Question

A 12 kg slender rod is placed against the wall at an angle of 45 degrees as shown. If it is released from rest, determine the angular velocity of the rod at the instant just before it hits the ground (angle = 0 degrees). Neglect friction. Moment of Inertia Ig of rod = 1 m 12 6 meters

Answer 1

Solution Applying T + principle of work and energy, Eurz = Te - 6 m=12 kg om 3m ks 45 MB since the rod is initially at rest, so Ti=0 Referring to FBD of sod NA and NB do no work while w does positive work, when 0-0 w displaces vertically through a distance of h= zsin45 h = 2:1213 m Thus, the work done by w is, Uw = mgh = 12x9.81x 2-1213 = 249.7218 Joule NB when the rod just hits the ground (o=0) I before IC - Instantaneous centre K (SAIIC - IC. frotect> B)2 (VG), = wz Toled CG)2 = W₂ (3) CA) W2 (VG)2 - 3M The mass moment of inertia about its mass centre is. In = 12 me² = 12x12x6² = 36 kg-m² This the final kinetic energy is, T2 = 1 m (G) + I GW² T2 = 2 x 12 x (36₂)2 + 2 x 36 xW₂ ² = 72 W2² substituting Ti, Tz and Uw in equation , T + Eura - Ta o + 249. 7218 = 72 W3 W2 = 10862 rad/s Hence angular velocity of rod just before it hits the ground is W = 1.862 radys