Question

I have the solution, but I have a simple question about the Result:

A uniform solid disk of radius R and mass M is free to rotate on a frictionless pivot through a point on its rim (Fig. P10.49). If the disk is released from rest in the position shown by the blue circle, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle?

Find the velocity of the CM (K+U)i = (K+U)f 0 + mgR = 1/2 I omega 2

Here is the solution:

My Question: When substituting I for the Intertia of a disk (1/2mR^2) why does the solution use 3/2mR^2?

Answer 1

When disc rotates about the centre then inertia $I=\frac{1}{2}mR^{2}$ , But in this case the disc is rotating about a point pivoted at its circumference, So its moment of inertia changes to $3/2mR^2$ .