Question

A small solid cylinder is released from a point at a height h on a rough as a track shown figure. Assuming that it does not slip anywhere, calculate its linear speed when it rolls on the horizontal part of the track .

Answer 1

R.E.F image

at A, the cylinder has only potential

energy, $u_1=Mgh$

at B, the cylinder gain a liner speed v

so it has rotational velocity $w=\frac{v}{R}$ (at pure rolling )

at B it has total kinetic energy :-

$k^{\prime }=k_{liner}+k_{rotational}$

$=\frac{1}{2}M{V}^2+\frac{1}{2}I{w}^2$ (for a cylinder, $I=\frac{M{R}^2}{2})$

$=\frac{1}{2}M{V}^2+\frac{1}{2}\left(\frac{M{R}^2}{2}\right)(\frac{V}{R}{)}^2=\frac{3}{4}M{V}^2$

From conversation of Energy

$u_1=k^1$

$\Rightarrow Mgh=\frac{3}{4}M{V}^2$ given $V=\sqrt{\frac{4}{3}gh}$ (Ans)