First slide

Moment of intertia

Question

The moment of inertia of a thin rod of mass M and length L, about an axis perpendicular to the rod at a distance ¼ th from one end

Moderate

A

$\frac{M L^{2}}{6}$
B

$\frac{3 M L^{2}}{8}$
C

$\frac{7 M L^{2}}{24}$
D

$\frac{7 M L^{2}}{48}$

Solution

$m o m e n t o f i n e r t i a a b o u t m i d p o i n t \frac{M L^{2}}{12} f r o m o n e e n d L / 4 m e a n s f r o m c e n t e r a l s o i t i s L / 4 a p p l y i n g p a r a l l e l a x i s t h e o r e m I = \frac{M L^{2}}{12} + M \frac{L^{2}}{16} = \frac{7 M L^{2}}{48}$

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