Question

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

Answer 1

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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