Question

The moment of intertia of a solid cylinder of mass $M$, length $2 \,R$ and radius $R$ about an axis passing through the centre of mass and perpendicular to the axis of the cylinder is $I_{1}$ and about an axis passing through one end of the cylinder and perpendicular to the axis of the cylinder is $I_{2}$, then

• A. $I_{2} < I_{1}$
• B. $I_{2}-I_{1}=M R^{2}$
• C. $\frac{I_{2}}{I_{1}}=\frac{19}{12}$
• D. $\frac{I_{2}}{I_{1}}=\frac{7}{6}$

Answer 1

Answer: (B)

Moment of inertia of a solid cylinder of mass $M$, length $2 \,R$ and radius $R$ about an axis passing through the centre of mass and perpendicular to the axis is
$I_{1}=M\left(\frac{L^{2}}{12}+\frac{R^{2}}{4}\right)$
$I_{1}=M\left[\frac{4 R^{2}}{12}+\frac{R^{2}}{4}\right]=M\left[\frac{7 R^{2}}{12}\right]$
Moment of inertia passing through one end of the cylinder and perpendicular to the axis of the cylinder
$ I_{2} =M\left(\frac{L^{2}}{3}+\frac{R^{2}}{4}\right) $
$I_{2} =M\left[\frac{4 R^{2}}{3}+\frac{R^{2}}{4}\right] $
$=M\left[\frac{19 R^{2}}{12}\right] $
$\therefore I_{2}-I_{1} =M R^{2} $
Hence, equations of the sides of the square are
$x=3, x=-4, y=3, y=-4$
So, the coordinates of the vertex are
$A(-4,-4)$, $B(3,-4), C(3,3)$ and $D(-4,3)$

$\therefore $ Equation of $A C: y+4=\frac{3+4}{3+4}(x+4)$
$\Rightarrow y+4=x+4 \Rightarrow y=x$
$\Rightarrow x-y=0$
$\therefore $ Equation $B D: y+4=\frac{3+4}{-4-3}(x-3)$
$\Rightarrow y+4=-(x-3)$
$\Rightarrow y+4=-x+3$
$\Rightarrow x+y+1=0$
Hence, combined equation of diagonals is
$(x-y)(x+y+1)=0$
$\Rightarrow x^{2}+x y+x-x y-y^{2}-y=0$
$\Rightarrow x^{2}-y^{2}+x-y=0$