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

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

So we can say that its $x_{c m}=0$
And its $z_{c m}=0$ as z-coordinate is zero for all particles of the semicircular ring.
Now we will calculate its $y_{c m}$ which is given by

$$
y_{c m}=\frac{\int y \cdot d m}{\int d m}
$$
 

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

As shown in the figure 

$\triangle A D E$ and $\triangle A B C$ will be similar
So $_3 \frac{r}{R}=\frac{H-y}{H}$

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

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

$$
\sigma=\frac{M}{R H}
$$

And, $\quad d m=\sigma d A=\sigma(2 r d y)$

$$
\begin{aligned}
& y_{c m}=\frac{\int y \sigma d A}{M} \\
& y_{c m}=\frac{\int_H^0 y \cdot \sigma d y \cdot 2\left(\frac{H-y}{H}\right) \cdot R}{M}=\frac{H}{3}
\end{aligned}
$$

Sa, $y_{\mathbf{c m}}=\frac{H}{3}$ from base