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A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2/3. Suppose a 2.50 m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position. The rod is released as the free end is allowed to fall. What is the angular acceleration as it is released?

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MathPhys

Answer:

6 rad/s²

Explanation:

Sum the torques about the hinge.

∑τ = Iα

mg(L/2) = mL²/3 α

g/2 = L/3 α

α = 3g/(2L)

α = 3 (10 m/s²) / (2 × 2.50 m)

α = 6 rad/s²

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