Question

Find the position of the centre of mass of the uniform lamina from AD​

Answer 1

Let the mass of the larger (original) triangle be M and that of the removed portion (i.e., the rightmost small triangle) be m.

Then the center of mass of the trapezium ABCD will be,

$\overline {x}=\dfrac {MX-mx}{M-m}$

where X and x are centers of masses of the larger and smaller triangles respectively.

Since it's a uniform lamina, the areal density of the lamina is same everywhere in it, thus.,

$\dfrac {M}{\left (\dfrac {(2a)^2\sqrt 3}{4}\right)}=\dfrac {m}{\left (\dfrac {a^2\sqrt 3}{4}\right)}\\\\\\m=\dfrac {M}{4}$

Since the larger triangle (as well as the smaller) is an equilateral one, its center of mass is $\dfrac {a}{\sqrt3}$ units right from the midpoint of AD.

Similarly, the center of mass is $\dfrac {a}{2\sqrt3}$ units right from midpoint of BC, i.e., $\dfrac {2a}{\sqrt3}$ units right from midpoint of AD.

Then the moment of inertia of the remaining portion is,

$\overline {x}=\dfrac {M\left(\dfrac {a}{\sqrt 3}\right)-m\left (\dfrac {2a}{\sqrt3}\right)}{M-m}\\\\\\\overline {x}=\dfrac {M\left(\dfrac {a}{\sqrt 3}\right)-\dfrac {M}{4}\left (\dfrac {2a}{\sqrt3}\right)}{\left (M-\dfrac {M}{4}\right)}\\\\\\\underline {\underline {\overline {x}=\dfrac {2a}{3\sqrt 3}}}$

That distance (value of $\overline {x}$) right from the midpoint of AD is the center of mass of the remaining portion.