Question

A pendulum consisting of a uniform rod and concentrated mass at its free end is pivoted to the inertial frame through a revolute joint. The rod has length L=1.2m and mass m=3 kg. The concentrated mass has magnitude M=5 kg. The system starts from rest with the rod making an angle 0=60° to the horizontal. A moment T=100 Nm acts on the rod. When the rod first makes an angle of ¢=300° to the horizontal, calculate: 1. Angular velocity of the pendulum 2. Reaction forces at the pivot M L m 0 DO

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Answer 1

Ciren, M m= 3kg m = 5kg • Mg e-1-2m T T = 100 Nm (Acw) 5som pafostussurs pluss moment of inertia (I) for tod I me + Mez 3 I- (3 + 5) (1.2) = 8.64 kg-m of tod moment due to mass mg. e cose (CW) moment due to concentrated mass= Mgl coso ((w) Net monent To ng ecoso Malcosa Thes-100-(3x9.8 x 1:2 (650) - (sx9 81x1-2) 630 76.44 COJO (ACW) 100 - Thet Angular accelesation That I That IX d = 11.574 - 8.84726030 мр dw.do do dt wdw => dt 11.574 8.8472 Caso wew do at = 1/3 at o=60, w=o 0 = 300, w=w wow = 561-574 - 8-8472030) de w 300 = 53/ 1/3

52/3 2 with w 2 (11.5740 - 2.8472 Sino) w 11.574.49 + 15.3238 2(63.8048) 127.609 Iw = 11.2964 rad's CACW) when angle = 300, angular velocity of pendulum is 11-2964 radls (ACW). (2) f Ry Rx mwilk mg mg mw²e Rx = (mwe + Mure) sinzo (m + M) we sino Rx - (+) (11.2964), (1-2) (b) Rx [R = 497.673 N by - mg + Mg + (mwl + mwl) cozzo (3+5)9-01 + ( + +5) (11-2964)*(1-2) ? 78.48 + 862 940.47N 940.47 N 1