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Pendulum bob velocity and acceleration; momentum

Short answer

Linear momentum = m v; angular momentum about O = r × m v. For the bob: T − W cos30° = m v²/L gives v = 5.66 m/s. Tangential acceleration = g sin30° = 4.91 m/s², normal = v²/L = 16.03 m/s²; total acceleration = 16.76 m/s².

Define the linear momentum and angular momentum. Find the velocity and the acceleration of the bob in the given position. The bob of a 2 m pendulum describes an arc of a circle in a vertical plane, which is shown in figure below. If the tension in the cord is 2.5 times the weight of the bob for the position shown. [Figure: pendulum with pivot O, cord length 2 m, cord at 30° to the vertical, bob B.]

Linear and angular momentum

  • Linear momentum of a particle of mass m and velocity v: L = m v (kg·m/s), a vector along the velocity.
  • Angular momentum about a point O: H_O = r × m v, with magnitude m v r sin φ (kg·m²/s); for motion in a circle of radius r, H_O = m v r.

Pendulum bob

The bob moves on a circle of radius L = 2 m. At the position shown the cord makes θ = 30° with the vertical and the tension is T = 2.5W. Let W = mg.

        O
        |\
        | \ 2 m
        |30\
        |   \ T
        |    B   (mass m, weight W)

Normal (toward O) direction: T − W cos 30° = m v²/L 2.5W − 0.866W = (W/g) v²/2 1.634 g × 2 = v², so v² = 1.634 × 9.81 × 2 = 32.06 v = 5.66 m/s (along the tangent to the circle, perpendicular to the cord)

Tangential direction: W sin 30° = m a_t, so a_t = g sin 30° = 4.905 m/s²

Normal acceleration: a_n = v²/L = 32.06/2 = 16.03 m/s² toward O

Total acceleration: a = √(16.03² + 4.905²) = 16.76 m/s², at tan⁻¹(4.905/16.03) = 17.0° to the cord.

Answer: v = 5.66 m/s; a = 16.76 m/s².

✓ Answer checked by Claude Opus 5.5 (high effort)

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