Question

The MI of the solid sphere about the tangent is $ 70kg{m^2} $ . It's MI about any diameter is –
(A) $ 25kg{m^2} $
(B) $ 20kg{m^2} $
(C) $ 50kg{m^2} $
(D) $ 15kg{m^2} $

Answer 1

Hint: From the parallel axis theorem, the moment of inertia about the tangent will be equal to the sum of the moment of inertia across a diameter and the product of the mass and the distance of the tangent from the diameter. From there we can calculate the moment of inertia about the diameter.

Formula Used: In the solution we will be using the following formula,
 $\Rightarrow MI = M{I_{cm}} + M{R^2} $
Where $ MI $ is the moment of inertia about any axis, and $ M{I_{cm}} $ is the moment of inertia about the center of mass of the body.

Complete step by step answer:
The parallel axis theorem is used to determine the moment of inertia of any rigid body about any axis using the moment of inertia of the body that is passing through the center of gravity of the body, and the perpendicular distance between the two axes.
In this problem, we are given the moment of inertia about a tangent to the sphere and we need to find the moment of inertia about any diameter of the sphere. Now the axis which is the diameter of the sphere will be the axis which is passing through the center of mass of the sphere.
The moment of inertia of a sphere about an axis that is passing through its center of mass is given as,
 $\Rightarrow M{I_{cm}} = \dfrac{2}{5}M{R^2} $
And the moment of inertia about the tangent is given as,
 $\Rightarrow MI = 70kg{m^2} $
The formula for the parallel axis theorem is given as,
 $\Rightarrow MI = M{I_{cm}} + M{R^2} $
So substituting the values in the formula we get,
 $\Rightarrow 70 = \dfrac{2}{5}M{R^2} + M{R^2} $
Therefore on taking common in the RHS we have,
 $\Rightarrow 70 = \left( {\dfrac{2}{5} + 1} \right)M{R^2} $
This gives us,
 $\Rightarrow 70 = \left( {\dfrac{{2 + 5}}{5}} \right)M{R^2} = \dfrac{7}{5}M{R^2} $
So we get,
 $\Rightarrow M{R^2} = 70 \times \dfrac{5}{7} $
Therefore we can now substitute this value in the formula for the moment of inertia about the diameter and get,
 $\Rightarrow M{I_{cm}} = \dfrac{2}{5} \times 70 \times \dfrac{5}{7} $
On calculating we have,
 $\Rightarrow M{I_{cm}} = 20kg{m^2} $ .
So the moment of inertia about the diameter is $ 20kg{m^2} $ .
Hence, the correct answer is option B.

Note:
The moment of inertia of a rigid body is the amount of torque that is required for a desired angular acceleration of the body about a rotational axis. It is similar to how the mass of a body determines the force required for desired acceleration.