In space of horizontal `EF (E = (mg)//q)` exist as shown in figure and a mass m is released at the end of a light rod. If mass m is releases from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position
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Explanation

Using conservation of mechanical energy,
`Delta K + Delta U = 0`
`Delta U = Delta U_(gravity) + Delta U_("electrical")`
`Delta U_("electrical") = q Delta V = - q(E l sin theta)`,
`Delta U_(gravity) = - mg (l - l cos theta)`
`Delta K = (1)/(2) m v^2 - 0` (i)
`q E l sin theta + mg (l - l cos theta) = (1)/(2) m v^2`
Fromn Eq. (i)
`mg l sin theta + mg l - mg l cos theta = (mv^2)/(2)`
`(g l)/(sqrt(2)) + g l - (g l)/(sqrt(2)) = (1)/(2) m v^2 or V= sqrt(2 g l)`
`omega = V/(l) = sqrt(2 g l)/(l) = sqrt((2 g)/(l)) or omega = sqrt((2 g)/(l))`.
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