In the picture below, a 3 kg bar 2.2 m long is set up to rotate about its center

Question

In the picture below, a 3 kg bar 2.2 m long is set up to rotate about its center (the dark circle). Each arrow represents a force that is being applied to the bar. FA = 34 N, FB = 43 N, FC = 29 N, & FD = 29 N. FB & FC are applied halfway between the center and the end while FA & FD are applied right at the ends. Using the center as the pivot point, answer the following questions.


a.) Determine the magnitude of the net torque on the bar. (Remember, magnitudes are entered as positive.)
__ m?N

b.) Determine the moment of inertia of the bar.
__ kg?m2

c.) Determine the magnitude of the angular acceleration of the bar.
__ rad/s2

Explanation / Answer

Torque = F * d
Hence Torque due to A = Fa * 2.3 =  82.8
Torque due to B = - Fb * 1.15 = -39.1
Torque due to C = Fc * 1.15 = 33.5
Torque due to D = - Fd * 2.3 = -62.1

Hence net torque = 82.8 - 39.1 + 33.5 - 62.1 = 15.1 Nm

Moment of inertia I = ML2/12 = 14.11 kgm2

Angular Acceleration = /I = 15.1/14.11 =1.07 rad/sec2

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