Question

Two masses of 8 kg & 4 kg are connected by a light rod of length 30 cm. Moment of inertia of system about an axis perpendicular to rod is I. Minimum value of I is

• A. $24\times {10}^{-2}kg{m}^2$
• B. $36\times {10}^{2}kg{m}^2$
• C. $48\times {10}^{-2}kg{m}^2$
• D. $108\times {10}^{-2}kg{m}^2$

Answer 1

The correct option is A $24\times {10}^{-2}kg{m}^2$

$I=I_{CM}+m{d}^2$ {Minimum when system rotates about axis through Center of Mass(CM) i.e. $d=0$}
$I_{CM}=m_1{x}_1^2+m_2{x}_2^2$
By def $8x_1=4x_2$
$8x_1=4(0.3-x_1)$
$2x_1=0.3-x_1$
$3x_1=0.3$
$x_1=0.1$

$I_{cm}=8\times (0.1{)}^2+4\times (0.2{)}^2$
$=8\times {10}^{-2}+16\times {10}^{-2}$
$=24\times {10}^{-2}kg{m}^2$