A small ring A is threaded on a rough wire bent in form of a circle with centre O and radius a. The wire is rotated about a fixed vertical diameter with angular speed w. The ring remains stationary relative to the wire when OA is at an acute angle @ to the downward vertical. The coefficent of friction between the ring and the wire is m. Prove that the ring will tend to slip upwards on the wire if w^2 > g/acos@ , but that it will not in fact slip, no matter how great w may be, if m >= cot@ .
If m < cot@ , find the smallest value of w which will cause the ring to slide up the wire.
Applied Maths - Mechanics 40
2009-08-11 11:57 pm
回答 (1)
2009-08-12 7:52 am
✔ 最佳答案
Let R be the normal reactionLet F be the friction
Let M be the mass of the ring
When the tube is rotating slowly, i.e. the friction is still pointing up slope, resolving forces,
Vertical: Rcosθ + Fsinθ = Mg … (1)
Horizontal: Rsinθ = Fcosθ + Mω2r … (2) and r = asinθ
(1) => Rcos2θ + Fsinθcosθ = Mgcosθ … (3)
(2) => Rsin2θ = Fsinθcosθ + Mω2rsinθ … (4)
(3) + (4) R = Mgcosθ + Mω2asin2θ … (5)
(1) => Rsinθcosθ + Fsin2θ = Mgsinθ … (6)
(2) => Rsinθcosθ = Fcos2θ + Mω2rcosθ … (7)
(6) – (7) => F = Mgsinθ – Mω2asinθcosθ … (8)
(8)/(5) => F/R = (Mgsinθ – Mω2asinθcosθ)/(Mgcosθ + Mω2asin2θ)
F/R = (gsinθ – ω2asinθcosθ)/(gcosθ + ω2asin2θ)
When g – ω2acosθ<0 the ring tends to slip upwards which is equivalent to
g < ω2acosθ or g/acosθ < ω2
In that case,
F/R = (ω2asinθcosθ – gsinθ)/(gcosθ + ω2asin2θ)
when ω → ∞, F/R → cotθ
If cotθ <= μ, F/R always <= μ, i.e. no slipping.
If cotθ > μ, then F/R = μ at max.
(ω2asinθcosθ – gsinθ)/(gcosθ + ω2asin2θ) = μ
ω2asinθcosθ – gsinθ = (gcosθ + ω2asin2θ)μ
ω2asinθ(cosθ – μsinθ) = gsinθ + μgcosθ
ω2 = g(sinθ + μcosθ) / [asinθ(cosθ – μsinθ)]
The minimum angular velocity is
ω = √{g(sinθ + μcosθ) / [asinθ(cosθ – μsinθ)]}
收錄日期: 2021-04-23 23:18:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090811000051KK01260