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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

A wheel having moment of inertia \(2 \mathrm{~kg} \mathrm{M}^{2}\) about its vertical axis, rotates at the rate of 60 rpm about this axis. The torque which can stop the wheels rotation in one minute will be.. \(\\{\mathrm{A}\\}(\pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{B}\\}(\pi / 18) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{C}\\}(2 \pi / 15) \mathrm{N}-\mathrm{m}\) \(\\{\mathrm{D}\\}(\pi / 12) \mathrm{N}-\mathrm{m}\)

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