Question

Consider a uniform horizontal solid cylinder of mass $10\, kg$ such that its length is $9$ times its radius. Let the radius be $40\, cm$. Calculate the moment of inertia of the cylinder about a line passing through its edge and perpendicular to its axis

• A. $213\,kg - m ^{2}$
• B. $18.7\, kg - m ^{2}$
• C. $23.6\, kg - m ^{2}$
• D. $10.9\, kg - m ^{2}$

Answer 1

Answer: (D)

Given, mass of solid cylinder, $M=10\, kg$
Radius, $R=40\, cm =0.4\, m$
$L=9 R=9 \times 0.4=36\, m$
Moment of inertia of the solid cylinder about the line passing through its centre and perpendicular to its axis is given as

$I_{C O M}=M\left[\frac{L^{2}}{12}+\frac{R^{2}}{4}\right]$
$=10\left[\frac{(36)^{2}}{12}+\frac{(0.4)^{2}}{4}\right]$
$=10[1.08+0.04]=10 \times 112$
$\Rightarrow I_{C O M}=11.2\, kg - m ^{2}$
According to parallel axis theorem, moment of inertia of cylinder about aline passing through its edge and perpendicular to its axis is given as
$I'=I_{\text {COM }}+M\left(\frac{L}{2}\right)^{2}=11.2+10\left(\frac{36}{2}\right)^{2}$
$=11.2+10(324)=11.2+324=436\, kg - m ^{2}$