Question

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

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

Answer 1

Answer: (D)

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

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