Two identical discs are moving with the same kinetic energy. One rolls and the other slides. The ratio of…
Two identical discs are moving with the same kinetic energy. One rolls and the other slides. The ratio of their speeds is
1 : 2
1 : 1
2 : 3
$\sqrt{2}: \sqrt{3}$
Solution
According to the question,
$\therefore$ Kinetic energy for the rolling disc,
or
$
\begin{aligned}
\mathrm{KE}_r= & \frac{1}{2} m v_1^2+\frac{1}{2} I \omega^2 \\
& \left(\therefore \text { Moment of inertia, } I=\frac{m R^2}{2}\right) \\
& =\frac{1}{2} m v_1^2+\frac{1}{2} \frac{m R^2}{2}\left(\frac{v_1}{R}\right)^2
\end{aligned}
$
Given, $\mathrm{KE}$ of rolling disc $=\mathrm{KE}$ of sliding disc or,
$
\begin{aligned}
\frac{3}{4} m v_1^2=\frac{1}{2} m v_2^2 & \text { or } \frac{v_1^2}{v_2^2}=\frac{2}{3} \\
\frac{v_1}{v_2} & =\sqrt{\frac{2}{3}} \\
or \quad v_1: v_2 & =\sqrt{2}: \sqrt{3}
\end{aligned}
$
