Question

The radius of gyration of a solid hemisphere of mass $M$ and radius $R$ about an axis parallel to the diameter at a distance $\frac{3}{4}R$ from this plane is given by (centre of mass of the hemisphere lies at a height $\frac{3R}{8}$ from the base):

• A. $\frac{3R}{\sqrt{10}}$
• B. $\frac{5R}{4}$
• C. $\frac{5R}{5}$
• D. $\sqrt{\frac{2}{5}}R$

Answer 1

The correct option is A $\frac{3R}{\sqrt{10}}$

Since the center of mass of a solid hemisphere is 3/8 R above the circular base; so reducing it to a one particle system, and the moment of inertia of that one particle system is$M(\frac{3}{8R}{)}^2=M\frac{9}{64}\times R^2$
Now moment of inertia of the particle about an axis parallel to the original one but at a distance of 3/4 R can be obtained using the parallel axis theorem
new M.I = $M{R}^2\frac{45}{64}$
so the radius of gyration =$R(\frac{45}{64}{)}^{\frac{1}{2}}$