Question

Find the centre of mass of a uniform disc of radius a from which a circular section of radius b has been removed. The centre of the hole is at a distance c from the centre of the disc​

Answer 1

Answer:

M = pi * a^2       original mass of disc    (denity = 1)

m = pi * b^2        mass removed by hole in disc

M' = M - m           mass of disc with hole in it

M Xcm = 0       Xcm    original center of mass

0 = M' xcm + m c

xcm = = m c / M' = pi b^2 c / pi ( a^2 - b^2)

xcm = b^2 * c / (a^2 - b^2)

Answer 2

Answer:

$x=\dfrac{ b^2\times c}{a^2- b^2}$

Explanation:

Lets take

Mass of disc before removing = M

Mass of  removing part = m

 Radius of disc = a

Radius of removed part = b

Lets take area density of disc =ρ kg/m²

We know that if there is no any external force on the system then the center of mass of the system will not move.

The center of mass of body is initially at point O and it shift left side after removing a disc of radius b.

So we can say that

$x=\dfrac{M\times 0-m\times c}{M-m}$

M= π ρ a²

m=π ρ b²

$x=\dfrac{-\pi\times\rho\timesb^2\times c}{\pi\time\rho\times a^2-\pi\times\rho\timesb^2}$

$x=\dfrac{-b^2\times c}{a^2- b^2}$

Negative sign indicates that center of mass will shift towards left side.

$x=\dfrac{ b^2\times c}{a^2- b^2}$