Flat plate, circular hole center of mass

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SUMMARY

The discussion focuses on calculating the x-coordinate of the center of mass (xcm) for a uniform flat plate with a circular hole. The relevant formula used is Xcm = (m1x1 + m2x2) / (m1 + m2), where m1 and m2 represent the areas of the rectangle and circle, respectively. Participants clarify that if density is not provided, area should be used in place of mass. The center of the circle is confirmed to be at (3,0), and the x1 coordinate is determined to be 1/2.

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gap0063
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Homework Statement


A uniform flat plate of metal with a circular hole, where it is a rectangle starting from -6 to 7 from the length, and the height is -3 to 3.

The circle has a radius of 2, and it starts at 1 to 5.
Calculate the x-coordinate of the center of
mass xcm of the metal plate.


Homework Equations


well Xcm= m1x1+m2x2/m1+m2


The Attempt at a Solution



is this the right equation?
and if so, how do I take out the circular part?
 
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hi gap0063! :wink:
gap0063 said:
well Xcm= m1x1+m2x2/m1+m2

is this the right equation?

yes :smile:
and if so, how do I take out the circular part?

it has negative mass! :biggrin:

(no, seriously, it really does … try it!)
 
tiny-tim said:
hi gap0063! :wink:


yes :smile:


it has negative mass! :biggrin:

(no, seriously, it really does … try it!)


this is probably a dumb question, but how do you find the mass of the rectangle and circle?

and is the x1 and x2 the coordinates?
 
hi gap0063! :smile:
gap0063 said:
this is probably a dumb question, but how do you find the mass of the rectangle and circle?

and is the x1 and x2 the coordinates?

if you're not given the density, then use area instead of mass in the formula

and yes, x1 and x2 are the x coordinates of the centre of mass and of the hole respectively (and similarly for y1 and y2) :wink:
 
tiny-tim said:
hi gap0063! :smile:


if you're not given the density, then use area instead of mass in the formula

and yes, x1 and x2 are the x coordinates of the centre of mass and of the hole respectively (and similarly for y1 and y2) :wink:

alright, is this right:

13*6(1/2)-(7)(22\Pi/ (13*6+ 4\Pi)
 
hi gap0063! :smile:

(have a pi: π :wink:)

gap0063 said:
alright, is this right:

13*6(1/2)-(7)(22\Pi/ (13*6+ 4\Pi)

for the total mass (on the bottom) you need minus 4π, don't you? :wink:

apart from that, it looks ok, except i don't understand where the (7) comes from

(and i didn't understand what this meant :confused: …)
gap0063 said:
The circle … starts at 1 to 5.
 
tiny-tim said:
hi gap0063! :smile:

(have a pi: π :wink:)



for the total mass (on the bottom) you need minus 4π, don't you? :wink:

apart from that, it looks ok, except i don't understand where the (7) comes from

(and i didn't understand what this meant :confused: …)


Yes I meant minus on the bottom


the 7 is from how far away the cicle is from the begin on the rectangle

sorry I don't have the picture but the shapes are on a coordinate system and inside the rectangle (that goes from (-6,0) to (7,0) ) the circle lies at (1,0) to (5,0).
 
so the centre of the circle is at (3,0)?

ok, then you must use x2 = 3 …

your x1 x2 and x must all be measured from the same origin :wink:
 
tiny-tim said:
so the centre of the circle is at (3,0)?

ok, then you must use x2 = 3 …

your x1 x2 and x must all be measured from the same origin :wink:

yes the center of the circle is (3,0)

so what is x1...? is it 1/2?
 
  • #10
yes, your x1 = 1/2 was correct :smile:
 

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