The objects in the figure are constructed of uniform wire bent into the shape shown. Find the position of the position of the centre of mass of each shape.

Explanation

(a) The centre of mass of each wire of length L will lie on its mid-point. Each rod will have same mass (say m) so, the centre of mass (C ) of the system will be at the mid-point of the line-segment joining `C_(1)` and `C_(2)`. Hence OC is the bisector of the angle `theta`.

From the figure, clearly,
`OC = (L)/(2)cos.(theta)/(2)`
(b) Figure (b), shows `C_(1), C_(2)` and `C_(3)` as the positions of the centre of masses of the wires. `C_(4)` is the position of the centre of mass of the two vertical rods. Hence at `C_(4)`, the mass 2m has been supposed to be placed.

Now, `C_(1)C=(2m)/(m+2m)(C_(1)C_(4))=(2)/(3).(L)/(2)=(L)/(3)`
(c ) Putting `theta = 90^(@)` in (a) we get the position of c.m.
`OC=(L)/(2)cos 45^(@)=(L)/(2sqrt(2))`

(d) The position of the centre of mass (C ) of the system is given by
`C_(1)C=(2m)/(m+2m)(C_(1)C_(4))`
`=(2)/(3)((L)/(2)sin 60^(@))`
`= (L)/(sqrt(3))`