Question

The disc of mass M with uniform surface mass density $\sigma$ is shown in the figure. The center of mass of the quarter disc (the shaded area) is at the position $\frac{xa}{3\pi },\frac{xa}{3\pi }$. $x$ is ____(Round off to the Nearest Integer) [a is an area as shown in the figure]

Answer 1

Step 1: Given data:

The mass of the disc$=M$

Surface mass density$=\sigma$

Step 2: Formula used:

x coordinate of the center of mass of the quarter disc.

$X_{cm}=\frac{\int xdm}{\int dm}$

Step 3: Calculation:

Mass of a segment of disc subtending angle $d\theta$

From the formula of density

$$\begin{gathered} massdensity=\frac{mass}{volume} \\ \sigma =\frac{dm}{{\displaystyle \frac{1}{2}}R.Rd\theta }....................\left(1\right) \end{gathered}$$

The volume is taken as the product of the area of a triangle having a base $R$ and arc subtended by an angle $d\theta$which is a height of a triangle having unit thickness

Volume came out as

$$\begin{gathered} \frac{1}{2}R\left(arc\right).1angle=\frac{arc}{radius} \\ \frac{1}{2}R\times \left(Rd\theta \right)\times 1 \end{gathered}$$

From equation (1)

$\begin{array}{rcl}dm& =& \sigma \left(\frac{1}{2}\right)R\times Rd\theta \\ & =& \frac{\left(\sigma R^{2}d\theta \right)}{2}\end{array}$

Therefore, $x$ coordinate of COM of quarter disc

$X_{cm}=\frac{\int xdm}{\int dm}$

In this case, x is the COM of dm mass's x-coordinate. Additionally, for a triangle section, the center of mass (COM) is at the cross-section of the median, which will be $2R/3$ from the vertex. Because of this, they chose to use $(2R/3)\cos \theta$ as the x-coordinate.

$X_{cm}=\frac{\int {\displaystyle \frac{2R}{3}}\cos \theta .{\displaystyle \frac{\sigma R^{2}d\theta }{2}}}{\int \frac{\sigma R^{2}d\theta }{2}}$

Taking $d\theta$ from zero to 90 degree

$\begin{array}{rcl}& & \begin{array}{l}{X}_{cm}=\frac{2R}{3}\frac{{\int }_0^{\pi /2}\cos \theta d\theta }{{\int }_0^{\pi /2}d\theta }\end{array}\end{array}$

$\begin{array}{rcl}\begin{array}{l}\begin{array}{l}{X}_{cm}=\frac{2R}{3}\frac{\sin {\theta }_0^{{\displaystyle \frac{\pi }{2}}}}{{\displaystyle \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\end{array}\end{array}& =& \frac{4R}{3\pi }\end{array}$

On comparing we get $x=4$

Thus, $x=4$