Applied Maths - Mechanics 35

Let T be the tension of the string
Let α be the angle made by the upper portion of the string with the horizontal while β the angle for the lower portion.
According to geometry,
b cosα = r (radius of circular motion) … (1)
Vertical component of force Tsinα – Tsinβ = mg … (2)
Horizontal component of force Tcosα + Tcosβ = mω2r … (3)
By cosine rule,
cos(90 – α) = (b2 + c2 – a2)/2bc
sinα = (b2 + c2 – a2)/2bc
(3)/(2) => (cosα + cosβ) / (sinα – sinβ) = ω2r / g
g(cosα + cosβ) = ω2r(sinα – sinβ)
g(cosα + b cosα/a) = ω2r[sinα – (c – bsinα)/a ]
g(acosα + b cosα) = ω2bcosα[asinα – (c – bsinα)] {use (1)}
g(a + b) = ω2b[(a + b)sinα – c]
g(a + b) = ω2b[(a + b)(b2 + c2 – a2)/2bc – c]
2gc(a + b) = ω2[(a + b)(b2 + c2 – a2) – 2bc2]
2gc(a + b) = ω2[ab2 + ac2 – a3 + b3 + bc2 – a2b – 2bc2]
2gc(a + b) = ω2(ab2 + b3 – a3 – a2b + ac2 – bc2)
2gc(a + b) = ω2[b2(a + b) – a2(a + b) + c2(a – b)]
2gc(a + b) = ω2[c2(a – b) + (b2 – a2)(a + b)]
2gc(a + b) = ω2[c2(a – b) – (a + b)(a – b)(a + b)]
2gc(a + b) = ω2(a – b)[c2 – (a + b)2]