Consider a circular disc of radius 20 cm wit | vedclass

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

Question

(D) Let the radius of the original disc be $R = 20 \ cm$ and the radius of the hole be $r = 5 \ cm$.The mass of the original disc is $M = \sigma \pi R

Vedclass · Accepted Answer

$1.0$

Vedclass · Answer

(D) Let the radius of the original disc be $R = 20 \ cm$ and the radius of the hole be $r = 5 \ cm$.The mass of the original disc is $M = \sigma \pi R^2$,where $\sigma$ is the surface mass density.The mass of the cut-out part is $m = \sigma \pi r^2$.Since $r = R/4$,the mass $m = \sigma \pi (R/4)^2 = M/16$.The centre of mass of the original disc is at the origin $(0, 0)$.The centre of the hole is at a distance $d = R - r = 20 - 5 = 15 \ cm$ from the origin.Let the centre of mass of the remaining part be $X_{	ext{com}}$. Using the formula for the centre of mass of a system with a cavity:$X_{	ext{com}} = \frac{M(0) - m(d)}{M - m}$$X_{	ext{com}} = \frac{0 - (M/16)(15)}{M - M/16} = \frac{-(15/16)M}{(15/16)M} = -1 \ cm$.The magnitude of the distance is $|X_{	ext{com}}| = 1 \ cm$.

Similar Questions

From a uniform square plate,one-fourth part is removed as shown. The centre of mass of the remaining part will lie on:

$A$ 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:

$A$ 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

$A$ semicircular portion of radius $r$ is cut from a uniform rectangular plate as shown in the figure. The distance of the centre of mass $C$ of the remaining plate from point $O$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module