Question

The mass of a circular ring is M and its radius is R. Find its moment of inertia is about an axis in the plane of ring at a perpendicular distance 2R​ from the centre of ring.
A) \[\dfrac{{{\rm{M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{4}}}\]
B) \[\dfrac{{{\rm{M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{2}}}\]
C) \[\dfrac{{{\rm{3M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{2}}}\]
D) \[\dfrac{{{\rm{3M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{4}}}\]

Answer 1

Hint:The moment of inertia of a point mass with respect to the axis is defined as the product of the mass multiplied by the distance from the axis square.

Step by step solution:
\[{\text{array}}{l}{{\rm{I}}_{{\rm{cm}}}}{\rm{ = }}\dfrac{{{\rm{M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{2}}}\\{\rm{\text{According to parallel axis theorem}}}\\
\Rightarrow {\rm{I = }}{{\rm{I}}_{{\rm{cm}}}}{\rm{ + M}}{{\rm{d}}^{\rm{2}}}\\{\rm{where d = }}\dfrac{{\rm{R}}}{{\rm{2}}}\\
\Rightarrow {\rm{I = }}\dfrac{{{\rm{M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ + }}\dfrac{{{\rm{M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{4}}}\\
\Rightarrow \dfrac{{{\rm{3M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{4}}}\\\therefore {\rm{\text{Moment Of Inertia} = }}\dfrac{{{\rm{3M}}{{\rm{R}}^{\rm{2}}}}}{{\rm{4}}}{array}\]

Option D is correct

Additional Information:The moment of inertia is the property where the object resists a change in the state of its rotational motion. The larger the moment of inertia, the greater the amount of torque required to bring about the same change in its angular momentum at a given time. Here, the angular analogs of torque and angular velocity force and velocity are similarly similar to the moments of inertia which are related to the masses of force and velocity.
Perpendicular axis theorem: The moment of inertia (MI) of a plane area about an axis which is perpendicular to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and passing through the given axis.This means that every moment of inertia.
$z = i_x + I_y$
Parallel axis theorem: The moment of area of an object about any axis parallel to the centroidal axis is the sum of MI about its centroidal axis and the product of area with the square of distance of from the reference ax
Moment of inertia = $I_{xx} + I_G +Ad^2$

Notes:Unlike inertia, the moment of inertia depends not only on mass, but also on the distribution of mass around the axis about which the inertia must be calculated. An object may have different moments of inertia about different axes. That is, to rotate an object about different axes with the same angular acceleration, different torque (or effort) is required. This concept is relevant and highly essential in whole mechanics.
While life will be simpler if nothing is rotated, realistically we need to have a way of dealing with both translation and rotation (often at the same time). This is an essential piece in analyzing more complex motions