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A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot in the vertical plane. The rod is held in the horizontal position and released from rest. Determine the rod’s linear density in SI units.
Determine the rod’s moment of inertia in SI Units.
Determine the initial angular acceleration of the rod. Determine the initial linear acceleration in the middle of the rod in SI Units.
Find the magnitude of torque in SI Units.

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Answer:

ρ = 1.8 kg / .90 m = 2 kg / m      linear density

Γ = m g L/2 = I α          torque producing acceleration  α

α = m g L/2 / I = m g L/2 / 1/3 m L^2 = 3 g / (2 *L)

α = 3 / 2 * 9.80 * / .90 s^2 = 16.3 / s^2      angular acceleration

a = L/2 α = .90 m / 2 * 16.3 / s^2 = 7.3 m/s^2     acceleration of middle of rod

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