Question

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be:
$\begin{align}
  & \left( A \right)\dfrac{{{K}^{2}}+{{R}^{2}}}{{{R}^{2}}} \\
 & \left( B \right)\dfrac{{{K}^{2}}}{{{R}^{2}}} \\
 & \left( C \right)\dfrac{{{K}^{2}}}{{{K}^{2}}+{{R}^{2}}} \\
 & \left( D \right)\dfrac{{{R}^{2}}}{{{K}^{2}}+{{R}^{2}}} \\
\end{align}$

Answer 1

Hint: Radius of gyration of a body is understood because the radial distance to some extent which might have a flash of inertia is equivalent because the body's actual distribution of mass, if the total mass of the body were concentrated. So first calculate the translational and rotational kinetic energy and then calculate the fraction with total energy.

Formula used:
Translational kinetic energy (TKE): $\dfrac{1}{2}m{{v}^{2}}$
Rotational kinetic energy (RKE): $\dfrac{1}{2}I{{\omega }^{2}}$
Where:
$m$- mass of the object
$v$- velocity
$I$- moment of inertia
$\omega $- angular speed

Complete answer:
Let the mass of the ball $m$ and velocity is $v$.
The moment of Inertia \[\left( MOI \right)~=m{{K}^{2}}\]
Since the Ball is rolling without slipping, ⇒\[v=R\omega \]
The Translational Kinetic Energy \[\left( TKE \right)~=\dfrac{1}{2}m{{v}^{2}}\]
Rotational Kinetic Energy \[\left( RKE \right)~~=\dfrac{1}{2}I{{\omega }^{2}}=\dfrac{m{{K}^{2}}{{v}^{2}}}{2{{R}^{2}}}\]
∴ Total Energy \[\begin{align}
  & =\text{ }TKE\text{ }+\text{ }RKE~ \\
 & =\dfrac{1}{2}m{{v}^{2}}+\dfrac{m{{K}^{2}}{{v}^{2}}}{2{{R}^{2}}} \\
\end{align}\]​
⇒ Fraction of RKE associated with total energy
 $\begin{align}
  & =\dfrac{\dfrac{m{{K}^{2}}{{v}^{2}}}{2{{R}^{2}}}}{\dfrac{1}{2}m{{v}^{2}}+\dfrac{m{{K}^{2}}{{v}^{2}}}{2{{R}^{2}}}} \\
 & \Rightarrow \dfrac{{{K}^{2}}}{{{K}^{2}}+{{R}^{2}}} \\
\end{align}$ ​

Additional Information:
The mechanical work required for or applied during rotation is that the torque times the rotation angle. The instant power of an angularity accelerating body is calculated as the torque times the angular velocity. For free-floating (unattached) objects, the axis of rotation is usually around its center of mass.

Note:
Mathematically the radius of gyration is that the root means the square distance of the object's parts from either its center of mass or a given axis, counting on the relevant application. it's actually the perpendicular distance from point mass to the axis of rotation. One can represent a trajectory of a moving point as a body. Then the radius of gyration is often wont to characterize the standard distance traveled by now.