Question

Find the coordinates of the centre of gravity of the uniform lamina whose area is bounded by the following curve and line respectively.$\begin{gathered} \rm \: y = \frac{1}{4} {x}^{2} \\ \\ \bf \: and \\ \\ \rm \: y= 1\end{gathered}$​​

Answer 1

Since the lamina is uniform and symmetric about y axis, the centre of gravity should lie along y axis so its x coordinate is zero.

• $\small\bar x=0$

To find the y coordinate, consider a rectangular element of length 2x and width dy at a distance of y units along y axis from the lamina.

Area of this element is,

• $\smalldA=2x\ dy$

Since the lamina is uniform, the density is constant, so the centre of gravity of the lamina will be given by,

$\small\longrightarrow \bar y=\dfrac{\displaystyle\int\limits_0^Ay\ dA}{\displaystyle\int\limits_0^AdA}=\dfrac{\displaystyle\int\limits_0^A2xy\ dy}{\displaystyle\int\limits_0^A2x\ dy}=\dfrac{\displaystyle\int\limits_0^Axy\ dy}{\displaystyle\int\limits_0^Ax\ dy}$

Here the dimensions of element depend on the graph $\smally=\dfrac{x^2}{4}$ as we see the coordinates (x, y) along this graph. Thus x and y are related as,

$\small\longrightarrow y=\dfrac{x^2}{4}$

$\small\longrightarrow dy=\dfrac{2x}{4}\ dx=\dfrac{x}{2}\ dx$

The points of intersection of the two graph $\smally=\dfrac{x^2}{4}$ and $\smally=1$ are (-2, 1) and (2, 1). If y and dy in the integrations are written in terms of x, then integration is done along x axis. As we see the intersection points are the limits of the lamina, so the integration ranges in between the x coordinates of these limits, i.e., from -2 to 2.

Thus,

$\small\text{ \longrightarrow \bar y=\dfrac{\displaystyle\int\limits_{-2}^2x\cdot\dfrac{x^2}{4}\cdot\dfrac{x}{2}\ dx}{\displaystyle\int\limits_{-2}^2x\cdot\dfrac{x}{2}\ dx} }$

$\small\text{ \longrightarrow \bar y=\dfrac{\dfrac{1}{8}\displaystyle\int\limits_{-2}^2x^4\ dx}{\dfrac{1}{2}\displaystyle\int\limits_{-2}^2x^2\ dx} }$

$\small\text{ \longrightarrow \bar y=\dfrac{1}{4}\cdot\dfrac{\frac{1}{5}\left[x^5\right]_{-2}^2}{\frac{1}{3}\left[x^3\right]_{-2}^2} }$

$\small\longrightarrow \bar y=\dfrac{1}{4}\cdot\dfrac{3}{5}\cdot\dfrac{64}{16}$

$\small\longrightarrow \bar y=\dfrac{3}{5}$

Hence the coordinates of the centre of gravity is (0, 3/5).

Shortcut:-

With similar integrations in the above method, one can prove the following:

"The coordinates of the centre of gravity of the uniform lamina whose area is bounded by the curve y = ax² and the line y = k (where a and k have same sign) is (0, 3k/5)."

In the question, k = 1. Hence the answer is (0, 3/5).

Answer 2

To find the center of gravity of the given lamina, we need to find the coordinates of its centroid. The centroid is the point (x_c, y_c) such that if the lamina is replaced by a point mass at (x_c, y_c), the resulting system is in equilibrium under gravity.

The x-coordinate of the centroid is given by:

$\rm x_c = \frac{\int\int_R x\ dA}{\int\int_R dA}$

where R is the region enclosed by the curve y = x²/4 and the line y = 1.

The y-coordinate of the centroid is given by:

$\rm y_c = \frac{\int\int_R y\ dA}{\int\int_R dA}$

Substituting the equation of the line y=1 into the equation of the parabola y=x²/4, we get the limits of integration for x:

$\rm 1 = \frac{1}{4}x^2 \Rightarrow x = \pm 2$

Therefore, the limits of integration for x are -2 and 2.

The limits of integration for y are 0 and 1, since the lamina is bounded above by the line y=1 and below by the x-axis.

Substituting x and y into the expressions for x_c and y_c, we get:

$\rm x_c = \frac{\int_{0}^{1}\int_{-2}^{2}x\ dy\ dx}{\int_{0}^{1}\int_{-2}^{2}1\ dy\ dx} = 0$

$\rm y_c = \frac{\int_{0}^{1}\int_{-2}^{2}y\ dy\ dx}{\int_{0}^{1}\int_{-2}^{2}1\ dy\ dx} = \frac{3}{4}$

Therefore, the coordinates of the centroid of the given lamina are (0, 3/4).