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Q7.16:
From a uniform disk of radius $R$, a circular hole of radius $R/2$ is cut out. The centre of the hole is at $R/2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Solution :

Step 1: Assigning Initial Parameters

Consider a uniform circular disk of radius $R$ and surface mass density $\sigma$. Let the center of the original disk be at the origin $(0, 0)$ in the $XY$-plane.

  • Mass of the original complete disk ($M_1$): $M_1 = \sigma \cdot \pi R^2$
  • Center of mass of the original disk ($C_1$): $(0, 0)$
  • Radius of the circular hole: $r = R/2$
  • Mass of the removed circular hole ($M_2$): $M_2 = \sigma \cdot \pi (R/2)^2 = \frac{1}{4} \sigma \pi R^2 = \frac{M_1}{4}$
  • Center of mass of the circular hole ($C_2$): $(R/2, 0)$

Step 2: Defining the Resultant Body

The resulting body is the original disk minus the mass of the hole. Let $M$ be the mass of the remaining body and $(X, Y)$ be its center of mass.

Mass of the remaining part ($M_{rem}$): $M_{rem} = M_1 - M_2 = M_1 - \frac{M_1}{4} = \frac{3M_1}{4}$

Step 3: Applying the Principle of Moments

Using the definition of the center of mass for a system of particles (or continuous mass distributions), the center of mass of the original disk can be expressed as the weighted average of the remaining part and the removed part:

$X_{CM} = \frac{M_{rem} \cdot X + M_2 \cdot X_2}{M_{rem} + M_2}$

Since the original center of mass is at $(0, 0)$, we have $X_{CM} = 0$. Substituting the values:

$0 = \frac{(\frac{3M_1}{4}) \cdot X + (\frac{M_1}{4}) \cdot (\frac{R}{2})}{\frac{3M_1}{4} + \frac{M_1}{4}}$

Step 4: Algebraic Derivation

Multiplying by the denominator and simplifying the $M_1$ terms:

$0 = \frac{3}{4} M_1 X + \frac{1}{8} M_1 R$

$\frac{3}{4} X = -\frac{1}{8} R$

$X = -\frac{1}{8} R \cdot \frac{4}{3} = -\frac{R}{6}$

Since the problem is symmetric about the $X$-axis, the $Y$-coordinate of the center of mass remains $Y = 0$.

O(0,0) C2(R/2,0) CM(-R/6, 0)

Step 5: Final Result

The negative sign indicates that the center of mass of the resulting body is shifted to the left of the original center, along the line connecting the centers, away from the hole.

Final Solution: The center of gravity of the resulting body is at a distance of $R/6$ from the centre of the original disk, in the direction opposite to the center of the hole.


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