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A thin rod of length 'b' is suspended horizontally using ideal strings tied to both ends of the rod. The length of the strings is 'a'. The rod is given an initial angular speed omega its central axis. Let deltay be the upwards displacement of rod's centre in a small time interval deltatt and deltaF be th total increment in the tension forces just after the rod was given the angular speed. |
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Answer» `delay=(b^(2))/(8a)(OMEGA delta t)^(2)` and `theta_(a)=(bphi)/2` `IMPLIES deltay=a/2((bphi)/2)^(2)=(b^(2))/(8a)phi^(2)` `impliesdeltay=(b^(2))/(8a)(omega delta t)^(2)` acc. of rod's CENTRE `=(b^(2))/(4a)omega^(2)` `implies delta F=(mb^(2))/(4a) omega^(2)` 
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