A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod...
A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.05 m and mass 23.4 kg , find the torque the person must exert on the ladder to give it an angular acceleration of 0.450 rad/s2 .
Homework Answers
Moment of inertia of the ladder whose axis is along the center is
I = mL^2 /12 = (23.4)(3.05)^2 /12
I = 18.14 kg m^2
Calculte the torque on the ladder .
T = I*angular acceleration
= 18.14*0.45
= 8.163 N-m
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