Question

The moment of intertia of a solid cylinder of mass $M$, length $2R$ 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 $2R$ and radius $R$ about an axis passing through the centre of mass and perpendicular to the axis is
$I_1=M\left(\frac{L^2}{12}+\frac{R^2}{4}\right)$
$I_1=M\left[\frac{4R^2}{12}+\frac{R^2}{4}\right]=M\left[\frac{7R^2}{12}\right]$
Moment of inertia passing through one end of the cylinder and perpendicular to the axis of the cylinder
$I_2=M\left(\frac{L^2}{3}+\frac{R^2}{4}\right)$
$I_2=M\left[\frac{4R^2}{3}+\frac{R^2}{4}\right]$
$=M\left[\frac{19R^2}{12}\right]$
$\therefore I_2-I_1=M{R}^2$
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 $AC:y+4=\frac{3+4}{3+4}(x+4)$
$\Rightarrow y+4=x+4\Rightarrow y=x$
$\Rightarrow x-y=0$
$\therefore$ Equation $BD:y+4=\frac{3+4}{-4-3}(x-3)$
$\Rightarrow y+4=-(x-3)$
$\Rightarrow y+4=-x+3$
$\Rightarrow x+y+1=0$
Hence, combined equation of diagonals is
$(x-y)(x+y+1)=0$
$\Rightarrow x^2+xy+x-xy-y^2-y=0$
$\Rightarrow x^2-y^2+x-y=0$