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Q6.15:
From a uniform disk of radius (R), a circular hole of radius (\frac{R}{2}) is cut out. The centre of the hole is at (\frac{R}{2}) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Solution :

Step 1: Establishing the Coordinate System and Mass Distribution

Let the original uniform disk have radius $R$ and mass $M$. We place the origin $(0,0)$ at the center of this original disk. The surface mass density ($\sigma$) is given by:

$\sigma = \frac{M}{\pi R^2}$

The hole has a radius $r = \frac{R}{2}$. Its area $A_{hole}$ is $\pi (\frac{R}{2})^2 = \frac{\pi R^2}{4}$. Thus, the mass of the material removed ($m$) is:

$m = \sigma \cdot A_{hole} = \left( \frac{M}{\pi R^2} \right) \cdot \left( \frac{\pi R^2}{4} \right) = \frac{M}{4}$

The center of the hole is located at coordinates $(\frac{R}{2}, 0)$ [Per problem statement].

Step 2: Defining the Principle of Superposition for Center of Mass

The center of mass of the remaining body ($X_{com}$) can be calculated by considering the original disk as the sum of the remaining body and the removed circular piece. By the principle of moments [By Definition of Center of Mass]:

$M \cdot X_{original} = m_{remaining} \cdot X_{com} + m \cdot X_{hole}$

Where $m_{remaining} = M - m = M - \frac{M}{4} = \frac{3M}{4}$.

Step 3: Calculating the Resultant Center of Mass

Given the symmetry, $X_{original} = 0$. Substituting the known values:

$M(0) = \left( \frac{3M}{4} \right) X_{com} + \left( \frac{M}{4} \right) \left( \frac{R}{2} \right)$

Rearranging the equation to solve for $X_{com}$:

$\left( \frac{3M}{4} \right) X_{com} = - \left( \frac{MR}{8} \right)$

$X_{com} = - \left( \frac{MR}{8} \right) \cdot \left( \frac{4}{3M} \right)$

$X_{com} = -\frac{R}{6}$

Step 4: Visualization of the Geometry

Origin (0,0) Hole (R/2, 0) CoM (-R/6, 0)

Analytical Summary

The calculation demonstrates that the center of mass shifts in the direction opposite to the center of the removed hole. Since the hole was removed from the positive $x$-axis, the shift towards the negative $x$-axis is consistent with the physics of mass deficiency [By Law of Conservation of Center of Mass].

Final Solution: The center of gravity of the resulting body is located at a distance of $\frac{R}{6}$ from the center of the original disk, in the direction opposite to the center of the hole.


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