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A uniform rigid rod of length l and mass m is hinged at one end O and can rotate in a vertical plane. Initially it is held vertically. When released from rest, it rotates about O (all surfaces are frictionless). Which of the following statements is/are correct? (A) The magnitude of the hinge force when the rod becomes horizontal is (sqrt(37)/4) mg. (B) The centre of mass undergoes uniform circular motion. (C) The hinge force is directed along the +y axis when the centre of mass is at its lowest position. (D) The mechanical energy of the rod is not conserved during the motion.
- (A)
- (B)
- (C)
- (D)
Correct answer: (A)
Solution
Statement A: When rod is horizontal, energy conservation: mg*(l/2) = (1/2)*(ml²/3)*omega² => omega² = 3g/l. Torque about O: tau = mg*(l/2) (weight acts at CM). I*alpha = mg*(l/2) => (ml²/3)*alpha = mgl/2 => alpha = 3g/(2l). At CM (l/2 from O): centripetal acceleration a_c = omega²*(l/2) = 3g/2 (directed toward O, i.e., horizontal, along -x if rod is along +x). Tangential acceleration aₜ = alpha*(l/2) = 3g/4 (directed downward when rod horizontal). Net force on rod: F_net = m*a. From Newton's 2nd law for rod: Hₓ - 0 = m*(-3g/2) [Hₓ is horizontal hinge force, centripetal toward O is in -x direction]. Wait: when rod is horizontal (along x-axis), centripetal acceleration is directed toward O which is at origin, so in -x direction. Tangential acceleration is perpendicular to rod and in -y direction (downward). Sum of forces on rod: Hₓ + 0 = m*aₓ = m*(-3g/2)... no. Let me use: Sigma Fₓ = m*a_CMₓ. a_CMₓ = -omega²*(l/2) = -3g/2 (centripetal, toward O when rod horizontal along +x). Sigma F_y = m*a_CM_y. a_CM_y = alpha*(l/2) downward? When rod is horizontal, alpha is counterclockwise (rod accelerating angularly), so tangential acc at CM is perpendicular to rod... In horizontal position, rod along +x. Tangential acc is in -y direction (downward, as gravity makes rod rotate downward). Sigma F_y = H_y - mg = m*(-3g/4). H_y = mg - 3mg/4 = mg/4. Hₓ = m*(-3g/2) (centripetal toward O) = -3mg/2? Then hinge force: Sigma Fₓ = Hₓ = m*a_CMₓ: the only horizontal force is hinge (weight is vertical). Hₓ = -3mg/2 (but we want magnitude). H_y: Sigma F_y = H_y - mg = m*a_CM_y = m*(-3g/4). H_y = mg - 3mg/4 = mg/4. |H| = sqrt(Hₓ² + H_y²) = sqrt((3mg/2)² + (mg/4)²) = mg*sqrt(9/4 + 1/16) = mg*sqrt(36/16 + 1/16) = mg*sqrt(37/16) = mg*sqrt(37)/4. Statement A is CORRECT. Statement B: The CM moves in a circle (constant radius l/2 from O), but the speed changes (non-uniform circular motion since alpha != 0). NOT uniform. B is incorrect. Statement C: Lowest position of CM means rod is horizontal. At this point, hinge force is not purely along +y (as shown: Hₓ = -3mg/2 != 0). C is incorrect. Statement D: Only gravity does work (hinge force does no work since hinge is fixed). No friction. Mechanical energy IS conserved. D is incorrect. Only A is correct.
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