Centre Of Mass Of A Triangle MCQ - Practice Questions & Answers

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Position of centre of mass for a triangular plate

Have a look at the figure of A triangular plate as shown in figure.

Since it is symmetrical about y-axis on both sides of the origin

So we can say that its $x_{cm} = 0$

And its $z_{cm} = 0$ as z-coordinate is zero for all particles of semicircular ring.

Now we will calculate its $y_{cm}$ which is given by

$y_{cm} = \frac{\int y.dm}{\int dm}$

For this take an elemental strip of mass dm and thickness dy at distance y from origin on y-axis

As shown in figure

$\Delta ADE \ and \ \Delta ABC$ will be similar

So, $\frac{r}{R}=\frac{H-y}{H}$

$r=(\frac{H-y}{H})R$

Take $\sigma =\frac{mass}{area}=\frac{M}{\frac{1}{2}*(2R)*H}$

$\sigma = \frac{M}{RH}$

And, $dm=\sigma dA=\sigma (2rdy)$

$y_{cm}=\frac{\int y\sigma dA}{M}$

$y_{cm}=\frac{\int_{H}^{0}y.\sigma dy.2(\frac{H-y}{H}).R}{M} = \frac{H}{3}$

So , $\mathbf{y_{cm}=\frac{H}{3}}$ from base

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Centre Of Mass Of A Triangle - Practice Questions & MCQ

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