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Answer

Let AB be the rod of mass M and length L at an angle θ with the axis of rotation z.

Assume an infinitesimally small piece of length dx at a distance x from A as shown in the figure.

To find the moment of inertia I of the rod about the axis of rotation, we need to find the moment of inertia dI (about the axis of rotation) of this small piece of length in terms of variable x & other given constants and integrate the result over the length AB of the rod.

Linear density of the rod ρ = M/L

Mass of the small piece dm = ρ dx = (M/L)dx

dI = dm (r2) = (M/L)dx (x cos θ)2

dI = [(M cos2 θ)/L] x2 dx

 Integrating with respect to x within limits 0 to L

I = (1/3) ML2 cos2 θ

Answer

Moment of inertia of a rod of length r = 1/3 (mr2)

Given, mass of rod = m, length = l and < of rotation = theta

Let, the radius of the circle of raotation be r.

Thus, r = l cos theta

and I = 1/3 (l cos theta)2

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