Question

Two identical discs are moving with the same kinetic energy. One rolls and the other slides. The ratio of their speeds is

• A. 1 : 2
• B. 1 : 1
• C. 2 : 3
• D. $\sqrt{2}:\sqrt{3}$

Answer 1

Answer: (D)

According to the question,
$\therefore$ Kinetic energy for the rolling disc,
or $K{E}_r=\frac{1}{2}m{v}_1^2+\frac{1}{2}I{\omega }^2$
$(\therefore$ Moment of inertia, $I=\frac{m{R}^2}{2})$
or $=\frac{1}{2}m{v}_1^2+\frac{1}{2}\frac{m{R}^2}{2}{\left(\frac{v_1}{R}\right)}^2$
Or $K{E}_r=\frac{3}{4}m{v}_1^2...(i)$
Now, $KE$ for the sliding disc,
$\therefore K{E}_s=\frac{1}{2}m{v}_2^2...(ii)$
Given, $KE$ of rolling disc$=KE$ of sliding disc
Or, $\frac{3}{4}m{v}_1^2=\frac{1}{2}m{v}_2^2$ or $\frac{v_1^2}{v_2^2}=\frac{2}{3}$
or $\frac{v_1}{v_2}=\sqrt{\frac{2}{3}}$
or $v_1:v_2=\sqrt{2}:\sqrt{3}$