Question

Prove the result that the velocity v of translation of a rolling body at the bottom of a inclined plane of height is given by V^2=2gh/1+k^2/R

Answer 1

$\boxed{ \huge{ \mathfrak{ \fcolorbox{red}{yellow}{\purple{Answer}}}}} \\ \\ \star \: \blue{ \mathfrak{To \: Prove}} \\ \\ \implies \rm \: the \: velocity \: of \: translation \: of \: a \\ \rm \: rolling \: body \: at \: the \: bottom \: of \: a \: inclined \\ \rm \: plane \: of \: height \: h \: is \: given \: by... \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \pink{ \bold{ {v}^{2} = \frac{2gh}{1 + \frac{ {k}^{2} }{ {r}^{2} } }}}} \\ \\ \star \: \blue{ \mathfrak{Concept }} \\ \\ \implies \rm \: law\: of \: energy \: conservation \: says \\ \rm \: that \: initial \: energy \: of \: system \: is \: always \: \\ \rm \: remain \: same \: as \: final \: energy \: of \: system. \\ \\ \implies \rm \: initial \: potential \: energy = mgh \\ \\ \implies \rm \: final \: kinetic\: energy = \frac{1}{2} m {v}^{2} \: (1 + \frac{ {k}^{2} }{ {r}^{2} }) \\ \\ \star \: \blue{ \mathfrak{Derivation}} \\ \\ \implies \rm \: initial \: energy = final \: energy \\ \\ \implies \rm \: mgh = \frac{1}{2} m {v}^{2} \: (1 + \frac{ {k}^{2} }{ {r}^{2} } ) \\ \\ \therefore \rm \: \boxed{ \rm{\orange{ {v}^{2} = \frac{2gh}{1 + \frac{ {k}^{2} }{ {r}^{2} } }}}}$