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 the time taken in the two motions is:
A. $\dfrac{1}{2}$
B. $1$
C. $\dfrac{1}{\sqrt{2}}$
D. $\dfrac{1}{4}$

Answer 1

Hint: We can solve this problem by using the formula for acceleration of the center of mass of the body when rolling and slipping down an inclined plane respectively. Then using these values, we can use equations for motion with constant acceleration relating the displacement with the time and can get the required ratio.

Formula used:
For a body moving down a smooth incline of angle $\theta $, the acceleration $a$ is
$a=g\sin \theta $
where $g$ is the acceleration due to gravity.
For a body rolling down a rough incline of angle $\theta $, the acceleration $a$ is
$a=\dfrac{g\sin \theta }{1+\dfrac{I}{M{{R}^{2}}}}$
where $g$ is the acceleration due to gravity, $I$ is its moment of inertia about an axis perpendicular to the acceleration, $M$ is the mass of the body and $R$ is the radius of the cross section of the body.
For motion with constant acceleration,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
where $s$ is the displacement of the body, $u$ is the initial velocity of the body, $a$ is the acceleration of the body and $t$ is the time period of the motion.

Complete step by step answer:
This problem can be solved by finding the respective accelerations of the center of mass for both the cases, slipping and rolling of the ring. Then, we can relate this acceleration with the time taken for the displacement by using the equation of motion for constant acceleration. Therefore, we can get the required ratio.
Therefore, let us analyze the question.
Let the mass of the ring be $M$ and its radius be $R$.
Let the incline angle be $\theta $ and the distance, covered by the ring in both cases be $s$.
Let the time of motion of the ring for slipping (translational motion) and rolling(rotational motion) be ${{t}_{1}}$ and ${{t}_{2}}$ respectively.

Let the acceleration of the center of mass in the case of slipping (translational motion) and rolling (rotational motion) be ${{a}_{1}}$ and ${{a}_{2}}$ respectively.
For a body moving down a smooth incline of angle $\theta $, the acceleration $a$ is
$a=g\sin \theta $ --(1)
where $g$ is the acceleration due to gravity.
For a body rolling down a rough incline of angle $\theta $, the acceleration $a$ is
$a=\dfrac{g\sin \theta }{1+\dfrac{I}{M{{R}^{2}}}}$ --(2)
where $g$ is the acceleration due to gravity, $I$ is its moment of inertia about an axis perpendicular to the acceleration, $M$ is the mass of the body and $R$ is the radius of cross section of the body.

Since, the body is a thin circular ring, the moment of inertia about the axis perpendicular to the plane will be $I=M{{R}^{2}}$ --(3)

Therefore, using (1), we get,
${{a}_{1}}=g\sin \theta $ -(4)
Using (2) and (3), we get,
${{a}_{2}}=\dfrac{g\sin \theta }{1+\dfrac{I}{M{{R}^{2}}}}=\dfrac{g\sin \theta }{1+\dfrac{M{{R}^{2}}}{M{{R}^{2}}}}=\dfrac{g\sin \theta }{1+1}=\dfrac{g\sin \theta }{2}$ --(5)
Now, for motion with constant acceleration,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ --(6)
where $s$ is the displacement of the body, $u$ is the initial velocity of the body, $a$ is the acceleration of the body and $t$ is the time period of the motion.
Using (4) and (6), we get,
$s=u{{t}_{1}}+\dfrac{1}{2}\left( g\sin \theta \right){{t}_{1}}^{2}$
Now, since, at the top of the incline, the body is considered to be at rest, $u=0$.
$\therefore s=0{{t}_{1}}+\dfrac{1}{2}\left( g\sin \theta \right){{t}_{1}}^{2}=\dfrac{g\sin \theta }{2}{{t}_{1}}^{2}$ --(7)
Similarly, using (5) and (6), we get,
$s=u{{t}_{2}}+\dfrac{1}{2}\left( \dfrac{g\sin \theta }{2} \right){{t}_{2}}^{2}$
Now, since, at the top of the incline, the body is considered to be at rest, $u=0$.
$\therefore s=0{{t}_{2}}+\dfrac{1}{2}\left( \dfrac{g\sin \theta }{2} \right){{t}_{2}}^{2}=\dfrac{g\sin \theta }{4}{{t}_{1}}^{2}$ --(8)
Now, since the displacement covered is the same, since the incline is geometrically identical, hence, we can equate (7) and (8). Therefore, equating, we get,
$\dfrac{g\sin \theta }{2}{{t}_{1}}^{2}=\dfrac{g\sin \theta }{4}{{t}_{2}}^{2}$
$\therefore \dfrac{{{t}_{1}}^{2}}{{{t}_{2}}^{2}}=\dfrac{\dfrac{g\sin \theta }{4}}{\dfrac{g\sin \theta }{2}}=\dfrac{2}{4}=\dfrac{1}{2}$
$\therefore {{\left( \dfrac{{{t}_{1}}}{{{t}_{2}}} \right)}^{2}}=\dfrac{1}{2}$
Square rooting both sides we get,
$\sqrt{{{\left( \dfrac{{{t}_{1}}}{{{t}_{2}}} \right)}^{2}}}=\sqrt{\dfrac{1}{2}}$
$\therefore \dfrac{{{t}_{1}}}{{{t}_{2}}}=\dfrac{1}{\sqrt{2}}$
Therefore, the required ratio is $\dfrac{1}{\sqrt{2}}$.
Hence, the correct option is C) $\dfrac{1}{\sqrt{2}}$.

Note: Students must be careful to write the correct value of the moment of inertia about the correct axis. The moment of inertia should be about the axis that is perpendicular to the plane of motion of the body. Writing the wrong value of moment of inertia will lead to a completely wrong answer, in general.
Students should also pay attention that rolling is not possible on a perfectly smooth surface. This is because the normal force and gravitational force for a uniform body passes through the center of mass and cannot provide a torque. However, on rough surfaces there is an extra force of friction, which can produce the necessary torque to initiate rolling.