Question

Suppose there is a uniform circular disc of mass M kg and radius r 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 $\frac{x{256} Mr^2$. The value of x is ___.

Hint:

Moment of Inertia is additive. For objects with holes, treat holes as negative mass: $I_{rem} = I_{whole} - I_{hole}$.

Answer 1

Correct Answer: 236

Let the original disc have mass $M$ and radius $r$. Area density $\sigma = \frac{M}{\pi r^2}$.
The cut-out regions appear to be two circles of radius $a = r/4$. Their centers are at distance $d = 3r/4$ from the axis.
Mass of one cut-out circle $m = \sigma \pi a^2 = \frac{M}{\pi r^2} \pi (\frac{r}{4})^2 = \frac{M}{16}$.
Moment of inertia of one cut-out about its own center $I_{cm} = \frac{1}{2} m a^2 = \frac{1}{2} (\frac{M}{16}) (\frac{r}{4})^2 = \frac{M r^2}{512}$.
Using Parallel Axis Theorem, MOI of one cut-out about the main axis $A$:
$I_{hole} = I_{cm} + m d^2 = \frac{M r^2}{512} + \frac{M}{16} (\frac{3r}{4})^2$.
$I_{hole} = \frac{M r^2}{512} + \frac{9 M r^2}{256} = \frac{M r^2 + 18 M r^2}{512} = \frac{19 M r^2}{512}$.
Total MOI removed for 2 holes = $2 \times \frac{19 M r^2}{512} = \frac{19 M r^2}{256}$.
MOI of original disc $I_{total} = \frac{1}{2} M r^2 = \frac{128 M r^2}{256}$.
MOI of remainder = $I_{total} - 2 I_{hole} = \frac{128 - 19}{256} M r^2 = \frac{109}{256} M r^2$.
Thus, $x = 109$.