
Moment of inertia of a circular loop of radius R about the axis in the above figure is:

A. \[M{R^2}\]
B. \[\dfrac{3}{4}M{R^2}\]
C. \[\dfrac{{M{R^2}}}{2}\]
D. \[2M{R^2}\]
Answer
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Hint:To calculate moment of inertia for rotation about an axis through an object, two theorems are used. One is the parallel axis theorem and another is the perpendicular axis theorem. Both are used depending upon the situations given in the question. In this question, we will use the parallel axis theorem to calculate the moment of inertia of this object about the given axis.
Formula Used:
The moment of inertia, \[I\] of the circular loop due to rotation about an axis passing through its centre O is given by:
\[{I_c} = \dfrac{{M{R^2}}}{2}\] ---- (1)
Theorem of parallel axis,
\[I = {I_C} + M{h^2}\]
Where, \[I\]= Moment of inertia about the given axis
\[{I_C}\]= Moment of inertia about the axis passing through the centre of the body
M = mass of the body
h = distance between the two axes
R = the radius of the object.
Complete step by step solution:
The moment of inertia, \[{I_c}\] of the circular loop due to rotation about an axis passing through its centre O is given by:
\[{I_c} = \dfrac{{M{R^2}}}{2}\] ---- (1)
From the theorem of parallel axis,
\[I = {I_C} + M{h^2}\]---- (2)
In the given question, the new reference axis is at a distance \[R/2\] from the axis passing through the centre O of the loop,
Hence \[h = R/2\]---- (3)
Substituting values from equations (3) and (1) in (2), we get,
\[I = \dfrac{{M{R^2}}}{2} + M{\left( {\dfrac{R}{2}} \right)^2}\]
\[\Rightarrow I = \dfrac{{M{R^2}}}{2} + \dfrac{{M{R^2}}}{4}\]
\[\Rightarrow I = \dfrac{{3M{R^2}}}{4}\]
\[\therefore I = \dfrac{3}{4}M{R^2}\]
Hence option B is the correct answer.
Note: If we have to calculate the moment of inertia about an axis parallel to the axis passing through the centre of the object, we use the theorem of the parallel axis. When the body has symmetry in about two out of three axes in a 3D plane, the perpendicular axis theorem is used. We have to calculate the perpendicular distance between the particle and the axis of rotation to determine the moment of inertia of a complete system of particles. In this way, a moment of inertia of a particular solid depends on the dimensions of the solid as well as the axis of rotation.
Formula Used:
The moment of inertia, \[I\] of the circular loop due to rotation about an axis passing through its centre O is given by:
\[{I_c} = \dfrac{{M{R^2}}}{2}\] ---- (1)
Theorem of parallel axis,
\[I = {I_C} + M{h^2}\]
Where, \[I\]= Moment of inertia about the given axis
\[{I_C}\]= Moment of inertia about the axis passing through the centre of the body
M = mass of the body
h = distance between the two axes
R = the radius of the object.
Complete step by step solution:
The moment of inertia, \[{I_c}\] of the circular loop due to rotation about an axis passing through its centre O is given by:
\[{I_c} = \dfrac{{M{R^2}}}{2}\] ---- (1)
From the theorem of parallel axis,
\[I = {I_C} + M{h^2}\]---- (2)
In the given question, the new reference axis is at a distance \[R/2\] from the axis passing through the centre O of the loop,
Hence \[h = R/2\]---- (3)
Substituting values from equations (3) and (1) in (2), we get,
\[I = \dfrac{{M{R^2}}}{2} + M{\left( {\dfrac{R}{2}} \right)^2}\]
\[\Rightarrow I = \dfrac{{M{R^2}}}{2} + \dfrac{{M{R^2}}}{4}\]
\[\Rightarrow I = \dfrac{{3M{R^2}}}{4}\]
\[\therefore I = \dfrac{3}{4}M{R^2}\]
Hence option B is the correct answer.
Note: If we have to calculate the moment of inertia about an axis parallel to the axis passing through the centre of the object, we use the theorem of the parallel axis. When the body has symmetry in about two out of three axes in a 3D plane, the perpendicular axis theorem is used. We have to calculate the perpendicular distance between the particle and the axis of rotation to determine the moment of inertia of a complete system of particles. In this way, a moment of inertia of a particular solid depends on the dimensions of the solid as well as the axis of rotation.
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