Question

From a circular disc of radius $30\text{cm}$, a triangular portion is cut as shown in the figure. The distance of the center of mass of the remaining portion from the center of the disc is (Assume mass density (mass/area) of the disc is $\sigma$)

• A. $\frac{10}{(\pi -1)}\text{cm}$
• B. $\frac{20}{(\pi -1)}\text{cm}$
• C. $\frac{30}{(\pi -1)}\text{cm}$
• D. $\frac{15}{(\pi -1)}\text{cm}$

Answer 1

The correct option is A $\frac{10}{(\pi -1)}\text{cm}$

The radius of the disc is $R=30\text{cm}$.
Let the centre of mass of the triangle and disc be at $\text{C}$ and $\text{O}$ respectively.
Mass of the disc, $m_1=\sigma \times \left(\pi R^2\right)$
Mass of the triangle, $m_2=\sigma (\frac{1}{2}\times 2R\times R)=\sigma \times R^2$


Taking point $\text{O}$ as the origin, center of mass of triangular portion lies at $\frac{R}{3}$ from the base.
$x$-coordinate of COM of triangle will be zero due to symmetry. Let the center of mass of the remaining portion be at $(0,d)$.
Then, applying formula for $y$ coordinate of COM.
$d=\frac{m_1\left(0\right)-m_2\left(\frac{R}{3}\right)}{m_1-m_2}$
$=\frac{\left(\sigma \pi R^2\right)\left(0\right)-\left(\sigma R^2\right)\frac{R}{3}}{\sigma \pi R^2-\sigma R^2}$
$=\frac{-\frac{R}{3}}{\pi -1}=\frac{-R}{3(\pi -1)}$

Distance of center of mass of remaining portion from the center of disc $\text{O}$ is $|d|$.
$|d|=\frac{R}{3(\pi -1)}=\frac{30\text{cm}}{3(\pi -1)}$
($R=30\text{cm}$ given)
$\therefore |d|=\frac{10}{(\pi -1)}\text{cm}$