Physics: What is the torque a person must exert on the ladder to give it an angular acceleration of 0.203 rad/s^2?

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A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.00 m and mass 6.88 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.203 rad/s^2.

1 Answer

  • Moment of Inertia for a rod= (MR^2)/12

    (6.88*3^2)/12=5.16

    I*angular acceleration= 5.16 * 0.203 = 1.05 Nm

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