Question

The moment of inertia of a solid cylinder of diameter $D$ and length $L$ about an axis perpendicular to its length and passing through the centre of gravity will be

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

Answer 1

The correct option is D $M[\frac{D^2}{16}+\frac{L^2}{12}]$


Given,
$R=\frac{D}{2}$
The moment of inertia of a solid cylinder
about an axis perpendicular to its length and passing through the centre of gravity,
$I=M(\frac{L^2}{12}+\frac{R^2}{4})$
According to given condition,
$I=M[\frac{L^2}{12}+\frac{{\left(\frac{D}{2}\right)}^2}{4}]$
$I=M[\frac{L^2}{12}+\frac{D^2}{16}]$
Final Answer: $\left(b\right)$