In the picture below, a 1 kg bar3.5 m long is set up to rotate about itscenter (

Question

In the picture below, a 1 kg bar3.5 m long is set up to rotate about itscenter (the dark circle). Each arrow represents a force that isbeing applied to the bar. FA = 48 N, FB = 29 N,FC = 27 N, & FD= 25 N. FB is applied halfwaybetween the center and the end and FC is applied rightat the end. Using the center as the pivot point, answer thefollowing questions. a.) Determine the net torque on the bar.
m·N

b.) Determine the moment of inertia of thebar.
kg·m2

c.) Determine the angular acceleration of thebar.
rad/s2

Explanation / Answer

(a)Torque due to FA & FD will be zero because they are passing through centre.    Total torque about centre,        =B + C   = FB* L/4 + FC * L/2         = 29*3.5/4 +27* 3.5/2 = 72.625Nm (b) I = M L2/12 =(3.5)2 /12 = 1.02kgm2   (c)  Use = I              = /I = 71.2rad/s2              = /I = 71.2rad/s2

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