Question

From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

Answer 1

Given, the radius of the uniform disk is R and the radius of the circular hole is R/2. The location of the centre of the hole is R/2 from the centre of the original disc.

Let M be the mass of the uniform disk, x 1 be the location of centre of gravity of the uniform disc with respect to centre of the uniform disk., m be the mass of the circular cut out part and x 2 be the location of centre of gravity of the cut out part with respect to centre of the uniform disk.

The expression for the location of the centre of gravity of the resulting flat body is,

x CG = M x 1 −m x 2 M−m …… (1)

The expression for the mass of the uniform disk is,

M=ρV =ρ( π R 2 t )

Here, ρ is the density of the disk material and t is the thickness of the material.

The centre of gravity of the uniform disk lies on its own centre. Therefore, x 1 =0 .

The expression for mass of the cut out circular part is,

m=ρ V ′ …… (2)

Here, V ′ is the volume of the cut out portion.

The volume of the cut out portion is,

V ′ =π ( R ′ ) 2 t

Here, R ′ is the radius of the cut out portion.

Substitute R/2 for R ′ in the above equation.

V ′ =π ( R ′ ) 2 t =π ( R 2 ) 2 t = π R 2 t 4

Substitute π R 2 t 4 for V ′ in the equation (2).

m=ρ V ′ =ρ( π R 2 t 4 ) = 1 4 ρπ R 2 t

Substitute the value of ρ( π R 2 t ) in the above equation.

m= 1 4 ( M ) = M 4

The location of centre of gravity of the circular cut part with respect to centre of the uniform disk is,

x 2 =R/2

Substitute the above calculated values in the equation (1).

x CG = M⋅0−( M 4 )( R/2 ) M−( M 4 ) = 0− MR 8 3M 4 =− R 6

Thus, the location of the centre of gravity of the resulting flat body is R/6 from the centre of the original uniform disk and in the opposite direction to the center of the cut portion.