A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of the rod about

Question

A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of the rod about this axis is given by (1/3)ML2. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the instant it is released?

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Thiên Ân 3 days 2021-07-22T03:43:56+00:00 1 Answers 0 views 0

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    2021-07-22T03:45:49+00:00

    Answer:

    Angular acceleration = 6.37rad/sec²

    Approximately, Angular acceleration =

    6.4 rad/sec²

    Explanation:

    Length of the rod = 2.0m long

    Inclination of the rod (horizontal) = 30°

    Mass of the rod is not given so we would refer to it as = M

    Rotational Inertia of the Rod(I) = 1/3ML²

    Angular Acceleration = ?

    There is an equation that shows us the relationship between Torque and Angular acceleration.

    The equation is :

    Torque(T) = Inertia × Angular Acceleration

    Angular acceleration = Torque ÷ Inertia

    Where:

    Torque = L/2(MgCosθ)

    Where M = Mass

    L = Length = 2.0m

    θ = Inclination of the rod (horizontal) = 30°

    g = Acceleration due to gravity = 9.81m/s²

    Inertia = 1/3ML²

    Angular Acceleration =  (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²

    Angular Acceleration =

    (3 × g × cos 30°) ÷ 2× L

    Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L

    Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m

    Angular Acceleration = 6.37rad/sec²

    Approximately Angular Acceleration =

    6.4rad/sec²

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