Question

A rod of length L and uniform mass density and variable positive charge density is kept in horizontal position and pivoted at point P. It can rotate about an axis passing through P in vertical plane. The mass of the rod is m and linear charge density varies as λ=λ0(1+xL), where λ0 is constant and x′ is distance from end P The magnitude of electric field is E0 towards left. Now if the rod is released from horizontal position then its angular speed in vertical position is

Answer 1

Well, when  you drop it, since, it is a rigid body and pivoted in the end it would fall just like a pendulum. Alright! Now consider it falling through anticlockwise direction making an angle ∅. (∅<90)

So if we draw a free body diagram the forces acting on it are centripetal force , Weight , and the electrostatic force.

We are going to calculate ω just from mω^2L remember!

Now lets finish off quickly the electric force part.

Since electric field is in the left , the net force on the rod would be on left.

its magnitude is given by,

$F_{net} = \int\limits^L_0 {E} \, dq$

Alright isn't dq seems to depend on dx? right??

λ = $\frac{dq}{dx} = Y(1+xL)$

Take dx to other side hence solve the integral to get

F =$EYL[1+ \frac{L^{2} }{2}]$

Now, You have a free body diagram, can you use lami's theorem, or just equate components, to obtain ω and plug in Ф=90 in the final equation.

If you need any further clarification, feel free to ask.