Question

From a uniform disk of radius R, a circular section of radius R/2 is cut out, the centre of the hole is at R/2, from the centre of the original disc. Locate the centre of mass of the resulting flat body.

Answer 1

Let mass per unit area of the original disc = σ

Thus mass of original disc = M = σπR² 

Radius of smaller disc = R/2.

Thus mass of the smaller disc = σπ(R/2)² = M/4

After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:

M(concentrated at O), and -M(=M/4) concentrated at O'

(The negative sign indicates that this portion has been removed from the original disc.)
Let x be the distance through which the centre of mass of the remaining portion shifts from point O.

The relation between the centres of masses of two masses is given as:

\(x = \frac{(m_1r_1 + m_2r_2)}{m_1 + m_2}\)

\(= (M \times 0 - \frac{M}{4}) \times \frac{\frac{R}{2}}{M - \frac{M}{4}} = \frac{-R}{6}\)

(The negative sign indicates that the centre of mass gets shifted toward the left of point O)