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Chapter 5: Problem 665

What is the moment of inertia of a solid sphere of density \(\rho\) and radius \(\mathrm{R}\) about its diameter? \(\\{\mathrm{A}\\}(105 / 176) \mathrm{R}^{5} \rho\) \(\\{\mathrm{B}\\}(176 / 105) \mathrm{R}^{5} \rho\) \(\\{C\\}(105 / 176) R^{2} \rho\) \(\\{\mathrm{D}\\}(176 / 105) \mathrm{R}^{2} \rho\)

Short Answer

Expert verified
The correct answer for the moment of inertia of a solid sphere of density \(\rho\) and radius R about its diameter is \(\frac{8\pi}{5} R^5 \rho\), which is not present in the given options.

Step by step solution

01

Moment of Inertia Formula

Moment of inertia (I) is found by integrating the square of the distance from the axis of rotation (r) multiplied by the mass (dm) over the entire volume of the object. For a sphere, we can use spherical coordinates to find the following formula for the moment of inertia about the z-axis: \[I = \int_{V} r^2 dm\]
02

Express dm in terms of the density and volume element

We know the density \(\rho\) and we can express mass (dm) in terms of density and volume element (dV): \[dm = \rho dV\]
03

Convert volume element from Cartesian Coordinates to Spherical Coordinates

We have to change the volume element from Cartesian coordinates to spherical coordinates. In spherical coordinates, the volume element is given by: \[dV = r^2 \sin\theta dr d\theta d\phi\]
04

Substitute dm and dV in the Moment of Inertia Formula

Now we can substitute for dm and dV in the moment of inertia formula: \[I = \int_{V} r^2 (\rho r^2 \sin\theta dr d\theta d\phi)\]
05

Integrate over the volume of the sphere

Performing the integration over the volume of the sphere with limits for r: 0 to R, for θ: 0 to π, and for φ: 0 to 2π: \[I = \rho \int_{0}^{\mathrm{R}} \int_{0}^{\pi} \int_{0}^{2 \pi} r^4 \sin\theta dr d\theta d\phi\] \[I = \rho \left[ \int_{0}^{\mathrm{R}} r^4 dr \right] \left[ \int_{0}^{\pi} \sin\theta d\theta \right] \left[ \int_{0}^{2 \pi} d\phi \right]\] After performing these integrations, we get: \[I = \rho \left[ \frac{1}{5} R^5 \right] \left[ -\cos\theta \Big|_{0}^{\pi} \right] \left[ \phi \Big|_{0}^{2\pi} \right] \] \[I = \rho \left[ \frac{1}{5} R^5 \right] [2] [2\pi] \] \[I = \frac{8\pi}{5} R^5 \rho\]
06

Compare the result with the given options

Comparing the result we obtained with the given options, none of the options MATCHES with our moment of inertia calculation result. If this is an exercise, it's possible that there is a minor mistake in the options or the provided statement. However, it's important to note that our calculated correct answer is: \[I = \frac{8\pi}{5} R^5 \rho\] One might argue that the closest option to the answer is option A with the calculated answer having the same power in the radius (\(R^5\)) and the appearance of scalar factors (\(\frac{105}{176}\) in A) resembling the \(\frac{8\pi}{5}\), but there is a mismatch due to the missing \(\pi\) factor.

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Most popular questions from this chapter

A cord is wound round the circumference of wheel of radius r. the axis of the wheel is horizontal and moment of inertia about it is I A weight \(\mathrm{mg}\) is attached to the end of the cord and falls from the rest. After falling through the distance \(\mathrm{h}\). the angular velocity of the wheel will be.... \(\\{B\\}\left[2 m g h /\left(I+m r^{2}\right)\right]\) $\\{\mathrm{C}\\}\left[2 \mathrm{mgh} /\left(\mathrm{I}+\mathrm{mr}^{2}\right)\right]^{1 / 2}$ \(\\{\mathrm{D}\\} \sqrt{(2 \mathrm{gh})}\)

Let I be the moment of inertia of a uniform square plate about an axis \(\mathrm{AB}\) that passes through its centre and is parallel to two of its sides \(\mathrm{CD}\) is a line in the plane of the plate that passes through the centre of the plate and makes an angle of \(\theta\) with \(\mathrm{AB}\). The moment of inertia of the plate about the axis \(\mathrm{CD}\) is then equal to.... \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{B\\} I \sin ^{2} \theta\) \(\\{C\\} I \cos ^{2} \theta\) \(\\{\mathrm{D}\\} I \cos ^{2}(\theta / 2)\)

The M.I. of a body about the given axis is \(1.2 \mathrm{kgm}^{2}\) initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \mathrm{~J}\(. an angular acceleration of \)25 \mathrm{rad} \mathrm{sec}^{2}$ must be applied about that axis for duration of \(\ldots\) \(\\{\mathrm{A}\\} 4 \mathrm{sec}\) \\{B \(\\} 2 \mathrm{sec}\) \(\\{C\\} 8 \mathrm{sec}\) \(\\{\mathrm{D}\\} 10 \mathrm{sec}\)

Match list I with list II and select the correct answer $$ \begin{aligned} &\begin{array}{|l|l|} \hline \text { List-I } & \begin{array}{l} \text { List - II } \\ \text { System } \end{array} & \text { Moment of inertia } \\ \hline \text { (x) A ring about it axis } & \text { (1) }\left(\mathrm{MR}^{2} / 2\right) \\ \hline \text { (y) A uniform circular disc about it axis } & \text { (2) }(2 / 5) \mathrm{MR}^{2} \\ \hline \text { (z) A solid sphere about any diameter } & \text { (3) }(7 / 5) \mathrm{MR}^{2} \\ \hline \text { (w) A solid sphere about any tangent } & \text { (4) } \mathrm{MR}^{2} \\ \cline { 2 } & \text { (5) }(9 / 5) \mathrm{MR}^{2} \\ \hline \end{array}\\\ &\text { Select correct option }\\\ &\begin{array}{|l|l|l|l|l|} \hline \text { Option? } & \mathrm{X} & \mathrm{Y} & \mathrm{Z} & \mathrm{W} \\\ \hline\\{\mathrm{A}\\} & 2 & 1 & 3 & 4 \\ \hline\\{\mathrm{B}\\} & 4 & 3 & 2 & 5 \\ \hline\\{\mathrm{C}\\} & 1 & 5 & 4 & 3 \\ \hline\\{\mathrm{D}\\} & 4 & 1 & 2 & 3 \\ \hline \end{array} \end{aligned} $$

Let \(\mathrm{Er}\) is the rotational kinetic energy and \(\mathrm{L}\) is angular momentum then the graph between \(\log \mathrm{e}^{\mathrm{Er}}\) and $\log \mathrm{e}^{\mathrm{L}}$ can be
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