Question

P3 (points) - The 15-kg slender rod is attached to a spring, which has an unstretched length of 3m. If the rod is released from rest when 0 = 45°, determine its angular velocity at the instant 0 = 60°. 5m B WWW 5m WW K= 50 N/m с

Answer 1

? 3 m 5 A Datum 45 mi (354 120° os 0 of Go m position 0 when o = 45° position ♡ when o = 60° k= 50 N/m length of road 5 m

The mass moment of inertia of the rod about A I A = 1 2 ME + M ( ²2 ) tex! x 15 x5 + 15x(5/2) = 125 me 1 = 1 1 10l = 1/2 x 125 wh = 6215 wr kg.m As the rod is released from rest, P=0. The gravitational potential energies of the rod at positions o no ® are (Vg), = mg (-4,) = 15*9.81*(-215 Sim 45') - - 260.13 J. (9) = mg (-y) = 15*9.81*(-2.5 sim 60). = - 318.59 T. The stretches of the spring spring when the rod is at positions 0 are

he = 2 [5 sin 22 (195.)] - 5 =4.2388m 2 [s sin (126)] -5 = 3.66 m. The initial and final elastic potential energies of the spring are (ve), - K2= * 504.2388= 449.18 ] I kag = 2 x 50 x 3.66 2 (ve), = 334.89 J

from conservation of energy 9, tv = 9 try = 0+ (-260. 13)+449.18 = 62.5 wit (-318.59) + 334.89 2 on = 2.764 1 to W = 1.66 rad's. The angular relocity of the rod at the instant- 0=60 is 1.66 rad/s.