Find the centre of mass for this sheet of paper with a cutout

[mrmerchant786]

Homework Statement

A uniform, rectangular sheet with sides of lengths a and b has a hole of dimensions a/4 by b/4 punched in it as shown below. Find the centre of mass of the sheet after the hole is made.

Homework Equations

CM= ∫xdm/∫dm

The Attempt at a Solution

I'm confused on what to input into the above equaiton
I do know that it will like on the axis of symetry so on the line y=b/2

many thanks :)

 

[ehild]

Homework Statement

View attachment 207632
A uniform, rectangular sheet with sides of lengths a and b has a hole of dimensions a/4 by b/4 punched in it as shown below. Find the centre of mass of the sheet after the hole is made.

Homework Equations

CM= ∫xdm/∫dm

The Attempt at a Solution

I'm confused on what to input into the above equaiton
I do know that it will like on the axis of symetry so on the line y=b/2

many thanks :)

You can consider the system as the big rectangle and the small rectangle, but with negative mass.

 

[mrmerchant786]

You can consider the system as the big rectangle and the small rectangle, but with negative mass.

thanks for the reply, so would the answer be
CM= ∫x_{the whole sheet}dm_{the whole sheet}/∫dm_{the whole sheet} - ∫x_{the cut out} dm_{the cut out}/∫dm _{the cut out}

 

[ehild]

thanks for the reply, so would the answer be
CM= ∫x_{the whole sheet}dm_{the whole sheet}/∫dm_{the whole sheet} - ∫x_{the cut out} dm_{the cut out}/∫dm _{the cut out}

No, it is not simply the difference between the CM-s.
Supposing you have two small balls at x1 and x2, of masses m1 and m2, how do you get the common center of mass?
And no need to integrate, the CM of a rectangle is in the centre.

 

[mrmerchant786]

No, it is not simply the difference between the CM-s.
Supposing you have two small balls at x1 and x2, of masses m1 and m2, how do you get the common center of mass?
And no need to integrate, the CM of a rectangle is in the centre.

hmm to answer your question x1*m1 + x2*m2 / m1 + m2 ?

in regards to my question so would it be the centre coordinates minus a (x_{centre of cut out}*ρ*area of cut out)/(ρ*area of cut out)

 

[ehild]

hmm to answer your question x1*m1 + x2*m2 / m1 + m2 ?

Do not forget the parentheses. Xcm=(x1*m1 + x2*m2) / (m1 + m2 )

in regards to my question so would it be the centre coordinates minus a (x_{centre of cut out}*ρ*area of cut out)/(ρ*area of cut out)

No.

 

[mrmerchant786]

Do not forget the parentheses. Xcm=(x1*m1 + x2*m2) / (m1 + m2 )

No.

okay then, what would it be then?

 

[ehild]

okay then, what would it be then?

Choose a system of coordinates, for example, that in the picture. Point P is the CM of the big rectangle, Q is the CM of the small one.
You can consider the mass of a rectangle compressed in the CM.
If the density and thickness of the plate is homogeneous, the mass is proportional to the area. The CM of a symmetric shape is on alll symmetry axis, in case of the rectangle, it is on the middle.
What are the positions of the CM-s of both rectangles?
What are the masses (areas)?
Then apply the formula X_CM =(x1m1+x2m2)/(m1+m2) for the masses at P and Q. ( consider m2 negative)