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Chapter 5: Problem 577

From a circular disc of radius \(\mathrm{R}\) and mass \(9 \mathrm{M}\), a small disc of radius \(\mathrm{R} / 3\) is removed from the disc. The moment of inertia of the remaining portion about an axis perpendicular to the plane of the disc and passing through \(\mathrm{O}\) is.... \(\\{\mathrm{A}\\} 4 \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(40 / 9) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\} 10 \mathrm{MR}^{2}\) \(\\{\mathrm{D}\\}(37 / 9) \mathrm{MR}^{2}\)

Short Answer

Expert verified
The short answer for the moment of inertia of the remaining portion of the circular disc is: \(\frac{161}{18} MR^2.\)

Step by step solution

01

Find the moment of inertia of the bigger disc

The formula for finding the moment of inertia of a uniform disc about an axis perpendicular to its plane and passing through its center is given by \(I = \frac{1}{2}MR^2\). Let's find the moment of inertia of the bigger disc. Given: mass of the bigger disc = 9M, radius = R \(I_{bigger} = \frac{1}{2}(9M)(R^2) = \frac{9}{2}MR^2\)
02

Find the moment of inertia of the smaller disc

Now, we will find the moment of inertia of the smaller disc using the same formula we used in Step 1. Given: mass of the smaller disc = M (since it's one-ninth the mass of the bigger disc), radius = R/3 \(I_{smaller} = \frac{1}{2}(M)(\frac{R}{3})^2 = \frac{1}{2} M (\frac{R^2}{9}) = \frac{1}{18} MR^2\)
03

Calculate the moment of inertia of the remaining portion

To find the moment of inertia of the remaining portion, we'll subtract the moment of inertia of the smaller disc from that of the bigger disc in the following manner: \(I_{remaining} = I_{bigger} - I_{smaller} = \frac{9}{2}MR^2 - \frac{1}{18}MR^2\) Now, we will simplify the expression: \(I_{remaining} = \frac{9}{2} MR^2 - \frac{1}{18} MR^2\\ = \frac{162}{18} MR^2 - \frac{1}{18} MR^2\\ = \frac{161}{18} MR^2\)
04

Compare the result with the options

Now let's compare our result with the given options to find the correct answer: \(I_{remaining} = \frac{161}{18} MR^2\) None of the given options (A, B, C, or D) matches the result we found. It seems there might be an error in the options, or the calculation needs to be rechecked.

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Most popular questions from this chapter

Particles of \(1 \mathrm{gm}, 1 \mathrm{gm}, 2 \mathrm{gm}, 2 \mathrm{gm}\) are placed at the corners \(A, B, C, D\), respectively of a square of side $6 \mathrm{~cm}$ as shown in figure. Find the distance of centre of mass of the system from geometrical centre of square. (A) \(1 \mathrm{~cm}\) (B) \(2 \mathrm{~cm}\) (C) \(3 \mathrm{~cm}\) (D) \(4 \mathrm{~cm}\)

Consider a two-particle system with the particles having masses \(\mathrm{M}_{1}\), and \(\mathrm{M}_{2}\). If the first particle is pushed towards the centre of mass through a distance \(d\), by what distance should the second particle be moved so as to keep the centre of mass at the same position? $\\{\mathrm{A}\\}\left[\left(\mathrm{M}_{1} \mathrm{~d}\right) /\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)\right]$ $\\{\mathrm{B}\\}\left[\left(\mathrm{M}_{2} \mathrm{~d}\right) /\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)\right]$ $\\{\mathrm{C}\\}\left[\left(\mathrm{M}_{1} \mathrm{~d}\right) /\left(\mathrm{M}_{2}\right)\right]$ $\\{\mathrm{D}\\}\left[\left(\mathrm{M}_{2} \mathrm{~d}\right) /\left(\mathrm{M}_{1}\right)\right]$

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

Statement \(-1\) - Friction is necessary for a body to roll on surface. Statement \(-2\) - Friction provides the necessary tangential force and torque. \(\\{\mathrm{A}\\}\) Statement \(-1\) is correct (true), Statement \(-2\) is true and Statement- 2 is correct explanation for Statement \(-1\) \(\\{B\\}\) 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 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 \(\\{\mathrm{A}\\} 6\) \(\\{\mathrm{B}\\} \overline{(1 / 4)}\) \(\\{C\\} 12\) \(\\{\mathrm{D}\\}(1 / 2)\)
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