Two circular rings of equal mass and radius are placed touching each other. The moment of inertia of the system about tangential axis in the plane of system passing through point of contact of rings is

The moment of inertia (I) of a single ring about an axis through its center and perpendicular to the plane of the ring is given by the formula I = MR^2, where M is the mass of the ring and R is its radius.

However, in this case, we are asked to find the moment of inertia about an axis that is tangential to the rings and passes through their point of contact. This axis is at a distance R from the center of each ring.

We can use the parallel axis theorem to find the moment of inertia about this new axis. The parallel axis theorem states that the moment of inertia about any axis parallel to and a distance d away from an axis through the center of mass is given by I = I_cm + Md^2, where I_cm is the moment of inertia about the axis through the center of mass.

For each ring, I_cm = MR^2, M is the mass, and d = R. So, the moment of inertia of each ring about the new axis is I = MR^2 + MR^2 = 2MR^2.

Since there are two rings, the total moment of inertia of the system about the tangential axis is 2I = 2(2MR^2) = 4MR^2.