The time period of a bar pendulum when suspended at distance `30 cm` and `50 cm` from its centre of gravity comes out to be the same. If the mass of the body is `2kg`. Find out its moment of inertia about an axis passing through first point.
Text Solution
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`I_(A) = MK^(2) + M_(1^(2))` but `l_(1) = 50 cm` & `(K^(2))/(l_(1)) = 30 cm rArr K^(2) = 30 xx 50 cm^(2)` `I_(A) = 2 xx(30 xx 50 xx 10^(-4)) +2 xx (50 xx 10^(-2))^(2) = 0.3 + 0.5 = 0.5 = 0.8 kg-m^(2)` Question based on calculation of time period by energy method.
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