Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^{2}$ . The value of $x$ is $\_\_\_\_$ .

Question

Suppose there is a uniform circular disc of mass

M \mathrm{~kg}

and radius

r \mathrm{~m}

shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis

A

of the disc is given by

\frac{x}{256} M r^{2}

. The value of

x

is

\_\_\_\_

.

Quantrex Academy · Accepted Answer

Correct answer: 0

The moment of inertia of the complete disc about axis A is

I_{disc} = \frac{1}{2}Mr^2 = \frac{128Mr^2}{256}

.
Each removed circular region has radius

r/4

and mass

m_i = M/(r^2/16r^2) = M/16

.
For each removed disc at distance

3r/4

from axis A, using the parallel axis theorem:

I_{removed} = \frac{1}{2}m_i(r/4)^2 + m_i(3r/4)^2 = \frac{Mr^2}{512} + \frac{9Mr^2}{256} = \frac{19Mr^2}{512}

.
For two removed discs:

I_{2removed} = \frac{38Mr^2}{512} = \frac{19Mr^2}{256}

.
Therefore:

I_{remainder} = \frac{128Mr^2}{256} - \frac{19Mr^2}{256} = \frac{109Mr^2}{256}

, so

x = 109

.

Suppose there is a uniform circular disc of mass M ~kg and radius r...