Question

A circular plate has a uniform thickness and has a diameter $56cm$ A circular disc of diameter $42cm$ is removed from one edge of the plate.The distance of the centre of mass of the remaining portion from the centre of mass of the plate is

• A. $18cm$
• B. $-9cm$
• C. $27cm$
• D. $4.5$

Answer 1

Answer: (B)

$-9cm$

Thickness of plate $=t$ Total mass of circular plate $=M$

Density of plate$=M/\left(\pi r₁²t\right)$ , $r₁=28cm$

$r₂=$ radius of cut out part $=21cm$

Mass of small circular portion cut out = volume \times density $=\pi r₂²t\times \left(M/\pi r₁²t\right)$

$m₂=Mr₂²/r₁²$

Mass of the remaining part =$m_3=M-Mr₂²/r₁²=M(r₁²-r₂²)/r₁²$

Let the center of mass of remaining portion be at a distance d

from the center of original full plate. From symmetry the center

of mass of remaining portion lies along the line joining the center

of original plate and the center of removed part.

Let Center of mass of total full plate = 0

$0=1/M$ ( d \times mass of remaining part $+C₁C₂\times$mass

of removed part)

$0=d\times M(r₁²-r₂²)/r₁²+7\times Mr₂²/r₁²$

$0=d\times (r₁²-r₂²)+7\times r₂²$

$d=-7\times r₂²/(r₁²-r₂²)=-7\times 21²/(28²-21²)=-7\times 21\times 21/49\times 7=-9cm$