A solid sphere and a solid cylinder roll down without slipping along an inclined plane. If they start from…
A solid sphere and a solid cylinder roll down without slipping along an inclined plane. If they start from rest from the top of the inclined plane, the ratio of the velocities of the solid sphere and solid cylinder when they reach the bottom of the inclined plane is
$\sqrt{25}: \sqrt{21}$
$\sqrt{3}: \sqrt{2}$
$\sqrt{25}: \sqrt{14}$
$\sqrt{15}: \sqrt{14}$
Solution
Apply the conservation of energy
$
\begin{aligned}
& \mathrm{mgh}=\frac{1}{2} \mathrm{~m} v^2+\frac{1}{2} \mathrm{I} \omega^2 \\
& \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{~m} \frac{\mathrm{k}^2 \mathrm{v}^2}{\mathrm{R}^2} \\
& \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2\left[1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right] \\
& \frac{2 g h}{1+\frac{K^2}{R^2}}=v^2 \\
&
\end{aligned}
$
For solid sphere
$
v_{\mathrm{ss}}^2=\frac{2 g h}{1+\frac{2}{5}}=\frac{5 \times 2 g h}{7}
$
for solid cylinder
$
\begin{aligned}
& v_{\mathrm{sc}}^2=\frac{2 \mathrm{gh}}{1+\frac{1}{2}}=\frac{2 \times 2 \mathrm{gh}}{3} \\
& \text { Ratio }=\frac{v_{\mathrm{ss}}^2}{v_{\mathrm{sc}}^2}=\frac{\frac{5 \times 2 \mathrm{gh}}{7}}{\frac{2 \times 2 \mathrm{gh}}{3}}=\frac{15}{14} \\
& \text { Ratio }=\frac{v_{\mathrm{ss}}}{v_{\mathrm{sc}}}=\sqrt{\frac{15}{14}}
\end{aligned}
$