Asked by calculs3
Find the center of gravity enclosed by y^2=4x,x=4,y=0 if its density is given by ä(x.y)=ky
Answers
Answered by
Damon
If we are dealing with real numbers x must be >/=0 becaue y^2 may not be negative
find vertical cg
y = 2 sqrt |x|
integrate from (0,0) to (4,4)
y of cg
= int dy y (4-x) (ky) / int dy(4-x) (ky)
numerator (the moment)
4 k y^2 dy - k dy y^2 (y^2/4)
4 k y^3/3 - k y^5/20
at y = 4
85.33 k - 51.2 k = 34.1 k
denominator (the mass)
dy(4-x) (ky) = 4k y dy - k y (y^2/4)dy
= 2 k y^2 - k y^4/16
at y = 4
32 k - 16 k = 16 k
so
34.1 k / 16 k = 2.13
You can do the stripes the other way for the Xcg
find vertical cg
y = 2 sqrt |x|
integrate from (0,0) to (4,4)
y of cg
= int dy y (4-x) (ky) / int dy(4-x) (ky)
numerator (the moment)
4 k y^2 dy - k dy y^2 (y^2/4)
4 k y^3/3 - k y^5/20
at y = 4
85.33 k - 51.2 k = 34.1 k
denominator (the mass)
dy(4-x) (ky) = 4k y dy - k y (y^2/4)dy
= 2 k y^2 - k y^4/16
at y = 4
32 k - 16 k = 16 k
so
34.1 k / 16 k = 2.13
You can do the stripes the other way for the Xcg
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.