Question

A simple pendulum of length l has a bob of mass m if it is released from horizontal position then find the velocity of bob at lowest position?

Answer 1

$\bf \underline{ \underline{ Answer}} : - \\ tension(T) = 3mg \\ \\ \underline{ \underline{\bf{step - by - step \: explanation}}} : - \\ \\ \underline{ \bf{according \: to \: the \: question}} : - \\ \\ length \: of \: pendulam \: = l \\ \\ mass \: of \: bob \: = m \\ \\ let \: v \: be \: the \: velocity \: at \: lowest \: point \: \\ \\ \star \: its \: potential \: energy \: converted \: into \: kinetic \\ \\ energy \\ \\ m \: g \: l \: = \frac{1}{2} m {v}^{2} \\ \\ The \: centrifugal \: acceleration \: = m \frac{ {v}^{2} }{l} \\ \\ \implies \: m \: \frac{2gl}{l} = 2mg \\ \\ Hence, \: the \: tension \: at \: lowest \: point \: \\ \\ \implies \: T= \:mg \: + 2mg \: = 3mg$

Answer 2

$\textsf{Given :}$

                 $\textsf{Length of Pendulum = l}\\\textsf{Mass of Pendulum = m}\\\textsf{Angle at which it is dropped = 90}$

$\textsf{To find :}$

                 $\textsf{Velocity of bob at lowest position}$

$\textsf{Solution :}$

$\textsf{We know from the work-energy theorem that work done by all the forces}\\\textsf{is equal to the change in the kinetic energy i.e. }$

                    $\boxed {W_{all forces} = {\Delta} K.E. }$

$\textsf{Applying this with the given values in the question, we have: }$

            $\textsf{Wgravity + Wtension = K.E.f - K.E.i }$

            $\textsf{Wg + Wt = 1/2 mv^{2} }$

$\textsf{Initial kinetic energy turns out to be zero as the bob was released from rest.}$

              $\textsf {mgl = \frac{1}{2} mv^{2} }$

                 $\textsf{v = \sqrt{2gl} }$

$\textsf{Hence,}$         $\boxed {\textsf{v = \sqrt{2gl} }}$