50.Torque of the given force about centre of the rod=Fr=5*(.25)Nm we know Torque=I@ I is moment of inertia of system through centre of rod perpendicular to plane I=Mr2+Mr2 I=2Mr2 M is mass of each ball and r is distance between the from center I=2*.5*(.25)2 On using the two equations we get @=20rad/s2 so use =0+@t t=.1 we get =12rad/s Answer Rate me if satisfied
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