Applied Maths - Mechanics 39

還不明白的薑

2009-08-11 10:50 pm

A particle which moves on the inner surface of a fixed, smooth hollow sphere of radius a is attached to the topmost point of the sphere by a light inextensible string of length b, were a*sqrt(2) < b < 2a . If the particle describes a circle with angular velocity w in a horizontal plane, with the string taut, prove that 2ga/b^2 <= w^2 <= 2ga/( b^2 - 2a^2 ) .

回答 (1)

用戶9112

2009-08-12 4:20 am

✔ 最佳答案

Let θ be the angle between the string and the vertical.
Let R be the normal reaction and T the tension
By geometry (cosine rule)
cosθ = (a2 + b2 – a2)/2ab = b/2a
Resolving forces,
Horizontal : Tsinθ + Rsin2θ = mω2r and r = bsinθ
Tsinθ + 2Rsinθcosθ = mω2bsinθ
T + 2Rcosθ = mω2b … (1)
Vertical : Tcosθ + Rcos2θ = mg … (2)
Sub (1) into (2)
(mω2b – 2Rcosθ)cosθ + Rcos2θ = mg
R(cos2θ – 2cos2θ) = mg – mω2bcosθ
R(2cos2θ – 1 – 2cos2θ) = mg – mω2bcosθ
R = mω2bcosθ – mg … (3)
R = mω2b2/2a – mg >= 0
ω2 >= 2ga/b2
Sub (3) into (1)
T = mω2b – 2Rcosθ
T = mω2b – (b/a)(mω2b2/2a – mg)
T = mω2b[1 – (b/a)b/2a] + mg(b/a)
T = mω2b[(2a2 – b2)/2a2] + mg(2ab)/(2a2) >=0
So 2mgab >= mω2b(b2 – 2a2)
2ga/(b2 – 2a2) >= ω2

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