Determine the position of the centre of mass for a uniform plate...

Let the surface mass density of the uniform plate be $\sigma$.

By symmetry, the centre of mass lies on the y-axis, perpendicular to the straight edge and passing through the centre.

Consider an elemental semicircular ring of radius $r$ and thickness $dr$.

The mass of this elemental ring is $dm = \sigma (\pi r dr)$.

The y-coordinate of the centre of mass of a semicircular ring of radius $r$ is $y_c = \dfrac{2r}{\pi}$.

The y-coordinate of the centre of mass of the entire plate is given by:

$y_{cm} = \dfrac{\int y_c dm}{\int dm}$

Substituting the expressions for $y_c$ and $dm$:

$y_{cm} = \dfrac{\int_{R_1}^{R_2} \left(\dfrac{2r}{\pi}\right) (\sigma \pi r dr)}{\int_{R_1}^{R_2} \sigma \pi r dr}$

$y_{cm} = \dfrac{\int_{R_1}^{R_2} 2 r^2 dr}{\int_{R_1}^{R_2} \pi r dr}$

Evaluating the integrals:

$y_{cm} = \dfrac{2 \left[ \dfrac{r^3}{3} \right]_{R_1}^{R_2}}{\pi \left[ \dfrac{r^2}{2} \right]_{R_1}^{R_2}}$

$y_{cm} = \dfrac{\dfrac{2}{3} (R_2^3 - R_1^3)}{\dfrac{\pi}{2} (R_2^2 - R_1^2)}$

$y_{cm} = \dfrac{4(R_2^3 - R_1^3)}{3\pi(R_2^2 - R_1^2)}$

Using the algebraic identities $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and $a^2 - b^2 = (a - b)(a + b)$:

$y_{cm} = \dfrac{4(R_2 - R_1)(R_2^2 + R_1 R_2 + R_1^2)}{3\pi(R_2 - R_1)(R_2 + R_1)}$

$y_{cm} = \dfrac{4(R_1^2 + R_1 R_2 + R_2^2)}{3\pi(R_1 + R_2)}$