of the thin disc formed at radial distance r from the apex and with thickness dr as \(dV = 蟺r虏dr\). Substitute these relations in the dm expression to get \(dm = \frac{3M}{蟺R虏h}r虏dr\)
03

Substitute dm into the moment of inertia formula

Substitute this value for dm into the moment of inertia formula. Given that the distance from the axis is r/2 (midpoint of the disc) we get \(I = \int_{0}^{h}(\frac{1}{2}r)虏 \frac{3M}{蟺R虏h} r虏dr = \frac{3M}{2蟺R^2h} \int_{0}^{R}\frac{1}{4} r^4 dr\).
04

Perform the integration

Now, perform the integration and simplify the expression. The antiderivative of \(r^4\) with respect to r is \(\frac{1}{5}r^5\). Evaluating the integral from 0 to R gives \(\frac{1}{5}R^5 - 0 = \frac{1}{5}R^5\). This leads to \(I = \frac{3M}{10蟺R^2h}\cdot \frac{1}{4}R^5 = \frac{3MR^2}{10}\).
05

Final result

The final moment of inertia of the cone about its axis through the center is \(I = \frac{3MR^2}{10}\).

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Key Concepts

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

Solid Cone
The concept of a solid cone is essential when dealing with exercises like finding the moment of inertia. A solid cone is a three-dimensional geometric shape that has a circular base tapering smoothly to a point called the apex. It can be visualized as a 'pyramid' with a round foundation. In any physical problem dealing with a cone, key dimensions include:
  • Base Radius (\( R \)): The distance from the center of the circular base to its edge.
  • Height (\( h \)): The distance from the base to the apex of the cone.
  • Mass (\( M \)): A measure of the amount of matter the cone contains.
These dimensions contribute to the overall understanding of the cone鈥檚 geometry, which is crucial for calculating properties like the moment of inertia. Understanding the shape and dimensions of a cone helps in visualizing how its mass is distributed, which directly influences its rotational dynamics.
Mass Distribution
In physics, understanding mass distribution within an object like a solid cone is critical when applying concepts like moment of inertia. Mass distribution refers to how mass is spread across an object. In a uniform solid cone, mass is evenly distributed throughout the volume.

To find mass distribution mathematically, we use the mass density (\( \rho \)), which is the mass per unit volume. For the cone, we calculate this density using the formula:
  • \[\rho = \frac{M}{\frac{1}{3}\pi R^2 h}\]
Here, we consider the total volume of the cone, which arises from its geometry. This density helps express the infinitesimal mass element (\( dm \)) as:
  • \[ dm = \rho dV = \frac{3M}{\pi R^2 h} r^2 dr \]
This relation allows for integration over the cone's volume to eventually find the moment of inertia.
Integration in Physics
Integration often appears in physics to calculate properties that span an entire object, such as mass, area, or, in this case, moment of inertia. For a solid cone, we use integration to account for the mass distributed along its volume.

The goal is to find a total moment of inertia (\( I \)) by integrating over small mass elements (\( dm \)) positioned at different distances (\( r \)) from the axis of rotation. The formula involves:
  • \[ I = \int r^2 dm \]
By substituting the expression for \( dm \) that includes density, we set up an integral that runs over the cone's height or radius. In calculus terms, the antiderivative of these terms helps us find a simplified, total value. Performing these integrals requires understanding limits of integration, which, for the cone, typically span from the base to the apex or similar geometric bounds.
Rotational Dynamics
Rotational dynamics involves studying objects in rotation, and the moment of inertia is a key parameter. It represents how difficult it is to change the rotational speed of an object.

For a solid cone, this inertia depends on its mass, shape, and how that mass is distributed about the axis of rotation. The formula derived earlier,
  • \[ I = \frac{3MR^2}{10} \]
shows that the moment of inertia is derived from these properties. This parameter plays a significant role in equations of motion, determining how much torque is needed to reach a desired angular acceleration.
Understanding this concept helps in designing and analyzing systems involving rotational motion, as the placement of mass can dramatically affect how an object spins. Whether in engineering or physics, grasping the idea of rotational dynamics is essential for analyzing and solving practical problems involving rotating bodies.

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Most popular questions from this chapter

A rotating flywheel has moment of inertia \(12.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for an axis along the axle about which the wheel is rotating. Initially the flywheel has \(30.0 \mathrm{~J}\) of kinetic energy. It is slowing down with an angular acceleration of magnitude \(0.500 \mathrm{rev} / \mathrm{s}^{2} .\) How long does it take for the rotational kinetic energy to become half its initial value, so it is \(15.0 \mathrm{~J} ?\)

\(\begin{array}{llll}\mathrm{In} & \mathrm{a} & \text { charming } & 19 \text { th-century }\end{array}\) hotel, an old-style elevator is connected to a counterweight by a cable that passes over a rotating disk \(2.50 \mathrm{~m}\) in diameter (Fig. E9.18). The elevator is raised and lowered by turning the disk, and the cable does not slip on the rim of the disk but turns with it. (a) At how many rpm must the disk turn to raise the elevator at \(25.0 \mathrm{~cm} / \mathrm{s} ?\) (b) To start the elevator moving, it must be accelerated at \(\frac{1}{8} g .\) What must be the angular acceleration of the disk, in rad/s \(^{2} ?\) (c) Through what angle (in radians and degrees) has the disk turned when it has raised the elevator \(3.25 \mathrm{~m}\) between floors?

The motor of a table saw is rotating at 3450 rev \(/\) min. A pulley attached to the motor shaft drives a second pulley of half the diameter by means of a V-belt. A circular saw blade of diameter \(0.208 \mathrm{~m}\) is mounted on the same rotating shaft as the second pulley. (a) The operator is careless and the blade catches and throws back a small piece of wood. This piece of wood moves with linear speed equal to the tangential speed of the rim of the blade. What is this speed? (b) Calculate the radial acceleration of points on the outer edge of the blade to see why sawdust doesn't stick to its teeth.

An electric turntable \(0.750 \mathrm{~m}\) in diameter is rotating about a fixed axis with an initial angular velocity of \(0.250 \mathrm{rev} / \mathrm{s}\) and a constant angular acceleration of \(0.900 \mathrm{rev} / \mathrm{s}^{2}\). (a) Compute the angular velocity of the turntable after \(0.200 \mathrm{~s}\). (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turntable at \(t=0.200 \mathrm{~s} ?\) (d) What is the magnitude of the resultant acceleration of a point on the rim at \(t=0.200 \mathrm{~s} ?\)

A safety device brings the blade of a power mower from an initial angular speed of \(\omega_{1}\) to rest in 1.00 revolution. At the same constant acceleration, how many revolutions would it take the blade to come to rest from an initial angular speed \(\omega_{3}\) that was three times as great, \(\omega_{3}=3 \omega_{1} ?\)

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