Question

A body of mass $m$ slides down a smooth inclined plane and reaches the bottom of the plane with a velocity $v$ . Now the smooth inclined plane is made rough and the same mass $m$ ,but in the form of a ring, rolls down this inclined plane. Find the velocity of the ring as it reaches the bottom of the plane.
A) $\sqrt {2v} $
B) $v$
C) $\left( {\sqrt {\dfrac{2}{5}} } \right)v$
D) $\dfrac{v}{{\sqrt 2 }}$

Answer 1

Hint: In the first case there is only translational motion and we ignore friction as the plane is smooth. In the second case as the body is a ring that rolls down the plane which is now rough, both translational and rotational motion take place. Use the conservation of energy principle to find the velocity.

Formula Used:
1. Conservation of energy gives, $\dfrac{1}{2}m{v_r}^2 + \dfrac{1}{2}I{\omega ^2} = mgh$ , where the term $\dfrac{1}{2}m{v_r}^2$ is the translational kinetic energy of the system, $\dfrac{1}{2}I{\omega ^2}$ is the rotational kinetic energy of the system and $mgh$ is the potential energy of the system; $m$ is the mass of the body, ${v_r}$ is its velocity with which it rolls down the plane, $I$ is the moment of inertia of the body, $\omega $ is its angular velocity, $g$ is the acceleration due to gravity and $h$ is the height at which the plane is inclined.
2. The moment of inertia $I$ of a ring of mass $m$and radius $r$ rotating along an axis perpendicular to the plane is given by, $I = m{r^2}$
3. Relation between linear velocity ${v_r}$ and angular velocity $\omega $ with $r$ as the radius of the ring is given as, ${v_r} = r\omega $

Complete step by step answer:



In the first case the body of mass $m$ slides down the smooth inclined plane. This is a translational motion. In the second case the body is a ring of the same mass $m$ but it rolls down this inclined plane. Here, there is translational motion as well as rotational motion since the body has a ring form and a ring rolls. To find the velocity ${v_r}$ of the body in the second case, we apply the conservation of energy principle which states that energy can neither be created nor destroyed; only a change in its form can occur, i.e., potential energy that exists before the body starts rolling gets converted to kinetic energy as the body rolls to reach the bottom.
Step 1: Express the kinetic energy of the system in both cases

Case 1: The body of mass $m$ slides down the inclined plane placed at a height $h$

The motion is translational.

It is given that $v$ is the velocity of the body as it reaches down the plane.

Therefore, the kinetic energy $ = \dfrac{1}{2}m{v^2}$

Case 2: The body in the form of a ring of mass $m$ rolls down the inclined plane kept at a height $h$

When the body rolls down the plane, the total kinetic energy consists of translational kinetic energy and rotational kinetic energy.

Let ${v_r}$ be the velocity of the ring as it reaches down.

Thus, the total kinetic energy of the body will be $\dfrac{1}{2}m{v_r}^2 + \dfrac{1}{2}I{\omega ^2}$

where $\dfrac{1}{2}m{v_r}^2$ is the translational kinetic energy and $\dfrac{1}{2}I{\omega ^2}$ is the rotational kinetic energy of the body of mass $m$.

The moment of inertia $I$ of a ring of mass $m$and radius $r$ rotating along an axis perpendicular to the plane is given by, $I = m{r^2}$

Moment of inertia of a system describes the resistance it offers to angular acceleration (i.e., to rotate).

Substituting the above relation for$I$, we have the kinetic energy of the system as $\dfrac{1}{2}m{v_r}^2 + \dfrac{1}{2}m{r^2}{\omega ^2}$

Now, we substitute $v = r\omega $ in the above equation.

Finally, kinetic energy of the system $ = \dfrac{1}{2}m{v_r}^2 + \dfrac{1}{2}m{v_r}^2$

Step 2: Express the potential energy of the system in both cases

Case 1 and 2: The potential energy of a body of mass $m$ on an inclined plane slanted at a height of $h$ is$mgh$ . This is the energy of the system before it starts to roll.

Step 3: Express the conservation of energy in both case

Case 1: The body of mass $m$ slides down the inclined plane placed at a height $h$

As the body moves, potential energy converts to kinetic energy.

By the conservation of energy principle, kinetic energy equals potential energy.

i.e., $mgh = \dfrac{1}{2}m{v^2}$

Case 2: The body in the form of a ring of mass $m$ rolls down the inclined plane kept at a height of $h$

When the ring rolls down the plane its potential energy converts to kinetic energy.

By the conservation of energy principle, kinetic energy equals potential energy.

i.e., $mgh = \dfrac{1}{2}m{v_r}^2 + \dfrac{1}{2}m{v_r}^2$ or $mgh = m{v_r}^2$

From case 1 we have, $mgh = \dfrac{1}{2}m{v^2}$ and from case 2 we have, $mgh = m{v_r}^2$

Thus, $m{v_r}^2 = \dfrac{1}{2}m{v^2}$

Cancel out the similar terms on both sides to get, ${v_r}^2 = \dfrac{1}{2}{v^2}$

Taking the square root of ${v_r}^2$ , we get ${v_r} = \dfrac{v}{{\sqrt 2 }}$

Therefore, the correct option is d) $\dfrac{v}{{\sqrt 2 }}$

Note: Theoretically, a body like a ring or sphere rolling without slipping over an inclined plane suffers no kinetic friction(because due to opposite and equal velocity of translational and rotational motion of the bottom most point) because at every instant, there is only one point of contact between the body and the plane and this point remains stationary relative to the plane. Hence, even though there is frictional force(static in nature) the net work done by it is zero.