What is the moment of inertia of a cylinder of diameter $D$ and length $L$ about an axis passing through its center of gravity and perpendicular to its length?

  • A
    $M \left[ \frac{D^2}{4} + \frac{L^2}{12} \right]$
  • B
    $M \left[ \frac{D^2}{16} + \frac{L^2}{12} \right]$
  • C
    $M \left[ \frac{D^2}{8} + \frac{L^2}{16} \right]$
  • D
    $M \left[ \frac{D^2}{4} + \frac{L^2}{6} \right]$

Explore More

Similar Questions

Four particles each of mass $M$ are placed at the corners of a square of side $L$. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

$5$ particles,each of mass $2 \ kg$,are attached to the rim of a circular disc of radius $0.1 \ m$ and negligible mass. The moment of inertia of this system about an axis passing through the center and perpendicular to its plane is ........... $kg \cdot m^{2}$.

Difficult
View Solution

The radius of gyration of a uniform rod of length $l$ about an axis passing through one of its ends and perpendicular to its length is

The analogue of mass in rotational motion is:

Four point masses,each of value $m$,are placed at the corners of a square $ABCD$ of side $l$. The moment of inertia of this system about an axis passing through $A$ and parallel to $BD$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo