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A thin uniform rod AB of mass m=1.0kg mo...

A thin uniform rod AB of mass `m=1.0kg` moves translationally with acceleration `w=2.0m//s^2` due to two antiparallel forces `F_1` and `F_2` (figure). The distance between the points at which these forces are applied is equal to `a=20cm`. Besides, it is known that `F_2=5.0N`. Find the length of the rod.

Text Solution

Verified by Experts

Since, motion of the rod is purely translational, net torque about the C.M. of the rod should be equal to zero.
Thus `F_1l/2=F_2(l/1-a)` or, `F_1/F_2=1-(a)/(l//2)` (1)
For the translational motion of rod.
`F_2-F_1=mv_e` or `1-F_1/F_2=(mw_e)/(F_2)`(2)
From (1) and (2)
`(a)/(l//2)=(mw_c)/(F_2)` or, `l=(2aF_2)/(mw_c)=1m`
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