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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. |
Answer» Solution :(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))` 
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