A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end is allowed t

Question

A 4.00-m long rod is hinged at one end. The rod is initially held in the horizontal position, and then released as the free end is allowed to fall. What is the angular acceleration as it is released

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Dulcie 5 years 2021-08-08T00:06:27+00:00 1 Answers 61 views 0

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  1. Vodka
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    2021-08-08T00:08:00+00:00
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    Answer:

    The angular acceleration α = 14.7 rad/s²

    Explanation:

    The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod  and L = length of rod = 4.00 m. α = angular acceleration of rod

    Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.

    So Iα = Wr

    Substituting the value of the variables, we have

    mL²α/12 = mgL/2

    Simplifying by dividing through by mL, we have

    mL²α/12mL = mgL/2mL

    Lα/12 = g/2

    multiplying both sides by 12, we have

    Lα/12 × 12 = g/2 × 12

    αL = 6g

    α = 6g/L

    α = 6 × 9.8 m/s² ÷ 4.00 m

    α = 58.8 m/s² ÷ 4.00 m

    α = 14.7 rad/s²

    So, the angular acceleration α = 14.7 rad/s²

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