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30 August, 03:58

Find the angular acceleration produced given the mass lifted is 12 kg at a distance of 27.1 cm from the knee joint, the moment of inertia of the lower leg is 0.959 kg m2 kg m 2, the muscle force is 1504 1504 N, and its effective perpendicular lever arm is 3.3 3.3 cm.

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  1. 30 August, 04:09
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    Angular acceleration, α = 26.973 rad/s²

    Explanation:

    Given dа ta:

    Lifted mass, M = 12 kg

    Distance of the lifted mass = 27.1 cm = 0.271 m

    Effective lever arm, d = 3.3 cm = 0.033 m

    Moment of inertia, I = 0.959 kg. m²

    Applied force, F = 1504 N

    Now,

    the torque (T) is given as:

    T = F * d

    also,

    T = I * α

    where,

    α is the angular acceleration

    Now,

    Total moment of inertia, I = 0.959 + 12 * (0.271) ² = 1.840 kg. m²

    Now equation both the torque formula and substituting the respective values, we get

    1504 * 0.033 = 1.840 * α

    ⇒ α = 26.973 rad/s²
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