The centre of mass of a thin rectangular plat | vedclass

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Question

(A) The mass per unit area $\sigma$ is constant in the $y$-direction,so the $y$-coordinate of the centre of mass is $y_{cm} = b / 2$.For the $x$-coord

Vedclass · Accepted Answer

$\left(\frac{2}{3} a, \frac{b}{2}\right)$

Vedclass · Answer

(A) The mass per unit area $\sigma$ is constant in the $y$-direction,so the $y$-coordinate of the centre of mass is $y_{cm} = b / 2$.For the $x$-coordinate,we consider a thin vertical strip of width $dx$ at a distance $x$ from the origin.The mass of this strip is $dm = \sigma dA = \left(\frac{\sigma_0 x}{ab}\right) (b dx) = \frac{\sigma_0}{a} x dx$.The $x$-coordinate of the centre of mass is given by:$x_{cm} = \frac{\int x dm}{\int dm} = \frac{\int_0^a x \left(\frac{\sigma_0}{a} x dx\right)}{\int_0^a \left(\frac{\sigma_0}{a} x dx\right)}$$x_{cm} = \frac{\int_0^a x^2 dx}{\int_0^a x dx} = \frac{[x^3 / 3]_0^a}{[x^2 / 2]_0^a} = \frac{a^3 / 3}{a^2 / 2} = \frac{2}{3} a$.Thus,the centre of mass is at $\left(\frac{2}{3} a, \frac{b}{2}\right)$.

The centre of mass of a thin rectangular plate (as shown in the figure) with sides of length $a$ and $b$,whose mass per unit area $(\sigma)$ varies as $\sigma = \frac{\sigma_0 x}{a b}$ (where $\sigma_0$ is a constant),would be . . . . . .

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