A uniform circular disc of radius a is taken. | vedclass

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

Question

(B) Let the original circular disc of radius $a$ be considered as the combination of the removed circular section of radius $b$ and the remaining part

Vedclass · Accepted Answer

$\frac{{ - c{b^2}}}{{\left( {{a^2} - {b^2}} \right)}}$

Vedclass · Answer

(B) Let the original circular disc of radius $a$ be considered as the combination of the removed circular section of radius $b$ and the remaining part.Let the line of symmetry joining the centres $O$ and $O_1$ be the $x$-axis with $O$ as the origin.The centre of mass of the original disc of radius $a$ is at the origin $O$,so $X_{CM} = 0$.The formula for the centre of mass is:$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \dots (i)$Let $\sigma$ be the surface mass density of the disc.The mass of the removed circular portion is $m_1 = \pi b^2 \sigma$ and its centre is at $x_1 = c$.The mass of the remaining part is $m_2 = \pi (a^2 - b^2) \sigma$ and its centre of mass is at $x_2$.The total mass is $M = m_1 + m_2 = \pi a^2 \sigma$.Substituting these values into equation $(i)$:$0 = \frac{(\pi b^2 \sigma)(c) + (\pi (a^2 - b^2) \sigma)(x_2)}{\pi a^2 \sigma}$$0 = \pi b^2 \sigma c + \pi (a^2 - b^2) \sigma x_2$$x_2 = \frac{-c b^2}{a^2 - b^2}$Thus,the distance $x_2$ of the centre of mass of the remaining part from the initial centre of mass $O$ is $\frac{-c b^2}{a^2 - b^2}$.

Similar Questions

From a homogeneous square plate,a triangle is cut out as shown in the figure. The side of the square is $a$,and the apex of the triangle is at the center of the square. The distance from the center of the square to the center of mass of the remainder of the plate is:

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:

$A$ circular hole of radius $\left(\frac{a}{2}\right)$ is cut out of a circular disc of radius $'a'$ as shown in the figure. The centroid of the remaining circular portion with respect to point $'O'$ will be:

$A$ carpenter has constructed a toy as shown in the adjoining figure. If the density of the material of the sphere is $12$ times that of the cone,the position of the centre of mass of the toy is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module