Question

The radius of gyration of a solid sphere of mass M and radius R about its diameter is K. The radius of gyration of a uniform circular disc of mass 2M and radius $\frac{R}{2}$ about its diameter is}

• A. $\frac{4K}{\sqrt{5}}$
• B. $\frac{K\sqrt{5}}{4}$
• C. $\frac{K\sqrt{10}}{8}$
• D. $\frac{8K}{\sqrt{10}}$ \bigskip

Hint:

Radius of gyration converts mass distribution into equivalent point mass distance.

Answer 1

The Correct Option is B

Concept: Radius of gyration is defined by: \[ I = Mk^2 \]

Step 1: For solid sphere about diameter.
\[ I_s = \frac{2}{5}MR^2 = MK^2 \Rightarrow K^2 = \frac{2}{5}R^2 \]

Step 2: For disc about diameter.
Moment of inertia of disc: \[ I_d = \frac{1}{4}MR^2 \] Here: Mass = $2M$, radius = $R/2$ \[ I = \frac{1}{4}(2M)\left(\frac{R}{2}\right)^2 \] \[ I = \frac{2M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{8} \]

Step 3: Find radius of gyration.
\[ 2M k^2 = \frac{MR^2}{8} \] \[ k^2 = \frac{R^2}{16} \Rightarrow k = \frac{R}{4} \]

Step 4: Express in terms of K.
\[ K = R\sqrt{\frac{2}{5}} \Rightarrow R = K\sqrt{\frac{5}{2}} \] \[ k = \frac{K}{4}\sqrt{\frac{5}{2}} = \frac{K\sqrt{5}}{4} \] Final Answer: \[ (B)\ \frac{K\sqrt{5}}{4} \]