Question

Q3) (10 Points) A rod with mass of 20 kg is rotating at 5 rad/s at the instant shown. A moment of 60 N·m is applied to the rod. Find: The angular acceleration a and the reaction at pin O when the rod is in the horizontal position. Draw Free body and kinetic diagrams. 60 Nm 10 = 5 rad/s

Answer 1

Given

Mass of rod (m) = 20 kg

Speed of rotation (w) = 5 rad/ sec

Moment applied a O (Mo) = 60 Nm

When the rod is in horizontal position then angular acceleration ($\alpha$) and reaction at O (Fo) is to be calculated.

In this condition the mass of rod acts in the middle of rod i.e. at 1.5 m as shown in figure.At this point mass have two components ,one component acting towards O and other component acting downward perpendicular to the rod.

Solution Free B Diagram and Kinematic Diagram GON-m LO A 1.5m 1.5m tone Y= 1.5m Normal Component the Efo = man = mrwh Efo = 20x15x52 Fo = 750W where an=rw? 0 Tangential component Itue done EF= =- mat =-mra Ro #mg) = -20x1.54 Ro +(-20x9.81) --20*1:54 we know that EMO = mixa 60 + mgx r = 2083 Xa 60 + 2049.81X5 = 60% 2 = 5.901 gad/s? from 0 - | Ro - 13 6. 2 = - 101:03 [Roz 19.17 N Answer