Q. The moment of inertia of a uniform horizontal cylinder of mass $M$ about an axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius $R$ is
A) $\frac{39}{4}MR^{2}$
B) $\frac{39}{4}MR^{\ }$
C) $\frac{49}{4}MR^{\ }$
D) $\frac{49}{4}MR^{2}$
Explanation
(d) Moment of inertia of cylinder about an axis through the centre and perpendicular to its axis is
$I_{c} = M\left( \frac{R^{2}}{4} + \frac{L^{2}}{12} \right)$
Using theorem of parallel axes, moment of inertia of the cylinder about an axis through its edge would be
$I = I_{c} + M\left( \frac{L}{2} \right)^{2} = M\left( \frac{R^{2}}{4} + \frac{L^{2}}{12} + \frac{L^{2}}{4} \right)$
$= M\left( \frac{R^{2}}{4} + \frac{L^{2}}{3} \right)$
When $L = 6\ R,\ I_{h} = \frac{49}{4}MR^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.