Question

Please provide fully worked solution

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 o T g

Answer 1

OM L A (mt mig 160) 300 A SOM (m+m)g IA = m22 + ML? 12 IA = (3) (1-2) + 5(1:2)? 12 JA = 0.36+ 7.2 = 7.56 kg.m? .com m()+ MIL) mt M Lcom - 3(12) + 5 (1-2) 3+5 .com = 0.975 m (1). W Torque - ARE + APE τ ΔΘ = & Inw? + (m+M)g (2.com Sinbo') 100f (500-609 m.] (750) + (364) (1 81[2* 0:976 sm 60} 418.88 = 3.78w? + 132.39 2 2 با - 75.79 w = 8.7 rad /8² (22 Lolcom 2 130 T- (MAM) g con 30 (MtM) 9 Sm 36 mmg (MAM)q Sm 36 (Lcom) = I a 100-(3+5) (9.8)(0.5)(0.975) = 7.560 a = 8.17 rud 1x² 2

N₂ - (m +M) g (ox 30 = (m +M) w².com N2 - (3+5)(9.8) CA 30 = (3+5) le (75.79) (0.975) N, - 591-162 + 67.896 Ny = 659.06N N, - (mtm Sm 30 = (MAM) (.com ) N, -(3+5)(9.8) (0.5) - (3+5)(0.975) ( B.1) N = 63.726 + 39.2 N = 102.93 N