Question

There are four rods A, B, C and D of same length l but different linear mass density d, 2d, 3d &
4d respectively. These are joined to form a square frame with sides C & D along x & y axis of
coordinate axes respectively. Find coordinate of centre of mass of structure.​

Answer 1

Answer:

We are given that the linear mass densities of four rods A, B, C & D are d, 2d, 3d & 4d respectively of same length.

So, we can say A, B, C & D have masses which are “m”, “2m”, “3m” & “4m” respectively.

Let the coordinates of the centre of mass of the square be “X”cm & “Y”cm.

Since we know that, a system consists of “n” no. of particles of masses m1, m2, ……, mn and their position vectors r1, r2, ……., rn respectively then the center of mass of the n particles is the weighted average of the position vectors of the n particles i.e.,

Vector r = [m1r1+m2r2+….+mnrn] / [m1+m2+……+mn] …….. (i)

Now, based on eq. (i), the below drawn figure and the values given in the question, we can find the values of X cm & Y cm,

X cm =  [m (l/2) + 2m (l) + 3m (l/2) + 4m (0)] / [m + 2m + 3m + 4m] = 4 (m*l)/10 m = 2l/5

And,  

Y cm = [m (l) + 2m (l/2) + 3m (0) + 4m (l/2)] / [m + 2m + 3m + 4m] = 4 (m*l)/10 m = 2l/5

Hence, the coordinates of the center of mass of the square (Xcm, Ycm) = (2l/5, 2l/5).

Answer 2

Answer:

 (2l/5,2l/5)

Explanation: Dear student,

As per given details center of mass can be given as ∑m$r^{2}$ , where r is the distance from center of mass and m is the mass of the rigid body.

Apply the above formula in the given question and do it as given in attachment.