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A uniform rod of mass m and length L is ...

A uniform rod of mass m and length L is free to rotate in the vertical plane about a horizontal axis passing through its end. The rod initially in horizontal position is released. The initial angular acceleration of the rod is:

Text Solution

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Applying law of conservation of energy
`(1)/(2) I omega^(2) = "mg" (l)/(2)(1-cos theta)`
Torque about hinge point is
`"mg"(l)/(2) sin theta = (ml^(2))/(3) alpha`
`a_(T) = (l//2)alpha`
`alpha_(N) = omega^(2) (l//2)`
`N_(x) = m omega^(2) (l)/(2) sin theta - (ml)/(2) alpha cos theta`
`N_(y) - mg = -(ml)/(2) alpha sin theta - m omega^(2) (l)/(2) cos theta`
Putting the value of, `theta = 90^(@)` in equation (1), (2), (3), (4)
`N_(x) = (3mg)/(2) (larr)`
`N_(y) = (mg)/(4) (uarr)`
Putting `theta = 180^(@)`
`N_(x) = 0`
`N_(y) = 4 mg (uarr)`
(a) `N_(x) = (3mg)/(2) larr, N_(y) = (mg)/(4) uarr`
(b) `N_(x) = 0, N_(y) = 4mg uarr`
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