The moment of inertia of a thin uniform circular disc of mass m and radius R about an axis, which is tangential to the circumference of the disc and parallel to the plane of disc, is

Solution:

To find the moment of inertia of a thin uniform circular disc about an axis tangential to its circumference and parallel to the plane of the disc, we can use the parallel axis theorem.

First, recall the moment of inertia of a thin uniform circular disc about an axis passing through its center and lying in its plane (i.e., along a diameter). This is given by:

$$I_{CM} = \frac{1}{4}mR^2$$

The axis about which we need to find the moment of inertia is tangential to the circumference and parallel to the plane of the disc. This means this axis is parallel to a diameter of the disc.

The distance (d) between the center of mass (CM) axis (a diameter) and the parallel tangential axis is equal to the radius of the disc, R.

According to the parallel axis theorem, if $I_{CM}$ is the moment of inertia of a body about an axis passing through its center of mass, then the moment of inertia (I) about another axis parallel to the CM axis at a distance 'd' from it is given by:

$$I = I_{CM} + Md^2$$

In this case, M = m (mass of the disc) and d = R (radius of the disc).

Substituting the values:

$$I = \frac{1}{4}mR^2 + mR^2$$

$$I = \left(\frac{1}{4} + 1\right)mR^2$$

$$I = \left(\frac{1+4}{4}\right)mR^2$$

$$I = \frac{5}{4}mR^2$$

Thus, the moment of inertia of the disc about the specified axis is $$\frac{5}{4}mR^2$$