Question

A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences
of the discs coincide. The centre of mass of the new disc is ar from the centre of the bigger disc. The
value of a is
[AIEEE 2007]​

Answer 1

Answer:

The value of a is $-\dfrac{1}{3}R$.

Explanation:

Given that,

Radius of circular disc = R

Radius of bigger circular = 2R

Let σ be the surface mass density of the disc.

The mass of bigger circular is

$M=\pi(2R)^2\sigma$

The mass of cut out section is

$m=\pi R^2\sigma$

The mass of the residue part

$M'=M-m$

$M'=\pi(2R)^2\sigma-\pi R^2\sigma$

$M'=3\piR^2\sigma$

According to figure,

$O''O=a$

$OO'=R$

The center of mass of bigger disc lies at the center.

The center of mass of remove disc lies O' at a distance R from O.

Let the center of mass of residue disc lies at O''

The moment of mass about center of mass is zero.

So, the moment of residue disc and cut of disc about O is zero.

Therefore,

The center of mass of the new disc is a from the center of the bigger disc

$m OO'+ M' OO''=0$

$OO''=-\dfrac{m}{M'}OO'$

$a=\dfrac{\pi R^2\sigma}{3\piR^2\sigma}\times R$

$a=-\dfrac{1}{3}R$

Hence, The value of a is $-\dfrac{1}{3}R$.

Answer 2

Answer:

Explanation:

Radius of circular disc = R

Radius of bigger circular = 2R

Let σ be the surface mass density of the disc.

The mass of bigger circular is

The mass of cut out section is

The mass of the residue part

According to figure,

The center of mass of bigger disc lies at the center.

The center of mass of remove disc lies O' at a distance R from O.

Let the center of mass of residue disc lies at O''

The moment of mass about center of mass is zero.

So, the moment of residue disc and cut of disc about O is zero.

Therefore,

The center of mass of the new disc is a from the center of the bigger disc

a=1/3