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Centre Of Mass Of A Triangle - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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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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Position of centre of mass for a triangular plate

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