Question

A long cylinderical rod is welded to a thin circular disc of diameter $0.5\, m$ at a point on its circumference. The rod is in the same plane as that of the disc and forms a tangent to the disc. The radius of gyration of the disc about the rod (in $m$ ) is

• A. $\frac{1}{4}$
• B. $\sqrt{\frac{5}{8}}$
• C. $\frac{1}{2}$
• D. $2 \sqrt{2}$

Answer 1

Answer: (B)

As we know, moment of inertia of disc tangential to the rim and parallel to diameter,
$I=\frac{5}{4} M R^{2}$
Now, radius of gyration,
$k=\sqrt{\frac{I}{M}}=\sqrt{\frac{\frac{5}{4} M R^{2}}{M}}=\sqrt{\frac{5}{4}} R$
As, given diameter of disc, $R=0.5\, m$
so, $k=\sqrt{\frac{5}{4}} \times 0.5$
$\Rightarrow k=\sqrt{\frac{5}{8}}$