Question

Mass center of truncated boldes
A truncated body is made by removing a portion of a body. Find mass center of the following truncated disk made by removing disk of radius equal to half of the original disk as shown in the figure. Radius of the original uniform disk is `r`.

Answer 1

To find mass center of truncated bodies we can make use of superposition principle that is. If we add the removed portion in the same place we obtain the original body. The idea is illustrated in the following figure.

The removed portion is added to the trucated body keeping their location unchanged relative to the coordinate frame.
Denoting massses of the truncated body, removed portion and the original body by `m_(tb), m_(rp)` and `m_(o)b` and location of their mass centres by `x_(tb), x_(rp)` and `x_(ob)`, we can write `m_(tb)x_(tb)+m_(rb)x_(rp)=m_(ob)x_(ob)`
From the above equation we obtain position coordinate `x_(tb)` of the mass center of the truncated body.
`x_(tb)=m_(ob)x_(ob)-(m_(rb)x_(rp))/m_(tb)`
Denoting mass per unit area by sigma, we can express the masses `m_(tb), m_(rp)` and `m_(ob)`.
Mass of truncated body `m_(tb)=sigma{pi(r^(2)-(r^(2))/(4))}=(3sigmapir^2)/4`
Mass of the removed portion `m_(rp)=(sigmapir^2)/4`
Mass of the original body `m_(ob)=sigmapir_2`
Mass center of the truncated body `x_(tb)`
Mass center of the removed portion `x_(rp)=r/2`
Mass center of the original body `x_(ob)=0`
Substituting the above values in equation (1), we obtain the mass the center of the truncated body.
`x_(tb)=(m_(ob)x_(ob)-m_(rb)x_(rp))/m_(tb)=((sigmapir^2)xx0-((sigmapir^2)/4)(r/2))/((3sigmapir^2)/4)=-r/6`
Mass center of the truncated body is at point `(-(r)/6,0)`