First slide

Moment of inertia

Question

The radius of gyration of a uniform rod of length l, about an axis passing through a point $\frac{L}{4}$ away from the center of the rod, and perpendicular to it is

Easy

A

$\sqrt{\frac{7}{48}} L$
B

$\sqrt{\frac{3}{8}} L$
C

$\frac{1}{8} L$
D

$\frac{1}{4} L$

Solution

$U \sin g P a r a l l e l A x i s T h e o r e m, I_{C} + M d^{2} = I_{P} \Rightarrow \frac{M L^{2}}{12} + M {(\frac{L}{4})}^{2} = M K^{2} h e r e K = R a d i u s o f G y r a t i o n$
$\Rightarrow \frac{L^{2}}{12} + \frac{L^{2}}{16} = K^{2} \Rightarrow K = \sqrt{\frac{7}{48}} L$

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