A circular disc of radius R is removed from a bigger uniform circular disc of radius 2R such that the edge of the removed part coincides on outer edge of the given disc. The centre of mass of the new disc is $aR$ from the centre of the bigger disc. The value of $a$ is

Assume the areal mass density of the disc is σ (mass per unit area).

Area: A_large = π(2R)² = 4πR²

Mass: M_large = σ × A_large = 4πσR²

Area: A_small = πR²

Mass: M_small = σ × A_small = πσR²

The larger disc has its center of mass at its center (O). The smaller disc is removed, and its center lies at a distance R from O.

A full larger disc with mass M_large.

A negative mass -M_small representing the removed disc.

Step 3: Position of the Centre of Mass

Using the formula for the center of mass:

x_CM = (Σm_ix_i) / Σm_i

For this system:

Larger disc: M_large at x = 0.

Removed smaller disc: -M_small at x = R.

Substitute:

x_CM = [(M_large × 0) + (-M_small × R)] / (M_large - M_small)

Substitute the values of M_large and M_small:

x_CM = [-(πσR²) × R] / [4πσR² - πσR²]

x_CM = [-πσR³] / [3πσR²]

x_CM = -R / 3

The center of mass of the remaining disc is at a distance R / 3 from the center of the larger disc, opposite to the center of the removed smaller disc.

Final Answer:

The value of a is  1/3