Applied Maths - Mechanics 51

r = a(1 – cosθ)
dr/dt = asinθω
d2r/dt2 = acosθω2 since ω is constant dω/dt = 0
Acceleration in radial direction = d2r/dt2 – rω2
= acosθω2 – a(1 – cosθ)ω2
= 2acosθω2 – aω2
Acceleration in angular direction = rd2θ/dt2 + 2(dr/dt)ω
= 2asinθω2
Component of acceleration parallel to initial line
= (2acosθω2 – aω2)cosθ – (2asinθω2)sinθ
= 2acos2θω2 – 2asin2θω2– aω2cosθ
= 2acos2θω2 – 2aω2 + 2acos2θω2– aω2cosθ
= 4acos2θω2 – 2aω2 – aω2cosθ
= cosθω2(4acosθ – a) – 2aω2
= -cosθω2(4a – 4acosθ – 3a) – 2aω2
= -cosθω2(4r – 3a) – 2aω2 … (1)
Component of acceleration perpendicular to initial line
= (2acosθω2 – aω2)sinθ + (2asinθω2)cosθ
= 4asinθcosθω2 – asinθω2
= sinθω2(4acosθ – a)
= -sinθω2(a – 4acosθ)
= -sinθω2(4a – 4acosθ – 3a)
= -sinθω2(4r – 3a) … (2)
Consider these two perpendicular vectors (1) & (2), they are like one vector 2aω2 parallel to the initial line and towards original, and another vector, ω2(4r – 3a) pointing to the origin and inclined at angle θ to the initial line.