# An Intuitive Understanding of ${{{k^2}} \over {{R^2}}}$

An Example Question-
Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

We can easily derive the expression for maximum velocity attained - $${\rm{v = }}\sqrt {{{{\rm{2gh}}} \over {\left( {{\rm{1 + }}{{{{\rm{k}}^{\rm{2}}}} \over {{{\rm{R}}^{\rm{2}}}}}} \right)}}}$$So the object that has the least k2/R2 value attains the maximum velocity.

We expect Moment of Inertia to be the factor that decides the velocity attained by an object in rolling motion. But two objects can have the same moment of inertia and still need not attain the same velocity after rolling certain distance. Its all dependent on the k2/R2 value of the particular object.
What does k2/R2 (k - Radius of Gyration and R - Radius of The Object) mean in real world and how to think about it intuitively?

Among Two Bodies having same $Radius\;of\;Gyration$, the one having greater radius attains the maximum velocity. Why is that?

The radius of gyration tells you something about the distribution of mass about an axis of rotation.
The "further" the mass is from the axis of rotation the larger is the radius of gyration.

As a body rolls down a slope without slipping it gains both translational kinetic energy $\frac 12 mv^2_{\rm com}$ and rotational kinetic energy $\frac 12 I\omega^2$ with the constraint that $v_{\rm com}=R\omega$.

The larger $k$ and hence the moment of inertia $mk^2$.
This would mean a greater proportion of the available energy (from the loss of gravitational potential energy decrease) would go into increasing the rotational kinetic energy rather than the translational kinetic energy.

For the ring $k=R$ as all the mass is distributed at a distance $R$ from the axis.

The cylinder of radius $R$ obviously has some mass closer to the axis than the ring so its radius of gyration will be less.
Then for a sphere there is even more mass nearer to the axis of rotation.

When a body is rotating the further the mass is away from the axis of rotation the faster it is moving and so its contribution to the rotational kinetic energy is greater.

The ratio $\frac kR$ is the factor which shows how the mass is distributed relative to the radius of the body.
If $k$ is kept constant and $R$ is increased this implies that more mass is concentrated nearer the axis of rotation.
Thus less rotational kinetic energy is gained compared to the gain in translational kinetic energy - the translational speed of the centre of mass of the body will be larger.

• 'If $k$ is kept constant and $R$ is increased this implies that more mass is concentrated nearer the axis of rotation.' - Isn't mass always assumed to be concentrated at the radius of gyration? If radius of gyration is constant, isn't distribution of mass also at a constant distance from the axis? – DoubtExpert Mar 27 '17 at 17:01
• @DoubtExpert I was referring to the original object's mass distribution which can of course be equated to a ring of mass with a radius equal to the radius of gyration. – Farcher Mar 27 '17 at 17:07
• My textbook defines radius of gyration to be the distance of a point where entire mass of the object is assumed to be concentrated such that its moment of inertia is same as that of the rigid object. – DoubtExpert Mar 27 '17 at 17:16
• @DoubtExpert For this example I wanted to choose a symmetrical shape about the axis which could roll so rather than choose a point mass I chose a ring. – Farcher Mar 27 '17 at 17:24
• Why do we require ${k \over R}$? Isn't $k$ enough to give an idea of average distribution of mass around the axis of rotation? And I don't think it's possible to have two objects having same $k$ but different $R$ because $k$ is dependent on $R$, so whenever $R$ changes, $k$ changes proportionally. – DoubtExpert Mar 28 '17 at 8:18

Rewrite your equation as $$\frac{m}{2}v^2+\frac{m}{2}\frac{k^2}{R^2}v^2=mgh$$ Clearly the right hand side is the potential energy. On the left, the term $T_{trans}=\frac{m}{2}v^2$ is the translational part of the kinetic energy. The term $\frac{m}{2}\frac{k^2}{R^2}v^2$ is the rotational part of the kinetic energy under the no-slip condition where $v=\omega R$ so this term is basically $$T_{rot}=\frac{1}{2}I\omega^2 \, ,\qquad mk^2=I$$ Thus, the body with the largest $v$ will win the race, which mean the body with the smallest $I$ since less of the potential energy will be converted into rotational kinetic energy.

• You do not answer my question! – DoubtExpert Mar 27 '17 at 17:08