# Centre of mass - Lamina

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#1
Please see attached picture for question

Used O as origin

So centre of mass of square is (2r,2r)

Centre of mass of quarter-circle is (4r/3pi + 2r, 4r/3pi)

mass of square is 32r^2 m (m is mass per unit area)
mass of quartercircle is 4pi r^2 m

so (32+4pi)x = 64r + 16/3r + 16pi r
xbar = 2.68 r
similarly for ybar, ybar = 1.56r

but the answer is (3.04r, 1.91r)

I have tried the question multiple times but I can't reach book answer Please help
0
3 weeks ago
#2
(Original post by golgiapparatus31)
Please see attached picture for question

Used O as origin

So centre of mass of square is (2r,2r)

Centre of mass of quarter-circle is (4r/3pi + 2r, 4r/3pi)

mass of square is 32r^2 m (m is mass per unit area)
mass of quartercircle is 4pi r^2 m

so (32+4pi)x = 64r + 16/3r + 16pi r
xbar = 2.68 r
similarly for ybar, ybar = 1.56r

but the answer is (3.04r, 1.91r)

I have tried the question multiple times but I can't reach book answer Please help
Should the quarter circle .com be +4r, not +2r?
0
#3
(Original post by mqb2766)
Should the quarter circle .com be +4r, not +2r?
Oops yes, it was typo.

I used +4r in working and answer I got is (2.68r, 1.56r)
0
3 weeks ago
#4
(Original post by golgiapparatus31)
Oops yes, it was typo.

I used +4r in working and answer I got is (2.68r, 1.56r)
Centre of mass of the quarter-circle is 4(radius)/3(pi) from B, horizontally and vertically. In this case, radius = 4r, not r.
1
3 weeks ago
#5
(Original post by golgiapparatus31)
Oops yes, it was typo.

I used +4r in working and answer I got is (2.68r, 1.56r)
R is 4r in your .com calcs for x and y
0
3 weeks ago
#6
(Original post by golgiapparatus31)
but the answer is (3.04r, 1.91r)
0
#7
(Original post by old_engineer)
Centre of mass of the quarter-circle is 4(radius)/3(pi) from B, horizontally and vertically. In this case, radius = 4r, not r.
(Original post by mqb2766)
R is 4r in your .com calcs for x and y
(Original post by ghostwalker)
Thanks! Got it now
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