The moment of inertia of a rod of mass 40 tex | vedclass

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(15) $1$. Moment of inertia of the rod about axis $AB$ (passing through one end and perpendicular to the length):
$I_{	ext{rod}} = \frac{1}{3}ML^2 =

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$3$

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(15) $1$. Moment of inertia of the rod about axis $AB$ (passing through one end and perpendicular to the length):
$I_{	ext{rod}} = \frac{1}{3}ML^2 = \frac{1}{3} 	imes 40 	imes (3)^2 = \frac{1}{3} 	imes 40 	imes 9 = 120 	ext{ kg m}^2$.
$2$. Moment of inertia of the solid sphere about an axis parallel to $AB$ at a distance $d = 3 	ext{ m}$ from its center:
Using the parallel axis theorem, $I_{	ext{sphere}} = I_{	ext{cm}} + md^2 = \frac{2}{5}mR^2 + md^2$.
Given $m = 10 	ext{ kg}$ and $d = 3 	ext{ m}$:
$I_{	ext{sphere}} = \frac{2}{5} 	imes 10 	imes R^2 + 10 	imes (3)^2 = 4R^2 + 90$.
$3$. Equating the two moments of inertia:
$120 = 4R^2 + 90$
$4R^2 = 30$
$R^2 = \frac{30}{4} = 7.5$.
$4$. Given $R = \sqrt{\frac{\alpha}{2}}$, so $R^2 = \frac{\alpha}{2}$.
Equating the values of $R^2$:
$\frac{\alpha}{2} = 7.5$
$\alpha = 15$.

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