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Chapter 5: Problem 661

The moment of inertia of a hollow sphere of mass \(\mathrm{M}\) and inner and outer radii \(\mathrm{R}\) and \(2 \mathrm{R}\) about the axis passing through its centre and perpendicular to its plane is \(\\{\mathrm{A}\\}(3 / 2) \mathrm{MR}^{2}\) \\{B \(\\}(13 / 32) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(31 / 35) \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(62 / 35) \mathrm{MR}^{2}\)

Short Answer

Expert verified
The moment of inertia of a hollow sphere with mass M and inner radius R and outer radius 2R about an axis passing through its center and perpendicular to its plane is \(\frac{14}{15} MR^2\).

Step by step solution

01

Find the moment of inertia of both spheres if they were solid

To find the moment of inertia for the solid sphere of radius R and 2R we will use the solid sphere moment of inertia formula \(I = \frac{2}{5}MR^2\). 1. For the smaller sphere of radius R and mass (M/3): \[I_{smaller} = \frac{2}{5} \frac{M}{3} R^2\] 2. For the larger sphere of radius 2R and mass (2M/3): \[I_{larger} = \frac{2}{5}\frac{2M}{3}(2R)^2\]
02

Find the moment of inertia for the hollow sphere

Now that we have the moments of inertia for both solid spheres, we can subtract the moment of inertia of the smaller solid sphere from that of the larger solid sphere: \[I_{hollow} = I_{larger} - I_{smaller}\] \[I_{hollow} = \frac{2}{5}\frac{2M}{3}(2R)^2 - \frac{2}{5} \frac{M}{3} R^2\]
03

Simplify the equation

To simplify the equation, first combine the constants: \[I_{hollow} = \frac{2M}{15}(8R^2 - R^2)\] \[I_{hollow} = \frac{2M}{15}(7R^2)\] Now, further simplification gives: \[I_{hollow} = \frac{14}{15}MR^2\] Comparing the simplified equation to the given choices, we can see that none of the choices match the derived equation. This implies that the provided choices are incorrect.

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Most popular questions from this chapter

A solid cylinder of mass \(\mathrm{M}\) and \(\mathrm{R}\) is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass \(\mathrm{m}\) is hung from the string as shown in figure the mass is released from rest then The angular speed of cylinder is proportional to \(\mathrm{h}^{\mathrm{n}}\), where \(\mathrm{h}\) is the height through which mass falls, Then the value of \(n\) is \(\\{\mathrm{A}\\}\) zero \(\\{\mathrm{B}\\} 1\) \(\\{\mathrm{C}\\}(1 / 2)\) \([\mathrm{D}] 2\)

A binary star consist of two stars \(\mathrm{A}(2.2 \mathrm{Ms})\) and \(\mathrm{B}\) (mass \(11 \mathrm{Ms}\) ) where \(\mathrm{Ms}\) is the mass of sun. They are separated by distance \(\mathrm{d}\) and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of \(\operatorname{star} B\). about the centre of mass is \(\\{\mathrm{A}\\} 6\) \(\\{\mathrm{B}\\} \overline{(1 / 4)}\) \(\\{C\\} 12\) \(\\{\mathrm{D}\\}(1 / 2)\)

The angular momentum of a wheel changes from \(2 \mathrm{~L}\) to $5 \mathrm{~L}$ in 3 seconds what is the magnitudes of torque acting on it? \(\\{\mathrm{A}\\} \mathrm{L}\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} \mathrm{L} / 3\) \(\\{\mathrm{D}\\} \mathrm{L} / 5\)

According to the theorem of parallel axis \(\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^{2}\) the graph between \(\mathrm{I} \rightarrow \mathrm{d}\) will be

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.... \(\\{B\\}\left[2 m g h /\left(I+m r^{2}\right)\right]\) $\\{\mathrm{C}\\}\left[2 \mathrm{mgh} /\left(\mathrm{I}+\mathrm{mr}^{2}\right)\right]^{1 / 2}$ \(\\{\mathrm{D}\\} \sqrt{(2 \mathrm{gh})}\)
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