Question

The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.490 kg and its length is 2.40 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown. M L/2 1.CG (a) Determine the speed of its center of gravity at its lowest position. m/s (b) When the rod reaches the vertical position, calculate the tangential speed of the free end of the rod. m/s

Answer 1

AnS :- mass m = 0.490 kg. Given: Length I = 2.40m applying energy conservation. E = E₂ mgh mU² + 1 IW2 h=42 C.G mol ) = _MU2 + 499 moment of inertia of rod about its end. r I= m² 3 de L2 U2 + 9.5 = ald o = 1/2=2.40 2 = 1,20 V2 +(2.40) 9.8 X 2140 = 12 (2,40/ 22 742 3 23.52 uztuva 3 - v=3,17 mis N2= 3x23.52 7 N 3.17 2.90/2 2 q angular velocity of C.G w = 2.69 rad's the angular reloury of G. G will be same as angular Velocity of Poee end and so tangenhol veloury od poee ond Row for free and R=4 = 2.40m 2.40 x 2.64 6.33 m/s IN