Question

A circular plate A of radius 1.5r is removed from one edge of a uniform circular plate B of radius 2r. The distance of centre of mass of the remaining portion from the centre of the plate B is

• A. \( \frac{5r}{12} \)
• B. \( \frac{9r}{14} \)
• C. \( \frac{3r}{4} \)
• D. \( \frac{7r}{8} \)

Hint:

Consider the original plate as a combination of the removed part and the remaining part. Use the principle of superposition for the center of mass: \( M_{original} \vec{R}_{cm, original} = M_{removed} \vec{R}_{cm, removed} + M_{remaining} \vec{R}_{cm, remaining} \). Set the center of the original plate as the origin.

Answer 1

The Correct Option is B

Let the center of the larger circular plate B be the origin (0, 0).
The radius of plate B is \( R_B = 2r \).
The area of plate B is \( A_B = \pi (2r)^2 = 4\pi r^2 \).
Let the mass per unit area of the uniform plate be \( \sigma \).
The mass of plate B is \( M_B = \sigma A_B = 4\pi r^2 \sigma \).
The center of mass of plate B is at the origin (0, 0).
The circular plate A of radius \( R_A = 1.
5r = \frac{3}{2}r \) is removed from one edge of plate B.
The center of plate A is at a distance \( 2r - 1.
5r = 0.
5r = \frac{r}{2} \) from the center of plate B along the x-axis.
So, the center of mass of the removed portion A is at \( (\frac{r}{2}, 0) \).
The area of plate A is \( A_A = \pi (\frac{3}{2}r)^2 = \frac{9}{4}\pi r^2 \).
The mass of plate A is \( M_A = \sigma A_A = \frac{9}{4}\pi r^2 \sigma \).
Let the remaining portion be C.
The mass of the remaining portion is \( M_C = M_B - M_A = 4\pi r^2 \sigma - \frac{9}{4}\pi r^2 \sigma = \frac{16 - 9}{4}\pi r^2 \sigma = \frac{7}{4}\pi r^2 \sigma \).
Let the center of mass of the remaining portion C be at \( (x_{cm}, y_{cm}) \).
Using the formula for the center of mass of a system of particles: \( (M_B) (0) = (M_C) (x_{cm}) + (M_A) (\frac{r}{2}) \) (for the x-coordinate) \( 0 = (\frac{7}{4}\pi r^2 \sigma) x_{cm} + (\frac{9}{4}\pi r^2 \sigma) (\frac{r}{2}) \) \( 0 = \frac{7}{4} x_{cm} + \frac{9}{8} r \) \( -\frac{9}{8} r = \frac{7}{4} x_{cm} \) \( x_{cm} = -\frac{9}{8} r \times \frac{4}{7} = -\frac{9r}{14} \) Due to symmetry about the x-axis, the y-coordinate of the center of mass is \( y_{cm} = 0 \).
The distance of the center of mass of the remaining portion from the center of the plate B is \( |x_{cm}| = |-\frac{9r}{14}| = \frac{9r}{14} \).