Mechanics
From the circular disk of diameter 100mm is cut out a circle whose diameter is the radius of the disk. F ind the center of gravity of remaining portion.
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From the circular disk of diameter 100mm is cut out a circle whose diameter is the radius of the disk. F ind the center of gravity of remaining portion.
Ah! You mean this?
Ah! You mean this?
To get the coordinates of centroid of the figure above, we need to use the formula
$$A \bar x = A_1x_1 + A_2x_2 + A_3x_3 +A_4x_4 +....A_n x_n$$ $$A \bar y = A_1y_1 + A_2y_2 + A_3y_3 +A_4y_4 +....A_n y_n$$
where:
$A = $ area of the whole figure
$x = $ distance of the centroid from the $x$-axis
$A_1 = $ area of one part of the figure
$x_1 = $ distance of its particular centroid from the $x$-axis
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$y = $ distance of the centroid from the $y$-axis
$A_1 = $ area of one part of the figure
$y_1 = $ distance of its particular centroid from the $y$-axis
The formula that is applicable to the figure above would be:
$$A \bar x = A_1x_1 + A_2x_2 + A_3x_3 +A_4x_4 $$ $$A \bar y = A_1y_1 + A_2y_2 + A_3y_3 +A_4y_4 $$
Redrawing the figure to make things clearer:
Now getting the A, A$_1$, A$_2$, A$_3$, A$_4$:
For A:
$$A = \pi R^2 - \pi r^2$$ $$A = \pi (5 \space cm)^2 - \pi (2.5 \space cm)^2$$ $$A = 25\pi \space cm^2 - 6.25\pi \space cm^2$$ $$A = 18.75\pi \space cm^2$$
For A$_1$:
$$A_1 = \frac{25}{4}\pi \space cm^2 - \frac{6.25}{2}\pi\space cm^2$$ $$A_1 = 3.125\pi \space cm^2$$
For A$_4$:
$$A_4 = \frac{25}{4}\pi \space cm^2 - \frac{6.25}{2}\pi\space cm^2$$ $$A_4 = 3.125\pi \space cm^2$$
For A$_2$:
$$A_2 = \frac{25}{4}\pi \space cm^2$$
For A$_3$:
$$A_3 = \frac{25}{4}\pi \space cm^2$$
Now getting the x, x$_1$, x$_2$, x$_3$, x$_4$:
For x:
We can get it after I got x$_1$, x$_2$, x$_3$, x$_4$
For x$_1$:
Then the centroidal coordinates of A$_1$ is:
$$ x_1 = 0.3488 r$$ $$ x_1 = (0.3488) (5 \space cm)$$ $$ x_1 = 1.744 \space cm$$
The real value of $\bar x$ is:
$$\bar x_1 = 5 + 1.744 \space cm = 6.744 \space cm$$
For x$_2$
The centroidal coordinates of a quarter circle is:
$$\bar x = \frac{4}{3\pi} r$$ $$\bar y = \frac{4}{3\pi} r$$
Then the centroidal coordinates of A$_2$
$$ x_2 = \frac{4}{3\pi} r$$ $$ x_2 = \frac{4}{3\pi} (5 \space cm)$$ $$ x_2 = 2.1 \space cm$$
The real value of $x_2$ is:
$$x_3 = 5-2.1 \space cm = 2.9 \space cm$$
For x$_3$:
The centroidal coordinates of a quarter circle is:
$$ x = \frac{4}{3\pi} r$$ $$ y = \frac{4}{3\pi} r$$
Then the centroidal coordinates of A$_3$
$$ x_3 = \frac{4}{3\pi} r$$ $$ x_3 = \frac{4}{3\pi} (5 \space cm)$$ $$ x_3 = 2.1 \space cm$$
The real value of $x_3$ is:
$$x_3 = 5-2.1 \space cm = 2.9 \space cm$$
For x$_4$:
Then the centroidal coordinates of A$_4$ is:
$$ x_4 = 0.3488 r$$ $$ x_4 = (0.3488) (5 \space cm)$$ $$ x_4 = 1.744 \space cm$$
The real value of $ x_4$ is:
$$ x_4 = 5 + 1.744 \space cm = 6.74 \space cm $$
We can now get the $\bar x$:
$$A \bar x = A_1x_1 + A_2x_2 + A_3x_3 +A_4x_4 $$ $$(18.75\pi \space cm^2) \bar x = (3.125\pi \space cm^2)(6.744 \space cm) + (\frac{25}{4}\pi \space cm^2)(2.9 \space cm) + (\frac{25}{4}\pi \space cm^2)(2.9 \space cm) + (3.125\pi \space cm^2)(6.744 \space cm) $$ $$\bar x = 4.18 \space cm$$
To get the $\bar y$, notice that the figure is symmetrical, so the $\bar y$ centroidal coordinate would be $0$
Therefore, the centroid $C(\bar x, \bar y)$ of the figure above would be $C(\bar x, \bar y) = C(4.18, 0)$
Alternate solutions are encouraged:-)