From a circle of radius a, an isosceles right | vedclass

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(C) Let the surface mass density of the disc be $\sigma$. The mass of the full disc is $M_1 = \sigma \pi a^2$,and its centre of mass is at the origin

Vedclass · Accepted Answer

$\frac{a}{3(\pi - 1)}$

Vedclass · Answer

(C) Let the surface mass density of the disc be $\sigma$. The mass of the full disc is $M_1 = \sigma \pi a^2$,and its centre of mass is at the origin $(0,0)$.The triangle has a hypotenuse equal to the diameter $2a$. The height of this isosceles right-angled triangle from the hypotenuse to the vertex is $a$. The area of the triangle is $A_2 = \frac{1}{2} 	imes (2a) 	imes a = a^2$. The mass of the removed triangle is $M_2 = -\sigma a^2$.The centre of mass of the triangle is located at a distance of $h/3 = a/3$ from the hypotenuse along the altitude. Thus,$y_2 = a/3$.The centre of mass of the remaining portion is given by $Y_{cm} = \frac{M_1 Y_1 + M_2 Y_2}{M_1 + M_2}$.Substituting the values: $Y_{cm} = \frac{(\sigma \pi a^2)(0) + (-\sigma a^2)(a/3)}{\sigma \pi a^2 - \sigma a^2} = \frac{-\sigma a^3 / 3}{\sigma a^2(\pi - 1)} = -\frac{a}{3(\pi - 1)}$.The magnitude of the distance is $\frac{a}{3(\pi - 1)}$.

From a circle of radius $a,$ an isosceles right-angled triangle with the hypotenuse as the diameter of the circle is removed. The distance of the centre of gravity of the remaining portion from the centre of the circle is

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