For the pivoted slender rod of length l as shown in figure, the angular velocity as the bar reaches the vertical position after being released in the horizontal position is
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a
gl
b
24g19l
c
24g7l
d
4gl
answer is C.
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Detailed Solution
when the systm is released loss in P.E for3 l4part is 3mg43l8 and gain in P.E. for l/4 part is mg4l8 ; net loss in P.E is 9mgl32-mgl32=mgl4= increase in roattional K.E ; 12Iω2=mgl4 here I about an axis l4 from the center is ml212+ml216=7ml248 ; soving for ω we get ω=24g7l