A circular disc of radius R has a uniform thi | vedclass

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

Question

(D) Let the original disc have mass $M$ and area $A_1 = \pi R^2$. Its centre of mass is at $O$ (origin).Let the removed circular hole have radius $r =

Vedclass · Accepted Answer

$\frac{R}{6}$

Vedclass · Answer

(D) Let the original disc have mass $M$ and area $A_1 = \pi R^2$. Its centre of mass is at $O$ (origin).Let the removed circular hole have radius $r = R/2$. Its area is $A_2 = \pi (R/2)^2 = \frac{\pi R^2}{4}$.The centre of mass of the removed hole is at a distance $x_2 = -R/2$ from $O$.The centre of mass of the remaining part is given by:$X_{cm} = \frac{A_1 x_1 - A_2 x_2}{A_1 - A_2}$Since $x_1 = 0$ (at $O$):$X_{cm} = \frac{(\pi R^2)(0) - (\pi R^2 / 4)(-R/2)}{\pi R^2 - \pi R^2 / 4}$$X_{cm} = \frac{(\pi R^2 / 4)(R/2)}{3 \pi R^2 / 4} = \frac{R/2}{3} = \frac{R}{6}$The distance from point $O$ is $\frac{R}{6}$.

$A$ circular disc of radius $R$ has a uniform thickness. $A$ circular hole of diameter equal to the radius of the disc has been cut out as shown. The distance of the centre of mass of the remaining disc from point $O$ is

Similar Questions

$A$ uniform square plate of side $a$ has a circular hole of diameter $a$ cut out from it as shown in the figure. The centre of the hole is at the centre of the square. The centre of mass of the remaining part is:

$A$ circular plate of uniform thickness with a diameter of $56\, cm$ has its center at the origin. $A$ circular part with a diameter of $42\, cm$ is removed from one edge. What is the distance of the center of mass of the remaining part in $cm$?

$A$ uniform square plate is shown in the figure. Four identical small squares are removed from its corners. If squares $1, 2,$ and $3$ are removed,where will the center of mass $(C.M.)$ be located?

$A$ uniform rectangular thin sheet $ABCD$ of mass $M$ has length $a$ and breadth $b$,as shown in the figure. If the shaded portion $HBGO$ is cut off,the coordinates of the centre of mass of the remaining portion will be

$A$ 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

Vedclass Test Series

Exam Paper Generator

Online Exam Module