Applied Maths - Mechanics 38

Let R be the normal reaction & F the friction
When the tube is stationary, F is pointing up slope
mgcosβ = R
mgsinβ = F
F/R = tanβ is the coefficient of friction
When the tube is rotating slowly, i.e. the friction is still pointing up slope, resolving forces,
Vertical: Rcosβ + Fsinβ = mg … (1)
Horizontal: Rsinβ = Fcosβ + mω2r … (2) and r = asinβ
(1)cosβ + (4)sinβ => R = mgcosβ + mω2acosβsinβ … (5)
(1)sinβ – (7)cosβ => F = mgsinβ – mω2asinβcosβ … (8)
(8)/(5) => F/R = (sinβ/cosβ)(mg – mω2acosβ)/(mg + mω2asinβ) < sinβ/cosβ
The turning point from slow to quick is when F = 0, i.e. mg – mω2acosβ = 0
Or g/(acosβ) = ω2


When the tube is rotating quickly [ω2 > g/(acosβ)], i.e. the friction is pointing down slope, resolving forces,
Vertical: Rcosβ – Fsinβ = mg … (1)
Horizontal: Rsinβ – Fcosβ = mω2r … (2) and r = asinβ
Solving, similar to above,
R(sin2β – cos2β) = mω2acosβsinβ – mgcosβ … (5) [sin2β – cos2β>0]
F(sin2β – cos2β) = mω2asinβcosβ – mgsinβ… (8)
(8)/(5) => F/R = (sinβ/cosβ)(mω2acosβ – mg)/(mω2asinβ – mg) … (9)
Since ω2 > g/(acosβ), both (mω2acosβ – mg) and (mω2asinβ – mg) are positive since sinβ > cosβ for β > π/4
And (9) is < sinβ/cosβ so in both cases, the particle can remain at P.


If β < π/4, when the tube turns too quick, (5) becomes
R(sin2β – cos2β) = mω2acosβsinβ – mgcosβ
Or R = (mgcosβ – mω2acosβsinβ)/(cos2β – sin2β)
If R is calculated to be negative, stability cannot be maintained.
Possible range is determined by mgcosβ – mω2acosβsinβ > 0
Or g – ω2asinβ > 0
g/(asinβ) > ω2

2009-08-11 21:45:29 補充：
There is a mistake in the calculation, solution re-attached here.
http://www.funp.net/89905

2009-08-12 00:08:26 補充：
This is the correct one after checking further errors
http://www.funp.net/127106

2009-08-12 00:08:39 補充：
This is the correct one after checking further errors
http://www.funp.net/127106