Question

If a ring rolls down from top to bottom of an inclined plane, it takes time \(t_1\). If it slides, it takes time \(t_2\). Then the ratio \(\dfrac{t_1}{t_2}\) is

• A. \(\dfrac{1}{3}\)
• B. \(\dfrac{1}{2}\)
• C. \(\dfrac{1}{4}\)
• D. \(\dfrac{2}{5}\)
• E. \(\dfrac{2}{3}\)

Hint:

Rolling motion takes more time than pure sliding because part of the energy goes into rotation.

Answer 1

Answer: (E)

For sliding without friction: \[ a_s=g\sin\theta \] For rolling ring: \[ a_r=\frac{g\sin\theta}{1+\frac{I}{mR^2}} \] For a ring: \[ I=mR^2 \] So, \[ a_r=\frac{g\sin\theta}{2} \] Since \[ t\propto \frac{1}{\sqrt{a}} \] \[ \frac{t_1}{t_2}=\sqrt{\frac{a_s}{a_r}} =\sqrt{\frac{g\sin\theta}{g\sin\theta/2}} =\sqrt{2} \] So the direct calculation gives: \[ \boxed{\sqrt{2}} \] which does not match the listed options.