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 T bo φ

Use Reaction forces and Energy methods. Please show all working. Thank you very much.

Answer 1

M ma Solution- 3 kg M-5 t-1.2 m 5 kg Img vmg T= loo N-m [ACW] T no 1] Moment of Inertia for rod and concentrated mass = I= me? + + M1² 3 = culo g+5) (1.2)" I= 8.64 kg-ma [cm] mass due to of rod - mg + coso + moment concentrated mass = moment due to Mge cosø ... [CH] COS Net mgel moment T- mgie cos o - Tnet - 100- (3 x 9.81 x 12 x Cos o) - (5 x 9.81x1-2 x roso) 2 Thet - 100 - 76.518 Cos o acceleration x Angular Tnet Ια .: Q- Thet I 11.574 - 8.8562 Coso d = 100- 76.518 COSO 8.64 Cs Scanned with CamScanner

x= dw dt dw.do do dt w du do . w dw11.574 - 8.8562 caso do at 60° 60'- IT 3 WO at Q = 300° W- W 300°: 5 TT cules 5/3 Gurdw : 511-574- 8-8562 cos o] de T/3 517/3 [11-574 - 8.8562 Sino) we T/3 w2 63.8204 2 W = 11.2978 rad/s ACW [angular velocity] OOO 2 ] Pg Roc :30 mwellz mg Mwel mg Cs Scanned with CamScanner

By the condition of equillibrium Fx po Rc = (m mw? + MW? 4) sin 30 2 Ra = (m + M) Wed sin 30 ( 3+5) Ra (11-2978)2 x (1-2) Sin30 . Rac = 497. 7971 N MW24 Eryto Ry - mg + Mg (3+5)*9.81 + (1 +5) (11-2978)* x1-2 x (os 30 Cos 30° mg + (mw? (mweļ + 124) Ry: 940. 940.6898 N CS Scanned with CamScanner