Question

The circular concrete culvert rolls with an angular velocity of w = 0.56 rad/s when the man is at the position shown. At this instant the center of gravity of the culvert and the man is located at point G, and the radius of gyration about G is kg = 3.2 ft. (Figure 1) PI Figure ( 1 oft 4 ft 0.5 ft

Answer 1

Use parralel axis theorem to calculate the mass moment of inertia at point O. Equate the net torque about point O with the product of mass moment of inertia at point O and angular acceleration at point O to calculate the angular acceleration of the culvert.

Solution Mass moment of inertia about point FG) >> Ms (kg) ² G IG » 500 (3.2)2 0.57 Ms 32.174 Is 7 159.135 lbm ft2 Fi motse weight met betretion due to gravity ] From parsellel axis theorem, inertia about o is, Io » In + Mg(OG)2 (0.5)2 To =) 159.135 t 500 32.194 To 163.02 lbm ft² Ws a Net moment about about point o is, torque To > Ws COG) To ) (500) (0.5) * 250 lb-ft بح weight of the system To ) To L (263) Angular acceleration about bout polnt o 250 ) 163.02 Lo Lo » 1053 rad/32 / Ay