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A uniform rod of length 0.7 m and mass 10 kg rotates freely about a horizontal axis passing through one end of the rod a bullet of mass 20 g traveling horizontally with velocity v=200 m/s penetrates the end of the rod. and hooked on the rod find the angular velocity of the rod when the bullet hits the bar

Respuesta :

Answer:

ω = 1.7 rad/s

Explanation:

Conservation of angular momentum

Assuming the rod is initially hanging vertically at rest.

Initial angular momentum is carried by the bullet only

L = Iω = (mR²)(v/R) = mvR = 0.020(200)(0.7) = 2.8 kg•m²/s

the same angular momentum exists after impact, only the moment of inertia has increased by that of the rod. I = ⅓mR²

2.8 = (⅓(10)(0.7²) + 0.020(0.7²))ω

2.8 = (1.64313333...)ω

ω = 1.70406134...

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