A body of mass 200 g oscillates about a horizontal axis at a distance of 20 cm from its center of gravity. If the length of the equivalent simple pendulum is 35 cm, find its moment of inertia about the point of suspension.

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Solution

$2 π \sqrt{\frac{l}{g}} = 2 π \sqrt{\frac{I}{m g l^{'}}}$
$∴$ $l = \frac{I}{m l^{'}}$
or $I = (m) (l) (l^{'})$
$= (200) (20) (35)$
$= 1.4 \times 10^{5}$ g-cm $^{2}$

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A body of mass 200 g oscillates about a horizontal axis at a distance of 20 cm from its center of gravity. If the length of the equivalent simple pendulum is 35 cm, find its moment of inertia about the point of suspension.

2π√lg=2π√Imgl′∴ l=Iml′or I=(m)(l)(l′) =(200)(20)(35) =1.4×105 g-cm2

$2 π \sqrt{\frac{l}{g}} = 2 π \sqrt{\frac{I}{m g l^{'}}}$
$∴$ $l = \frac{I}{m l^{'}}$
or $I = (m) (l) (l^{'})$
$= (200) (20) (35)$
$= 1.4 \times 10^{5}$ g-cm $^{2}$