Question

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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 0=300° to the horizontal, calculate: 1. Angular velocity of the pendulum 2. Reaction forces at the pivot M L m Ꮎ T g

Answer 1

1 M = 5 kg m = 3 kg L=1.2 m T = 100 N.m 0,= 60° - =300 2.

In the image , 0 initial position is marked as and final position as , = 60 Initially Finally the rotation is 300° .. o O2 = 300 Initially the rod is at rest si Wi = 0 Finally suppose the anguler velocity - Wz Rod mass moment of inertia about the pivot point IR - 1 x mxt = 1.44 kg-m of the concentrated mass Mass moment of inextin about pivot point Ic = My ♡ 7.2 kg-m

Total mass moment of innertia I Ip t I c 8.64 kg-m' Initially w,=0 .: Initial anguler kinetic energy O Exi The center of gravity (ca) of the rod is at its midspan 는 Loo o.b : Initally the potential energy of the rod - mxgx (Lea & sino) = 3x9.81 x (0.6 X sin bo') 15.3 J Initially the potential energy of the mass - Mx gx (LX sin o) 50.97 J Total initial potential energy Ep, 15.3 * 50.97 - =66.27 J

At final state The angular kinetic energy / IX Wz = 4.32 WY Ekz J The potential energy of rod m xgx (Lea sing, = - 15.3 J [ re comes because it is below the horizontal Potential energy of the mass datum line Mx gx (L sin o) 50.97 J :. To tull potential energy Ep = (-15.9)+(-50.07) 66.27 J The total evork done on the system during this period W = TX [Anguler displacement in radian] Auguler displacement O, 240 240 rad 10 180 x .: 40 = 4.189 rad.

W = 110 x 4.1897 = 4180) J ². Energy balance Ek + Ep, tW = Ekzt Ep, +(-66.27) of 66.27 +418.9= 4.32 wy 4.32 W 551.44 W2 11.3 rad/s .. The auguer relocity at final position is rad/s. Am Wz = 11:3 ② At final position, there will be only 2 force acting on the pirot. & Weight of the system =) Centrifugal force of the system Weight of the system acting downward Ws (m + M) xq = 78. 48 Centrifugal force long the rod mx L ca x wa 229.77 N Centrifugal force by the mass

Mx L x W 765.89 N .: Total centrifugal forc. F = 228.77 + 715.80 =Dg5.66 N As vector, the 2 forces are 270 Ø = 62- 270 30 Ws F Horizontal component magnetude. Fc Fc Fer Fc sino = 497.83 1 vertical component magnetude For = Fc cos oso = 862:27 Fch Total Horizontal component force FA 407.83 M Total vertical component force Fra 940.75 N Wst For

. Reactim forces on the pivot FH = 497.83 N Ev=940.75 m If you have in the comments. any confusim with the solution, ask me