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Hinged rod with a rope

A homogeneous rod AB (M = 8 kg, L = 1.2 m) is hinged at A and held horizontal by a rope at B making an angle α = 40° with the rod. Find: rope tension, vertical reaction at the pivot, and the angular velocity when — once the rope is cut — the rod passes through the vertical.
AB T mg α
Given data
M = 8 kgL = 1.2 mα = 40°
Review the theory: Dinamica
Steps
0 / 3
  1. The rod is in equilibrium: the sum of torques about pivot A is zero. I take torques about A so the pivot reaction (applied at A) does not appear. The weight, applied at the center, tends to rotate the rod downward; the rope holds it back. Equating the two torques: T·L·sinα = Mg·(L/2), hence T = Mg/(2sinα).
  2. Besides torques, the forces must balance too. Vertically the weight pulls down, while it is held up by the vertical component of the tension (T·sinα) and the vertical pivot reaction. From vertical equilibrium I get N_y = Mg − T·sinα.
  3. Once the rope is cut the rod falls rotating about A. I use energy conservation: the center of mass drops by L/2, so the lost potential energy Mg·(L/2) becomes rotational kinetic energy ½Iω², with I = ML²/3 (rod rotating about one end). From Mg·(L/2) = ½·(ML²/3)·ω² I get ω = √(3g/L).
Full worked solution
  1. The rod is in equilibrium: the sum of torques about pivot A is zero. I take torques about A so the pivot reaction (applied at A) does not appear. The weight, applied at the center, tends to rotate the rod downward; the rope holds it back. Equating the two torques: T·L·sinα = Mg·(L/2), hence T = Mg/(2sinα).
    T=Mg2sin⁡α=8⋅9.812⋅sin⁡40°T = \frac{Mg}{2\sin\alpha} = \frac{8\cdot9.81}{2\cdot\sin40°}
    T=78.51.286≈61.0 NT = \frac{78.5}{1.286} \approx \mathbf{61.0\,N}.
  2. Besides torques, the forces must balance too. Vertically the weight pulls down, while it is held up by the vertical component of the tension (T·sinα) and the vertical pivot reaction. From vertical equilibrium I get N_y = Mg − T·sinα.
    Ny=Mg−Tsin⁡α=78.5−61.0⋅sin⁡40°N_y = Mg - T\sin\alpha = 78.5 - 61.0\cdot\sin40°
    Ny=78.5−39.2=39.3 NN_y = 78.5 - 39.2 = \mathbf{39.3\,N}.
  3. Once the rope is cut the rod falls rotating about A. I use energy conservation: the center of mass drops by L/2, so the lost potential energy Mg·(L/2) becomes rotational kinetic energy ½Iω², with I = ML²/3 (rod rotating about one end). From Mg·(L/2) = ½·(ML²/3)·ω² I get ω = √(3g/L).
    ω=3gL=3⋅9.811.2\omega = \sqrt{\frac{3g}{L}} = \sqrt{\frac{3\cdot9.81}{1.2}}
    ω=24.53≈4.95 rad/s\omega = \sqrt{24.53} \approx \mathbf{4.95\,rad/s}.
Result:T≈61T \approx 61 N, Ny≈39.3N_y \approx 39.3 N, ω≈4.95\omega \approx 4.95 rad/s
TheoryDinamicaFormula sheetPhysics 1

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