A simple pendulum is released from rest at th | vedclass

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(B) Let $v$ be the velocity of the bob at position $	heta$ after being released from the horizontal position.From the law of conservation of energy,t

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$\phi=	an ^{-1}\left(\frac{	an 	heta}{2}ight)$

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(B) Let $v$ be the velocity of the bob at position $	heta$ after being released from the horizontal position.From the law of conservation of energy,the potential energy lost equals the kinetic energy gained:$mgh = \frac{1}{2}mv^2$Since the height $h$ dropped from the horizontal is $l \cos 	heta$,we have:$mg(l \cos 	heta) = \frac{1}{2}mv^2$$\Rightarrow \frac{v^2}{l} = 2g \cos 	heta$The radial (centripetal) acceleration is $a_c = \frac{v^2}{l} = 2g \cos 	heta$.The tangential acceleration is due to the component of gravity perpendicular to the string:$a_t = g \sin 	heta$.If the total acceleration vector $\vec{a}$ makes an angle $\phi$ with the string (which is the direction of the radial acceleration),then:$	an \phi = \frac{a_t}{a_c}$Substituting the expressions for $a_t$ and $a_c$:$	an \phi = \frac{g \sin 	heta}{2g \cos 	heta} = \frac{	an 	heta}{2}$Therefore,$\phi = 	an^{-1}\left(\frac{	an 	heta}{2}ight)$.

$A$ simple pendulum is released from rest at the horizontally stretched position. When the string makes an angle $\theta$ with the vertical,the angle $\phi$ which the acceleration vector of the bob makes with the string is given by

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