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# A torque of $400\,Nm$ acting on a body of mass $40\,Kg$ produces an angular acceleration of $20\,rad\,{\sec ^{ - 2}}$. What is the moment of inertia of the body? And find the radius of gyration of the body. Verified
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Hint:When a body is in translational motion it moves with a linear acceleration while body rotates due to torque acting on it it produce an angular acceleration and the relation between torque and moment of inertia is related as $\vec \tau = I\vec \alpha$.

Let us assume that $I$ is the moment of inertia of the body and it’s given that:
The magnitude of angular acceleration of the body is $\alpha = 20\,rad\,{\sec ^{ - 2}}$.
The magnitude of torque acting on the body is $\tau = 400Nm$.
The mass of the given body is $m = 40\,Kg$.
Now, using the relation $\vec \tau = I\vec \alpha$ we get,
$400 = I \times 20$
$\therefore I = 20\,Kg{m^2}$
So, the moment of inertia of the body is $I = 20\,Kg{m^2}$.

Now, as we know that the general formula of Moment of inertia is written as,
$I = M{K^2}$
where $K$ denotes the radius of gyration
Now we have, the magnitude of moment of inertia is $I = 20\,Kg{m^2}$
Mass of the body is $m = 40\,Kg$
Putting these value in equation $I = M{K^2}$
We get,
$20 = 40 \times {K^2}$
$\Rightarrow K = \dfrac{1}{{\sqrt 2 }}$
$\therefore K = 0.707\,m$
So, the radius of gyration of the body is $K = 0.707m$

Hence, the moment of inertia of the body is $I = 20\,Kg\,{m^2}$ and the radius of gyration of the body is $K = 0.707\,m$.

Note: It should be remembered that, the radius of gyration of the body is the distance from the body to the axis of rotation which have same moment of inertia if whole mass of body considered to be act at that particular point and produce same moment of inertia with particular distance, this radius is called radius of gyration.