Question

Starting from the rest, at the same time, a ring, a coin and a solid ball of the same mass roll down an incline without slipping. The ratio of their translational kinetic energies at the bottom will be
(A) $1:1:1$
(B) $10:5:4$
(C) $21:28:30$
(D) None

Answer 1

To solve this question, we need to use the formulae for the translational and the kinetic energy to find out the total energies of the given bodies at the bottom. Then using the conservation of energy, we can obtain the translational kinetic energy for each body in terms of the radius of gyration. Finally, we can obtain the required ratio from these values.

Complete Step-by-Step solution:
Let us suppose that the height of the given inclined plane be $h$ . Also, let us take the potential energy zero at the bottom of the plane.
So, if the mass of the ring, the coin and the solid ball is $m$ each, then the potential energy of each of these at the top of the incline must be
$U = mgh$ ......................(1)
Therefore, the change in the potential energy is given by
Now, at the bottom of the incline, they will only have the rotational and the kinetic energies. Let the translational velocity at the bottom be $v$ and the angular velocity be $\omega$ .
If the radius of each of the three objects is equal to $r$ , then the translational and the angular velocities can be related by
$v = \omega r$ ......................(2)
We know that the translational kinetic energy is given by
${K_T} = \dfrac{1}{2}m{v^2}$ ......................(3)
Also, we know that the rotational kinetic energy is given by
${K_R} = \dfrac{1}{2}I{\omega ^2}$
We know that the moment of inertia is given by
$I = m{k^2}$ ( $k$ is the radius of gyration)
So the rotational kinetic energy becomes
${K_R} = \dfrac{1}{2}m{k^2}{\omega ^2}$
So the total energy at the bottom becomes
$K = {K_T} + {K_R}$
Putting (3) and (4) in the above equation, we get
$K = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{k^2}{\omega ^2}$
Taking $\dfrac{1}{2}m{v^2}$ common on the right hand side
$K = \dfrac{1}{2}m{v^2}\left( {1 + {k^2}{{\left( {\dfrac{\omega }{v}} \right)}^2}} \right)$
Putting (2) in the above expression, we get
$K = \dfrac{1}{2}m{v^2}\left( {1 + {k^2}{{\left( {\dfrac{\omega }{{\omega r}}} \right)}^2}} \right)$
$\Rightarrow K = \dfrac{1}{2}m{v^2}\left( {1 + {{\left( {\dfrac{k}{r}} \right)}^2}} \right)$
From (3) we can write the above expression as
$K = {K_T}\left( {1 + {{\left( {\dfrac{k}{r}} \right)}^2}} \right)$ ......................(4)
By conservation of energy, the total energy at the top is equal to the total energy at the bottom, that is,
$K = U$
Putting (1) and (4)
${K_T}\left( {1 + {{\left( {\dfrac{k}{r}} \right)}^2}} \right) = mgh$
$\Rightarrow {K_T} = \dfrac{{mgh}}{{\left( {1 + {{\left( {k/r} \right)}^2}} \right)}}$
Now, we know that the radius of gyration for a ring is equal to $r$ . Substituting this above, we get the translational kinetic energy for the ring as
${K_{TR}} = \dfrac{{mgh}}{{\left( {1 + {{\left( {r/r} \right)}^2}} \right)}}$
$\Rightarrow {K_{TR}} = \dfrac{{mgh}}{2}$ ......................(5)
Similarly, we get the translational kinetic energies for the coin and the solid ball respectively as
${K_{TC}} = \dfrac{{2mgh}}{3}$ ......................(6)
${K_{TC}} = \dfrac{{5mgh}}{7}$ ......................(7)
From (5), (6), and (7) we can write that
${K_{TR}}:{K_{TC}}:{K_{TB}} = \dfrac{{mgh}}{2}:\dfrac{{2mgh}}{3}:\dfrac{{5mgh}}{7}$
$\Rightarrow {K_{TR}}:{K_{TC}}:{K_{TB}} = \dfrac{1}{2}:\dfrac{2}{3}:\dfrac{5}{7}$
On simplifying, we finally get
${K_{TR}}:{K_{TC}}:{K_{TB}} = 21:28:30$
Thus, the required ratio of the translational kinetic energies of the given bodies is equal to $21:28:30$ .
Hence, the correct answer is option C.

Note:
We must not forget the fact that along with having translational kinetic energies, the bodies will also possess rotational kinetic energies at the bottom of the incline. If we forget this fact, then we will obtain the ratio as $1:1:1$ , which is incorrect.