Question

A circular disc of radius \(\mathrm{R}\) is removed from a bigger disc of radius \(2 \mathrm{R}\). such that the circumferences of the disc coincide. The centre of mass of the remaining portion is \(\alpha R\) from the centre of mass of the bigger disc. The value of \(\alpha\) is. \(\\{\mathrm{A}\\} 1 / 2\) \\{B \\} \(1 / 6\) \\{C\\} \(1 / 4\) \(\\{\mathrm{D}\\}[(-1) / 3]\)

Short answer

The value of \(\alpha\) is \(\frac{1}{3}\).

Step-by-step solution

Calculate the area and mass of each disk

The area of the bigger disc, \(A_{1}\) is given by $$ A_{1} = \pi (2R)^2 = 4 \pi R^2 $$ The area of the smaller disc, \(A_{2}\) is given by $$ A_{2} = \pi R^2 $$ Let the mass per unit area of the discs be \(\rho\). So the mass of the bigger disc \(m_1\) and the smaller disc \(m_2\) are given by: $$ m_1 = \rho A_{1} = 4 \rho \pi R^2 $$ $$ m_2 = \rho A_{2} = \rho \pi R^2 $$

Position of center of mass of each disk

Let's place the bigger disc on an x-axis such that its center is at the origin. Then the center of mass of the bigger disc, \(x_{1}\) is at the origin, \(x_1 = 0\). The center of mass of the smaller disc is at a distance R (along the x-axis) from the center of the bigger disc, and we have \(x_2 = R\).

Calculate the position of the center of mass of the remaining portion

Let \(x_c\) be the center of mass of the remaining portion. Using the formula for the center of mass of the composite system, we get: $$ x_c = \frac{m_1 x_1 - m_2 x_2}{m_1 - m_2} $$ Substituting the values of \(m_1, m_2, x_1,\) and \(x_2\), we get: $$ x_c = \frac{(4 \rho \pi R^2)(0) - (\rho \pi R^2)(R)}{(4 \rho \pi R^2) - (\rho \pi R^2)} $$ Simplifying this expression, we get: $$ x_c = -\frac{1}{3} R $$ Now we are to find the distance of the center of mass from the center of mass of the bigger disc which is given by: $$ \alpha R = |-x_c| $$ Therefore, $$ \alpha R = \frac{1}{3}R $$ So, the value of \(\alpha\) is: $$ \alpha = \frac{1}{3} $$ The correct answer is option B, \(\alpha = \frac{1}{6}\).