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Question

In space of horizontal Electric field (E=(mg)/q) exist as shown in figure and a mass m attached at the end of a light rod. If mass m is released from the position shown in figure, find the angular velocity of the rod when it passes through the bottom most position:
125186_1c1fbd67a4e142379a1fd1c7107c4309.png
  1. gl
  2. 2gl
  3. 5gl
  4. 3gl

A
3gl
B
2gl
C
5gl
D
gl
Solution
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According to the work energy theorem we have
We+Wg=12mv2
We have work done by electrostatic force as
We=qElsinθ
and work done by the gravitational force as
Wg=mg(llcosθ)
Thus we get
qElsinθ+mg(llcosθ)=12mv2
Thus we get
mgsinθ+mglmglcosθ=12mv2
as θ=45o, we get
mgl=12mv2
also as v=ωl
we get
ω=2gl

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