A rod of mass M and length L is placed in a horizontal plane with one end hinged about the vertical axis. A horizontal force of F=Mg/2 is applied at a distance 5L/6 from the hinged end. The angular acceleration of the rod will be
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The angular acceleration of the rod will be 5 g / 4 l.
Explanation:
T = Iα
T=Mg x 5l / 2 x 6
= 5 Mgl / 12
I = Ml^2 / 3
α = TI
= 5Mgl / 12 x 3 / Ml^2
= 5 g / 4 l
Hence the angular acceleration of the rod will be 5 g / 4 l.
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