Question

A simple pendulum with a bob of mass m =1 kg, charge q =5 $\mu$C and string length l= 1 m is given a horizontal velocity u in a uniform electric field $E=2\times {10}^6$ V/ m at its bottom most point A, as shown in the figure. It is given, that the speed u is such that the particle leaves the circle at point C. Find the speed u in m/s. (Take g = 10 m/$s^2$) The answer is $\sqrt{xy}$ Then x+y =

Answer 1

Let the speed of the particle at C is $vm/s.$

then,

centripital force $=mg\cos {30}^0-qe\cos {60}^0$

$1\times \frac{10\sqrt{3}}{2}-5\times 2\times \frac{1}{2}$

$5\sqrt{3}-5$

$f_C=\frac{{mv}^2}{e}=5\sqrt{3}-5$

$\frac{1\times v^2}{1}=5\sqrt{3}-5\Rightarrow v^2=5\sqrt{3}-5⟶\left(1\right)$

By energy conservation,

$\Delta K.E.=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

Work done by gravity $=-mgl(1+\sin \theta )\theta ={60}^0$

Work done by electric field $=qel\cos \theta$

Work done by all force $=\Delta K$

so$-mgl(1+\sin {60}^0)+qel\cos {60}^0=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2\frac{v^2-u^2}{2}=-10(1+\frac{\sqrt{3}}{2})+\frac{10\times 1}{2}{v}^2-u^2=2(-5-5\sqrt{3})u^2=v^2+2(5+5\sqrt{3})u^2=5\sqrt{3}-5+10+10\sqrt{3}{u}^2=5(1+3\sqrt{3})u^2=5(1+3\times 1.73)u^2=31u=\sqrt{31}m/s.$