Question

A uniform rod of mass m is bent into the form of a semicircle of radius R. The moment of inertia of the rod about an axis passing through A and perpendicular to the plane of the paper is

Answer 1

Hint: We will assume a full circle and find the moment of inertia for that circle and then as both the halves will be symmetrical, we can halve the value of the moment of inertia that we will get for it. First, we will calculate the moment of inertia for an axis at the centre of the circle and perpendicular to the page and then use the law of parallel axis to calculate the moment of inertia at A.
Formula used:
Parallel axis theorem
\[{{I}_{p}}={{I}_{cm}}+M{{x}^{2}}\]

Complete answer:
When we assume it to be a complete circle, the mass of the whole ring will be double the mass of the semicircle. So, the moment of inertia about the centre of the ring and perpendicular to the page will be equal to
\[{{I}_{cm}}=M{{R}^{2}}\]
As M = 2m, we get the moment of inertia as \[{{I}_{cm}}=2m{{R}^{2}}\]. Now we will use the parallel axis theorem. The distance between the new parallel axis and the centre of the circle is equal to the radius of the circle. So, the moment of inertia about the point A will be
\[{{I}_{A}}={{I}_{cm}}+M{{R}^{2}}=2m{{R}^{2}}+2m{{R}^{2}}=4m{{R}^{2}}\]. As both the upper and the lower half are the same the moment inertia for one will be half of the total. So, the moment of inertia of the semicircle will be
\[I=\dfrac{{{I}_{A}}}{2}=\dfrac{4m{{R}^{2}}}{2}=2m{{R}^{2}}\]. Hence the correct answer will be \[2m{{R}^{2}}\]

Note:
Students must take care that the parallel axis theorem can only be applied when the moment of inertia is given along the axis passing through the centre. Here we can take half of the complete circle as for every point on one of the halves there will be some point on the other half that is at the same distance from point A.