Question

The diagram shows a thin rod of uniform mass distribution plvoted 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. (a) (a) Determine the speed of its center of gravity at its lowest position in m/s. Consider the conservation of energy of the center of mass of the rod. What is the relationship between angular and tangential velocities? m/s (b) When the rod reaches the vertical position, calculate the tangential speed of the free end of the rod in m/s. Check your calculations. m/s

Answer 1

Conservation of energy. Conversion of Gravitational PE into Rotational Kinetic energy.
By law of conservation of energy, Ghange in gravitational P.C. & Gain in K.B. (portational) malha-h.) - Į I w2 ha, h, are initial and final height of centre of mass. Moment of inertia (1) about the linge = ₂ MLL yang yun Box Maxx wz 39

Now, speed of centre of gravity v = wxr = ا المه - Jos y Tv - Jag = 4.2 m/s Speed of lowest point vewr = Bul Soo x L Iraq.4 m/s