In the picture above, a 9 kg bar 4.1 m long is set up to rotate about its center

Question

In the picture above, a 9 kg bar 4.1 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 = 24 N, FB = 44 N, FC = 28 N, & FD = 28 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. PHY110 Torque 2 Picture

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

a>Net torque= Fa*l/2 -Fb*l/4 +Fc*l/4 -Fd*l/2 -----------------taking clockwise direction positive T=24.6 m-N in anticlockwise direction b>Moment of Inertia of bar about center I = (m*L^2)/12 = 12.6 kg-m^2 c>Let angular acceleration = a I*a=T so a= 1.95 rad/s^2 in anti clockwise direction

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