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7 August, 10:24

A 0.250-kg wooden rod is 1.35 m long and pivots at one end. It is held horizontally and then released. Part A What is the angular acceleration of the rod after it is released?

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  1. 7 August, 12:50
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    We assume that the rod's weight is evenly distributed, making its center of gravity 0.675 m from the end.

    First, we calculate the moment present on the rod:

    τ = F*d

    τ = m*g*d

    τ = 0.25 * 9.81 * 0.675

    τ = 1.66

    Next, in the case of rotational motion, Newton's second law is:

    τ = Iα, where I is moment of inertia and α is the angular acceleration

    The moment of inertia for a rod is:

    I = (ML²) / 12

    I = (0.25*1.35²) / 12

    I = 0.038

    Now, we use the formula given by Newton's law:

    α = τ / I

    α = 1.66 / 0.038

    α = 43.7 rad/s²

    The angular acceleration is 43.7 radians per seconds squared.
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