Moment of inertia of a 'non-uniform' ring of mass M and radius R about axis through centre and normal to its plane is
To find the moment of inertia of a non-uniform ring with respect to an axis through its center and normal to its plane, you need to consider an element of mass on the ring and integrate over the entire ring.
Let's say the ring has a thickness dr and is located at a distance r from the center. The mass of this element can be expressed as dm = λdl, where λ is the linear mass density (mass per unit length) of the ring and dl is a length element along the circumference.
Now, the moment of inertia of this small element about the axis is given by dI = dm * r^2. Substituting dm = λdl and r^2 = r^2, we get dI = λr^2 dl.
To find the moment of inertia of the entire ring, we need to integrate this expression over the entire ring (from 0 to R for the radius), taking into account the linear mass density distribution along the circumference.
The linear mass density λ can be expressed in terms of the total mass M and the length of the ring L using λ = M / L.
Therefore, the moment of inertia I can be calculated as follows:
I = ∫(0 to R) λr^2 dl
= ∫(0 to R) (M / L) r^2 dl
To proceed further, we need to know the specific distribution of linear mass density along the circumference of the non-uniform ring. Without this information, we cannot provide a numerical answer.