Question

Find moment of inertia of a solid hemisphere of mass M shown in figure, about an axis AA' passing through its centre of mass.

• A. $\frac{19}{320}MR^2$
• B. $\frac{21}{320}MR^2$
• C. $\frac{83}{320}MR^2$
• D. $\frac{1}{20}MR^2$

Answer 1

Answer: (C)

(C)

Answer 2

1. Moment of Inertia of Solid Hemisphere about its Base:

   The moment of inertia of a solid sphere of mass $M_{sphere}$ and radius R about its diameter is given by $I_{sphere} = \frac{2}{5} M_{sphere} R^2$.

   A solid hemisphere of mass M can be considered as half of a solid sphere of mass $2M$.

   Therefore, the moment of inertia of the solid hemisphere about an axis passing through the center of its flat base and perpendicular to the base (let's call this $I_0$) is:

   $I_0 = \frac{1}{2} \times \left( \frac{2}{5} (2M)R^2 \right) = \frac{2}{5}MR^2$

2. Center of Mass of Solid Hemisphere:

   The center of mass (CM) of a solid hemisphere of radius R is located at a distance $h = \frac{3R}{8}$ from its flat base, along the axis of symmetry. The axis AA' shown in the figure passes through this center of mass and is parallel to the axis $I_0$.

3. Applying Parallel Axis Theorem:

   The Parallel Axis Theorem states that $I = I_{CM} + Md^2$, where I is the moment of inertia about an axis, $I_{CM}$ is the moment of inertia about a parallel axis passing through the center of mass, M is the mass, and d is the perpendicular distance between the two parallel axes.

   In our case, $I = I_0 = \frac{2}{5}MR^2$, and the distance $d$ between the axis through the base and the axis through the center of mass (AA') is $h = \frac{3R}{8}$.

   So, we have:

   $I_0 = I_{CM} + M\left(\frac{3R}{8}\right)^2$

   $\frac{2}{5}MR^2 = I_{CM} + M\left(\frac{9R^2}{64}\right)$

4. Calculating $I_{CM}$:

   Now, solve for $I_{CM}$:

   $I_{CM} = \frac{2}{5}MR^2 - \frac{9}{64}MR^2$

   To subtract the fractions, find a common denominator, which is $5 \times 64 = 320$:

   $I_{CM} = \left(\frac{2 \times 64}{5 \times 64} - \frac{9 \times 5}{64 \times 5}\right)MR^2$

   $I_{CM} = \left(\frac{128}{320} - \frac{45}{320}\right)MR^2$

   $I_{CM} = \left(\frac{128 - 45}{320}\right)MR^2$

   $I_{CM} = \frac{83}{320}MR^2$

The moment of inertia of the solid hemisphere about an axis AA' passing through its center of mass is $\frac{83}{320}MR^2$.