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Dynamic equilibrium; moment of block B on rod

Short answer

In dynamic equilibrium, an imaginary force -m a is added so that ΣF - m a = 0. For the block, transverse acceleration a_θ = rα + 2ṙω = 1.45 m/s², transverse force F_θ = 8.87 N. Moment exerted about O is r F_θ = 0.89 N·m, which the block exerts on the rod.

Discuss dynamic equilibrium with an example. In figure below, the motion of block B on a smooth horizontal table is controlled by the rod which rotates about a vertical axis at O. At the position shown, B has the given values of velocity and acceleration relative to the rod which is rotating with the given values of angular velocity ω and angular acceleration α. If B weighs 60 N, what moment does it exert about O? [Figure: rod from O at 30° to the horizontal, block B at 10 cm from O on the rod, relative velocity V = 22.5 cm/sec, relative acceleration a = 15 cm/sec², α = 1 rad/sec², ω = 3 rad/sec.] OR: see the next question.

Dynamic equilibrium

When a body is accelerating, adding an imaginary inertia force (−m a) to the actual forces makes the body appear to be in equilibrium (D'Alembert's principle): ΣF − m a = 0. Example: a lift moving upward with acceleration a carries a person of mass m. The real forces are the normal reaction N and the weight mg. With the inertia force m a added downward, N − mg − m a = 0, so N = m(g + a); the person feels heavier. A car taking a curve is similar: the centrifugal (inertia) force m v²/r balances the friction force at the tyres.

Moment of block B about O

Block B (W = 60 N) slides on a smooth horizontal table along the rod. The vertical forces cancel, and the horizontal forces on B have radial and transverse components. Data in polar coordinates: r = 10 cm = 0.1 m, ṙ = 22.5 cm/s = 0.225 m/s, r̈ = 15 cm/s² = 0.15 m/s², ω = θ̇ = 3 rad/s, α = θ̈ = 1 rad/s². Mass m = 60/9.81 = 6.116 kg.

  • a_r = r̈ − rω² = 0.15 − 0.1(9) = −0.75 m/s²
  • a_θ = rα + 2ṙω = 0.1(1) + 2(0.225)(3) = 1.45 m/s²
  • Transverse force (from the rod) F_θ = m a_θ = 6.116 × 1.45 = 8.87 N
  • Radial force F_r = m a_r = −4.59 N (it passes through O, so it has no moment about O)

Moment about O:

M_O = r F_θ = 0.1 × 8.87 = 0.887 N·m

Check using H_O = m r²θ̇: dH_O/dt = m(r²θ̈ + 2rṙθ̇) = 6.116(0.01 + 0.135) = 0.887 N·m ✓.

Answer: M_O ≈ 0.89 N·m. The rod exerts this moment on B in the sense of the rotation; B therefore exerts an equal moment of 0.89 N·m on the rod about O in the opposite sense (opposing the rotation).

✓ Answer checked by Claude Opus 5.5 (high effort)

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