A solid cylinder of mass 'M' and radius 'R' rolls down a smooth inclined plane about its own axis and…
A solid cylinder of mass 'M' and radius 'R' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' $\mathrm{v}^{\prime}$. The height of the inclined plane
is $(\mathrm{g}=$ acceleration due to gravity $)$
$\frac{3 v^{2}}{4 g}$
$\frac{4 v^{2}}{5 g}$
$\frac{7 v^{2}}{9 g}$
$\frac{2 v^{2}}{3 g}$
Solution
Work-energy theorem, \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^2+\frac{1}{2} \mathrm{Iw}^2\)
For solid cylinder, \(\mathrm{I}=\frac{1}{2} \mathrm{MR}^2\)
Also \(\mathrm{v}=\mathrm{Rw}\) (due to pure rolling)
Thus \(\mathrm{Mgh}=\frac{1}{2} \mathrm{Mv}^2+\frac{1}{2} \times \frac{1}{2} \mathrm{MR}^2 \mathrm{w}^2\)
\(\begin{aligned}
& \Rightarrow v=\sqrt{\frac{4}{3}} \mathrm{gh} \\
& \Rightarrow>h=3 v^{2} / 4 g
\end{aligned}\)
.