Engineering Mechanics: Dynamics

Engineering Mechanics: Dynamics Newton’s 2nd Law & Linear Momentum of a Particle • When a particle of mass m is acted upon by a force the acceleration of the particle must satisfy r r F = ma

• Replacing the acceleration by the derivative of the velocity yields r r dv ∑F = m dt r d r dL = (m v ) = dt dt r L = linear momentum of the particle 12 - 1

Engineering Mechanics: Dynamics Equations of Motion • Newton’s second law provides r r F = m a ∑ • Solution for particle motion is facilitated by resolving vector equation into scalar component equations, e.g., for rectangular components, r r r r r r ∑ (Fx i + Fy j + Fz k ) = m(a x i + a y j + a z k )

∑ Fx = ma x ∑ Fy = ma y ∑ Fz = ma z ∑ Fx = m&x& ∑ Fy = m&y& ∑ Fz = m&z& • For tangential and normal components,

∑ F t = mat dv = F m ∑ t dt

∑ F n = man ∑Fn = m

v2

ρ

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