Find the centre of mass of a uniform $L-$shaped lamina $($a thin flat plate$)$ with dimensions as shown. The mass of the lamina is $3 \ kg.$
Example-(6.3)
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Choosing the $X$ and $Y$ axes as shown in Fig. $6.11$ we have the coordinates of the vertices of the $L-$shaped lamina as given in the figure. We can think of the $L-$shape to consist of $3$ squares each of length $1 m$. The mass of each square is $1 \ kg$, since the lamina is uniform. The centres of mass $C _1, C _2$ and $C _3$ of the squares are, by symmetry, their geometric centres and have coordinates $(1/2,1/2), (3 / 2,1 / 2),(1 / 2,3 / 2)$ respectively. We take the masses of the squares to be concentrated at these points. The centre of mass of the whole $L$ shape $(X, Y)$ is the centre of mass of these mass points.
Hence
$X=\frac{[1(1 / 2)+1(3 / 2)+1(1 / 2)] \ kg m }{(1+1+1) \ kg }=\frac{5}{6} m$
$Y=\frac{[1(1 / 2)+1(1 / 2)+1(3 / 2)] \ kg m }{(1+1+1) \ kg }=\frac{5}{6} m$
The centre of mass of the $L-$shape lies on the line $OD.$ We could have guessed this without calculations. Can you tell why? Suppose, the three squares that make up the $L$ shaped lamina of Fig. $6.11$ had different masses. How will you then determine the centre of mass of the lamina?
Hence
$X=\frac{[1(1 / 2)+1(3 / 2)+1(1 / 2)] \ kg m }{(1+1+1) \ kg }=\frac{5}{6} m$
$Y=\frac{[1(1 / 2)+1(1 / 2)+1(3 / 2)] \ kg m }{(1+1+1) \ kg }=\frac{5}{6} m$
The centre of mass of the $L-$shape lies on the line $OD.$ We could have guessed this without calculations. Can you tell why? Suppose, the three squares that make up the $L$ shaped lamina of Fig. $6.11$ had different masses. How will you then determine the centre of mass of the lamina?
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