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Chapter 5: Problem 568

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... $\begin{array}{llll}\\{\mathrm{A}\\}-0.4 \mathrm{~cm} & \\{\mathrm{~B}\\}-2.4 \mathrm{~cm} & \\{\mathrm{C}\\}-1.8 \mathrm{~cm} & \\{\mathrm{D}\\} & 1.2 \mathrm{~cm}\end{array}$

Short Answer

Expert verified
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 by step solution

01

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#
02

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.

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Most popular questions from this chapter

The M.I. of a body about the given axis is \(1.2 \mathrm{kgm}^{2}\) initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \mathrm{~J}\(. an angular acceleration of \)25 \mathrm{rad} \mathrm{sec}^{2}$ must be applied about that axis for duration of \(\ldots\) \(\\{\mathrm{A}\\} 4 \mathrm{sec}\) \\{B \(\\} 2 \mathrm{sec}\) \(\\{C\\} 8 \mathrm{sec}\) \(\\{\mathrm{D}\\} 10 \mathrm{sec}\)

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