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Angular momentum rate; 5 kg particle under F

Short answer

Differentiating angular momentum H_O = r×m v gives dH_O/dt = v×m v + r×m a = r×F, so rate of change equals moment of force. For x-components: a_x = F_x/m = 2.4t, at t=5 s a_x=12 m/s²; velocity v_x = 1.2t² = 30 m/s; position x = 0.4t³ = 50 m.

Show that, "rate of change of angular momentum about a point is equal to moment of the force about the same point." The resultant external force acting on a 5 kg particle in space is F = (12t i − 24t² j + 40t³ k) N, where t is in seconds. The particle is initially at rest at origin. Determine the x component of acceleration, velocity and position at the instant of 5 sec.

Proof: rate of change of angular momentum = moment of the force

Let a particle of mass m have position vector r from a fixed point O and velocity v = dr/dt. Its angular momentum about O is

H_O = r × m v

Differentiate with respect to time (product rule):

dH_O/dt = (dr/dt) × m v + r × m (dv/dt) = (v × m v) + r × m a

  • v × m v = 0, because the cross product of a vector with itself is zero.
  • m a = F (Newton's second law), where F is the resultant force.

Hence dH_O/dt = r × F = M_O: the rate of change of angular momentum about O equals the moment of the resultant force about the same point O.

Numerical part (m = 5 kg, F = 12t i − 24t² j + 40t³ k N, at rest at the origin)

  • x-component of acceleration: aₓ = Fₓ/m = 12t/5 = 2.4t, so at t = 5 s: aₓ = 12 m/s²
  • Velocity: vₓ = ∫2.4t dt = 1.2t², so at t = 5 s: vₓ = 1.2(25) = 30 m/s
  • Position: x = ∫1.2t² dt = 0.4t³, so at t = 5 s: x = 0.4(125) = 50 m

Answer: aₓ = 12 m/s², vₓ = 30 m/s, x = 50 m at t = 5 s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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