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Asked by eknathmote50 | 02 Jun, 2019, 14:23: PM
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When the mass is released at P, it moves towards O as shown in figure.
When it is moved upto the point O, change in gravitational potential energy and workdone by the electric field
to move the ball of charge q to the distance l sinθ are converted to kinetic energy of the ball at O
 
change in gravitational potential energy = m g l ( 1 - cosθ ) ..............................(1)
 
workdone by the electric field to move the ball of charge q  = q×( mg/q )× l sinθ  ..................(2)
 
By adding (1) and (2) with substitution θ = 45°, we get total energy = mg l  ......................(3)
 
hence kinetic energy,  (1/2)mv2 = m g l    or  begin mathsize 14px style v space equals space square root of 2 space g space l end root end style ...........................(4)
angular velocity = ω = v / l    or   begin mathsize 14px style omega space equals space square root of fraction numerator 2 g over denominator l end fraction end root end style  .
Answered by Thiyagarajan K | 02 Jun, 2019, 16:47: PM

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