Question

A circular plate of uniform thickness has diameter $56cm$. A circular part of diameter $42cm$ is removed from one edge. What is the position of the center of mass of the remaining part

• A. $3cm$
• B. $6cm$
• C. $9cm$
• D. $12cm$

Answer 1

The correct option is C $9cm$

New center of mass is given by:

$x_{com}=\frac{m_1{x}_1-m_2{x}_2}{m_1-m_2}$ where,

$m_1$ is mass of original disc

$m_2$ is mass of removed disc

$x_1$ is center of mass position of original disc

$x_2$ is center of mass position of removed disc

If mass per unit area=$m$ then,

mass of original disc=$m\times \pi \times (28\times {10}^{-2}{)}^2$

mass of removed disc=$m\times \pi \times (21\times {10}^{-2}{)}^2$

$x_{com}==\frac{m\times \pi \times (28\times {10}^{-2}{)}^2\times 0-m\times \pi \times (21\times {10}^{-2}{)}^2\times (7\times {10}^{-2})}{m\times \pi \times (28\times {10}^{-2}{)}^2-m\times \pi \times (21\times {10}^{-2}{)}^2}=\frac{-m\pi {(28\times {10}^{-2})}^2(7\times {10}^{-2})}{m\pi \left[(28\times {10}^{-2}{)}^2-(21\times {10}^{-2}{)}^2)\right]}=\frac{-0.003087}{0.0343}=-0.09m=-9cm$

Center of mass shifts by 9 cm.