A uniform rod of length 1.8 ,m suspended by a | vedclass

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(B) Given, length of the rod, $l = 1.8 \,m$.The time period of oscillation of a uniform rod of length $l$ pivoted at one end is given by $T = 2 \pi \s

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$1.2$

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(B) Given, length of the rod, $l = 1.8 \,m$.The time period of oscillation of a uniform rod of length $l$ pivoted at one end is given by $T = 2 \pi \sqrt{\frac{I}{mgR}}$, where $I$ is the moment of inertia about the pivot and $R$ is the distance from the pivot to the center of mass.For a rod pivoted at one end, $I = \frac{ml^2}{3}$ and $R = \frac{l}{2}$.Substituting these values into the formula:$T = 2 \pi \sqrt{\frac{ml^2/3}{mg(l/2)}} = 2 \pi \sqrt{\frac{2l}{3g}} = 2 \pi \sqrt{\frac{2 	imes 1.8}{3g}} = 2 \pi \sqrt{\frac{1.2}{g}}$.The time period of a simple pendulum of length $l'$ is $T' = 2 \pi \sqrt{\frac{l'}{g}}$.Since $T = T'$, we have $2 \pi \sqrt{\frac{l'}{g}} = 2 \pi \sqrt{\frac{1.2}{g}}$.Squaring both sides, we get $l' = 1.2 \,m$.

$A$ uniform rod of length $1.8 \,m$ suspended by an end is made to undergo small oscillations. Find the length of the simple pendulum having the mass and time period equal to that of the rod. (in $\,m$)

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