Question

A solid sphere and a ring have equal masses and equal radius of gyration. If the sphere is rotating about its diameter and ring about an axis passing through and perpendicular to its plane, then the ratio of radius is \(\sqrt{\frac{x}{2} }\) then find the value of x.

Answer 1

Correct Answer: 5

\((\frac{2}{5})mR_1^2 = mK_1^2 and   R_2^2 =K_2\)
\(K_1 = \sqrt{(\frac{2}{5})R_1}\)
\(K_2=R_2\) 
\(K_1 = K_2\)
\(\sqrt{(\frac{2}{5})} R_1=R_2\)
\(\frac{R_1}{R_2}= \sqrt{\frac{5}{2}}\)
Therefore, the value of x is 5.

Answer 2

Given radius of gyration is same for ring and solid sphere: \[ K_R = K_{ss} \] For ring and sphere, the radius of gyration is: \[ R_R = \sqrt{\frac{2}{5}} R_{ss} \] Thus, \[ \frac{R_R}{R_{ss}} = \sqrt{\frac{2}{5}} \] Therefore, \(x = 5\).