From a uniform disc of radius R,an equilatera | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the disc have its center at $(0, R)$ and radius $R$. The origin $(0, 0)$ is at the bottom of the disc.An equilateral triangle of side $a = \sq

Vedclass · Accepted Answer

$(0, R)$

Vedclass · Answer

(B) Let the disc have its center at $(0, R)$ and radius $R$. The origin $(0, 0)$ is at the bottom of the disc.An equilateral triangle of side $a = \sqrt{3}R$ is inscribed in the disc. The height of this triangle is $h = \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2}(\sqrt{3}R) = 1.5R$.The centroid of the triangle coincides with the center of the disc at $(0, R)$.Since the disc is uniform and the equilateral triangle is symmetric about the $y$-axis and centered at $(0, R)$,the removal of the triangle maintains the symmetry of the remaining mass about the $y$-axis.However,the mass distribution is no longer symmetric about the horizontal line passing through $(0, R)$.Let $M$ be the mass of the disc and $m$ be the mass of the triangle. The center of mass of the remaining part is given by $Y_{cm} = \frac{M Y_{disc} - m Y_{triangle}}{M - m}$.Since $Y_{disc} = R$ and $Y_{triangle} = R$,the center of mass remains at $(0, R)$ only if the triangle is centered at the disc's center. Looking at the figure,the triangle's top vertex is at $(0, 2R)$ and its base is at $y = 0.5R$. The centroid of this triangle is at $y = \frac{2R + 0.5R + 0.5R}{3} = R$. Thus,the triangle is indeed centered at the disc's center $(0, R)$.Therefore,the center of mass of the remaining portion remains at $(0, R)$.

From a uniform disc of radius $R$,an equilateral triangle of side $\sqrt{3}R$ is cut as shown. The new position of the centre of mass is:

Similar Questions

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

$A$ circular disc of radius $R$ is removed from one end of a bigger circular disc of radius $2R$. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

The coordinates of the centre of mass of a uniform $L$-shaped plate of mass $3 \ kg$ shown in the figure is:

As shown in the figure,when a spherical cavity (centred at $O$) of radius $1$ is cut out of a uniform sphere of radius $R$ (centred at $C$),the centre of mass of the remaining (shaded) part of the sphere is at $G$,i.e.,on the surface of the cavity. $R$ can be determined by the equation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module