A thin uniform rod (mass = 0.420 kg) swings about an axis that passes through one...

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A thin uniform rod (mass = 0.420 kg) swings about an axis thatpasses through one end of the rod and is perpendicular to the planeof the swing. The rod swings with a period of 1.45 s and an angularamplitude of 10.6

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3.7 Ratings (530 Votes)

Mass of the rod M = 0.42 Kg

Time period T = 1.45 s

Angular displacement ? = 10.6 degree

The rotational inertia of a uniform rod with pivot point at its end is:

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â I = 1/3ML2

T = 2? Sqrt[(1/3ML^2)/Mg(L/2)]

Â Â Â = 2? Sqrt[(2/3) (L/g)]

Squaring on both sides we get

T^2 = 4?^2 (2/3)(L/g)

Â Â Â

Â Â Â Â Â Â Â = (8?^2)(L/3g)

Length of the pendulum L = 3gT^2 / 8?^2

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3(9.8 m/s^2)(1.45s)^2 / 8?^2

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.784 m

b) Maximum Kinetic energy ofÂ Â the rod = Mg(L/2)(1-cos?)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (0.42 Kg)*(9.8 m/s^2)*(0.784 m/2)*(1-cos10.6)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0275 J

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