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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

Statement \(-1\) -If the cylinder rolling with angular speed- w. suddenly breaks up in to two equal halves of the same radius. The angular speed of each piece becomes \(2 \mathrm{w}\). Statement \(-2\) - If no external torque outs, the angular momentum of the system is conserved. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement \(-1\) \(\\{\mathrm{B}\\}\) Statement \(-1\) is true, statement \(-2\) is true but statement- 2 is not the correct explanation four statement \(-1 .\) \(\\{\mathrm{C}\\}\) Statement \(-1\) is true, statement- 2 is false \\{D \(\\}\) Statement- 2 is false, statement \(-2\) is true

A uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The acceleration of the centre of mass of the disc is \(\\{\mathrm{A}\\} \mathrm{g} \sin \theta\) \(\\{B\\}[(2 g \sin \theta) / 3]\) \(\\{\mathrm{C}\\}[(\mathrm{g} \sin \theta) / 3]\) \(\\{\mathrm{D}\\}[(2 \mathrm{~g} \cos \theta) / 3]\)

A wheel is rotating at \(900 \mathrm{rpm}\) about its axis. When power is cut off it comes to rest in 1 minute, the angular retardation in $\mathrm{rad} / \mathrm{sec}$ is \(\\{\mathrm{A}\\}(\pi / 2)\) \(\\{\mathrm{B}\\}(\pi / 4)\) \(\\{\mathrm{C}\\}(\pi / 6)\) \(\\{\mathrm{D}\\}(\pi / 8)\)

Three particles of the same mass lie in the \((\mathrm{X}, \mathrm{Y})\) plane, The \((X, Y)\) coordinates of their positions are \((1,1),(2,2)\) and \((3,3)\) respectively. The \((X, Y)\) coordinates of the centre of mass are \(\\{\mathrm{A}\\}(1,2)\) \(\\{\mathrm{B}\\}(2,2)\) \(\\{\mathrm{C}\\}(1.5,2)\) \(\\{\mathrm{D}\\}(2,1.5)\)

A thin uniform rod \(A B\) of mass \(M\) and length \(L\) is hinged at one end \(\mathrm{A}\) to the horizontal floor. Initially it stands vertically. It is allowed to fall freely on the floor in the vertical plane. The angular velocity of the rod when its end \(B\) strikes the floor is \(\\{\mathrm{A}\\} \sqrt{(\mathrm{g} / \mathrm{L})}\) \(\\{\mathrm{B}\\} \sqrt{(2 \mathrm{~g} / \mathrm{L})}\) \(\\{C\\} \sqrt{(3 g / L)}\) \(\\{\mathrm{D}\\} 2 \sqrt{(\mathrm{g} / \mathrm{L})}\)
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