Question

A wheel of mass $10\;kg$ has a moment of inertia of $160kgm^{2}$ about its own axis. The radius of gyration is:
\[\begin{align}
  & A.10m \\
 & B.4m \\
 & C.5m \\
 & D.6m \\
\end{align}\]

Answer 1

Hint: Moment of inertia is the property of a body to resist angular acceleration. It is also the sum of the product of the masses of every particle with the square of the distances from the chosen axis of rotation. It is also known as the angular mass or rotational inertia.

Formula used:
 $I=mk^{2}$

Complete step-by-step answer:
Moment of inertia is the resistance offered by an object against rotational acceleration. Let us consider the given wheel of mass of the to be $m$, and $r$ its radius. Then the moment of inertia $I$ experienced by the wheel during a rotation along its own axis is given by $I=mk^{2}$, where $k$ is the radius of gyration.
We know that the moment of inertia depends on the density of the material, the axis of rotation and the dimensions of the body, i.e. the shape and the size of the body. We also know that the moment of inertia depends mainly on the mass of the object and its distance from the axis of rotation.
Given that the mass of the wheel is $m=10\;kg$ and the moment of inertia experienced by the given wheel is $I=160kgm^{2}$.
Then substituting the values, we get $k^{2}=\dfrac{I}{m}=\dfrac{160}{10}=16$
$\implies k=\sqrt{16}=4m$
Thus the answer is option \[B.4m\]

So, the correct answer is “Option B”.

Note: Rotational inertia of wheel is $I=mk^{2}$, where $k$ is the radius of gyration, this value is unique for each object. Thus it is useful to remember the value of $k$ or $I$ for all the objects. Also, the question might look complex, but it can be solved easily, if the formulas are known. Not that here, the radius of the wheel is not considered.