Question

A thin circular ring first slips down a smooth incline then rolls down a rough incline of identical geometry from the same height. The ratio of time taken in the two motion is
A. $\dfrac{1}{2}$
B. $1$
C. $\dfrac{1}{{\sqrt 2 }}$
D. $\dfrac{1}{4}$

Answer 1

In the case of a rough inclined plane, the body will be in general motion i.e, it will be in both translational as well as rotational motion. The acceleration due to gravity always works in the perpendicular direction and if the plane is inclined it will be divided into two components.

Complete step by step answer:

Given that a thin circular ring first rolls on a smooth inclined surface so in this case, as we know there is no friction so there will be no opposing force at the point of contact of the circular ring and the inclined plane so the ring will simply slide down without rotating, si it means it is the simple case of translation motion
Here acceleration of the ring will be caused by gravity but gravity acts perpendicularly downwards and our plane is inclined

Let us assume that our plane is inclined at angle 0, Refer figure
So the acceleration of the ring, in this case, will be given by
${a_1} = g\sin \theta$
$\theta$is the inclined angle

Now, in the second case, it is said that the inclined plane is not smooth but it is rough
From this statement, we are sure that now there will be an opposing force at the point of contact and due to this force, the ring will rotate so acceleration, in this case, will be due to general motion.

So acceleration will be,
${a_2} = \dfrac{{g\sin \theta }}{{\left[ {1 + \left( {\dfrac{I}{{m{R^2}}}} \right)} \right]}}$
Where
$m$ is the mass of the ring
$I$ is the moment of inertia of a hollow ring and its value is given by
$I = m{R^2}$
$R$ mean radius of the ring
$\therefore {a_2} = \dfrac{{g\sin \theta }}{2}$

Now, as the object is falling from the same height and the rough and smooth plane are identical so the distance covered by the ring in both cases will be the same.
According to newton's formula
$S = ut + \dfrac{1}{2}a{t^2}$
Here,
$S$ is the distance covered by the ring
$t$ is the time taken
$a$ is the acceleration
$u$ is the initial velocity

Distance travelled in the first case will be
${S_1} = u{t_1} + \dfrac{1}{2}{a_1}{t_1}^2$ --------(1)
And distance covered in the second case will be
${S_2} = u{t_2} + \dfrac{1}{2}{a_2}{t_2}^2$ -------(2)
Here $S$ will be same in both the cases
$\therefore {S_1} = {S_2}$
Equating equations, we get

As initially ring was no moving in both the cases so its initial velocity $u = 0$ in both the cases,
$u{t_1} + \dfrac{1}{2}{a_1}{t_1}^2 = u{t_2} + \dfrac{1}{2}{a_2}{t_2}^2 \\\ \Rightarrow \dfrac{{{t_1}^2}}{{{t_2}^2}} = \dfrac{{{a_2}}}{{{a_1}}} \\\$
Now substituting values of accelerations
$\dfrac{{{t_1}^2}}{{{t_2}^2}} = \dfrac{{\dfrac{{g\sin \theta }}{2}}}{{g\sin \theta }} \\\ \Rightarrow \dfrac{{{t_1}^2}}{{{t_2}^2}} = \dfrac{{g\sin \theta }}{{2g\sin \theta }} \\\ \Rightarrow \dfrac{{{t_1}^2}}{{{t_2}^2}} = \dfrac{1}{2} \\\ \Rightarrow \dfrac{{{t_1}}}{{{t_2}}} = \dfrac{1}{{\sqrt 2 }} \\\$

Hence, the correct answer is option (C).

Note: In absence of friction i.e, on a smooth surface, the ring will slide down the slope without rotating and this will be the pure translational motion.
In case of rough surface, the ring will experience a motion opposing force at the point of contact so the lower point of the ring tends to stop it while the body is in the effect of gravity will try to move forward this will induce rotation in-ring and the motion will be general.