Curvilinear Motion of Particles [Kinematics]
December 8, 2016
Categorised in: 1st Year Mechanics
$\table Direction,Velocity-Component,Acceleration-Component;x,v_x={dx}/{dt},a_x={dv_x}/{dt};y,v_y = {dy}/{dt},a_y = {dv_y}/{dt};Tangential,v_T = V, a_T = {dv}/{dt};Normal, v_N = 0, a_N = {v^2}/{ρ};Radial, v_r = r’,a_r = r” – r θ’^2;Transverse, v_θ = r θ’, a_θ = 2r’θ’ + r θ”$
Note:
⋆ $a_T$ represents change in speed while $a_N$ represents change in direction.
⋆ At point of inflection ρ -> ∞ so $a_N$ = 0
⋆ Always use kinematic equations with $a_T$ in curvilinear motion
Methods to find Radius of curvature
(1) When speed v and $a_N$ are known
$|ρ| = {v^2}/{a_N}$
(2) When path equation y = f(x) is given,
$|ρ| = {[a + ({dy}/{dx})^2]^{3/2}}/{{d^2 y}/{dx^2}}$
(3) When rectangular components in 2D motion are known
$|ρ| = {[{v^2_x} + {v^2_y}]^{3/2}}/{v_x a_y – a_x v_y} = {v^3}/{v_x a_y – a_x v_y}$
(4) When $\ov{v}$ and $\ov{a}$ are known
(a) Find magnitude of tangential acceleration
$a_T = {\ov{v} * \ov{a}}/{|v|}$
(b) Then normal acceleration
$a_N = √{a^2 – {a^2_T}}$
(c) Then $ρ = {v^2}/{a_N}$
Note when $\ov{v}$ is perpendicular to $\ov{a}$
$\ov{v} * \ov{a} = 0$
$a_T = 0$
$a_N$ = a
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