Question

from a square sheet of uniform density, a portion is removed as shown . find the centre of mass of the remaining portion if the side of the square is a​

Answer 1

square can be divided into four parts let mass of each part is m.

as it is clear shape of each part is a right angled triangle.

so, centre of mass of each part = centroid of triangle.

see figure,

centre of mass of ∆COD = (x1, y1) = [(a/2 + 0 + a)/3, (a/2 + 0 + 0)/3 ]

= (a/2 , a/6)

centre of mass of ∆AOB = (x2, y2) = [(a/2 + 0 + a)/3, (a/2 + a + a)/3 ]

= (a/2, 5a/6)

centre of mass of ∆AOC = (x3, y3) = [(0 + a/2 + 0)/3, (a + a/2 + 0)/3 ]

= (a/6, a/2)

now, centre of mass of remaining part of square , x = (mx1 + mx2 + mx3)/(m + m + m)

= (m × a/2 + m × a/2 + m × a/6)/3m

= 7a/18

y = (my1 + my2 + my3)/(m + m + m)

= (m × a/6 + m × 5a/6 + m × a/2)/3m

= (a/6 + 5a/6 + a/2)/3

= a/2

hence, centre of mass of remaining part is (7a/18, a/2)

Answer 2

Answer:

Explanation:the above answer is wrong the right answerr is a /9.

We use formula x*m remaining. =.m cut *x. From 0 to c2

M-m/4* x=m/4* a/3

3m/4 x = m/4. A/3.

X =a/9

Thanks