A thin uniform rod (length = 125 cm,mass = 250 g)is pivoted about a horizontal frictionless pin through one end of the rod.The moment of inertia of the rod through this axis is (1/3)mL².The rod is released when it is 20^o below the horizontal.What is the angular acceleration of the rod at the instant it is released? A)7.36 rad/s² B)14.7 rad/s² C)11.1 rad/s² D)22.1 rad/s² E)16.7 rad/s²

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Quizplus · Accepted Answer

The torque τ about the pivot due to gravity is τ = m*g*(L/2)*sin(θ). The moment of inertia I is (1/3)mL². Angular acceleration α is given by α = τ/I. Substituting the values, α = (250g * 9.8m/s² * (1.25m/2) * sin(20°)) / ((1/3) * 250g * (1.25m)²) = 11.1 rad/s².

A Thin Uniform Rod (Length = 125 Cm,mass = 250