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A particle of mass 200g is attached to an ideal string of length 1.30 m whose upper end is fixed to ceiling. The particle is made to revolve in a horizontal circle of radius 50 cm. The tension in the string is (g = 10 m/s^2) |
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Answer» Solution :Moment of INERTIA of the particle `L = NR^(2)` Given `m = 200 = 0.200 KG, r=2 cm = 0.20 m` `l=mr^(2) =0.200 xx 0.20 = 0.04 kg m^(2)` |
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