Question

A ring rolls down an inclined plane. The ratio of the rotational kinetic energy to translational kinetic energy is:
(A) $1:3$
(B) $1:1$
(C) $3:1$
(D) $2:1$

Answer 1

Hint: Rolling motion is defined as the motion in which the object undergoes both rotational as well as translational motion simultaneously. Rolling motion is not possible without friction.

Complete step by step solution:
Translational motion: Translation motion is defined as the motion in which the object moves along the x-axis, y-axis or the z-axis or along all of them. Conventional Newtonian mechanics is based upon this type of motion.
Rotational motion: Rotational motion is defined as the motion in which the loci of all the particles of the object form complete circles with one common center. In rotational motion all the particles of the object undergo circular motion except the center. Its center remains at rest.
Rolling motion: Rolling motion is defined as the motion in which all the particles of the object undergo circular motion except the center. The center undergoes translational motion. So we can say that in a rolling body, both the rotational motion and the translational motion take place simultaneously.
For an easier understanding between rotational and translational motion, the physical parameters which define the existence of rotational motion are correlated with the physical parameters which define the translational motion. For example, velocity is correlated with angular velocity, acceleration is correlated with angular acceleration, mass is correlated with moment of inertia and force is correlated with torque.
Now we know that Translational Kinetic Energy can be calculated by,
${K_T} = \dfrac{1}{2}m{v^2}$
Where,
${K_T} = $Translational Kinetic Energy,
$m = $Mass of body, and
$v = $Velocity of body.
Also, the Rotational Kinetic Energy can be calculated by replacing the terms in the above formula with their rotational equivalents. So,
${K_R} = \dfrac{1}{2}I{\omega ^2}$
Where, ${K_R} = $Rotational Kinetic Energy,
$I = $Moment of Inertia, and
$\omega = $Angular velocity.
Now for a ring,
$I = m{R^2}$
Where, $R = $Radius of ring.
And, $\omega = \dfrac{v}{R}$
So,
${K_R} = \dfrac{1}{2}\left( {m{R^2}} \right){\left( {\dfrac{v}{R}} \right)^2}$
$ \Rightarrow {K_R} = \dfrac{1}{2} \times m{R^2} \times \dfrac{{{v^2}}}{{{R^2}}}$
$ \Rightarrow {K_R} = \dfrac{1}{2}m{v^2} = {K_T}$
$\therefore \dfrac{{{K_R}}}{{{K_T}}} = \dfrac{1}{1}$

Therefore, option B is the correct answer.

Note: The moment of inertia is different for different types of objects and for different axes at which the object is being rotated. Moment of Inertia depends on the mass distribution in the object and how that distributed mass undergoes motion.