Problem 568 - JEE Physics

Question

A small disc of radius \(2 \mathrm{~cm}\) is cut from a disc of radius $6 \mathrm{~cm}\(. If the distance between their centres is \)3.2 \mathrm{~cm}$, what is the shift in the centre of mass of the disc... $\begin{array}{llll}\\{\mathrm{A}\\}-0.4 \mathrm{~cm} & \\{\mathrm{~B}\\}-2.4 \mathrm{~cm} & \\{\mathrm{C}\\}-1.8 \mathrm{~cm} & \\{\mathrm{D}\\} & 1.2 \mathrm{~cm}\end{array}$

Step-by-Step Solution

Answer

The calculated shift in the center of mass is approximately \(-4.27 \mathrm{~cm}\), which does not match any of the given options. There might be a mistake in the problem or the options.

Step 1: First, let's assume that the discs have the same thickness and uniform density. Then, their masses will be directly proportional to their areas. Let's calculate the areas: - Area of the smaller disc: \(A_1 = \pi r_1^2 = \pi (2)^2 = 4\pi\) - Area of the larger disc: \(A_2 = \pi r_2^2 = \pi (6)^2 = 36\pi\) Let the masses of the two discs be - \(m_1 = k (4\pi)\), where k is a constant of proportionality. - \(m_2 = k (36\pi - 4\pi) = k (32\pi)\), as the larger disc has the smaller disc cut from it. #Step 2: Calculate the positions of their centers of mass relative to a reference point#

Let's use the center of the larger disc as the reference point. - Position of the center of mass of the smaller disc: \(x_1 = 3.2\) cm - Position of the center of mass of the larger disc: \(x_2 = 0\) cm #Step 3: Calculate the position of the center of mass of the system#

Step 2: Using the formula for the position of the center of mass of the system, we get: \(x_M = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{4\pi k (3.2) + 0}{4\pi k + 32\pi k} = \frac{12.8\pi k}{36\pi k} = \frac{12.8}{36}\) #Step 4: Calculate the shift in the center of mass#

Using the formula mentioned in the analysis, we get: \(\Delta x = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} - \frac{M x_M}{M} = \frac{12.8\pi k}{36\pi k} - \frac{(4\pi k + 32\pi k) \left(\frac{12.8}{36}\right)}{36\pi k}\) Simplifying, we get: \(\Delta x = \frac{12.8}{36} - \frac{(36\pi k) \left(\frac{12.8}{36}\right)}{36\pi k}\) \(\Delta x = \frac{12.8}{36} - \frac{12.8}{36}\) \(\Delta x = 0 - 12.8 \left(\frac{1}{36}\right)\) \(\Delta x = - \frac{1}{3} \times 12.8 \) \(\Delta x = - \frac{12.8}{3}\) \(\Delta x \approx - 4.27\) However, note that we cannot find a matching answer in the options provided. There might be a mistake in the problem or options.