Question

Mass of thin long metal rod is $2$ kg and its moment of inertia about an axis perpendicular to the length of the rod and passing through its one end is $0.5kg{m}^2$. Its radius of gyration is:

• A. $25$ cm
• B. $40$ cm
• C. $50$ cm
• D. $1$ m

Answer 1

The correct option is A $25$ cm

Solution we know that,

$I_{end}=\frac{M{L}^2}{3}=\frac{1}{2}$ [given]

$I_{com}=\frac{M{L}^2}{12}=\frac{1}{4}\left(\frac{M{L}^2}{3}\right)=\frac{1}{8}$

Also,

radius of gyration $=\sqrt{\frac{I_{com}}{m}}$

$=\sqrt{\frac{1}{8\times 2}}=\frac{1}{\sqrt{16}}=25cm$