Banked Circular Tracks question (1 Viewer)

[vds700]

hi im having trouble with parts 2 and 3 of this questio0n, im fine with part 1.

Any help much appreciatred

 

[Trebla]

ii) Since F ranges from -μN to μN, the maximum value of F is -μN (i.e. in the opposite direction as shown, to keep it from going up the bank) for it to travel without slipping:
So sub F = - μN
v²/Rg = (Nsin θ + μNcos θ)/(N cos θ - μNsin θ)
Multiply RHS by (sec θ/sec θ)
v²/Rg = (Ntan θ + μN)/(N - μNtan θ)
v²/Rg = N(tan θ + μ)/N(1 - μtan θ)
Hence v²/Rg = (tan θ + μ)/(1 - μtan θ) at v max

 

[Trebla]

iii) The particle only slides down when F > μN, if -μN < F < μN, then it will not slide.
So assume the particle does slide down, then
F/N > μ (N > 0)
but mv²/r = Nsin θ - Fcos θ > 0
=> F/N < tan θ
So μ < F/N < tan θ
=> μ < tan θ
But we actually DON'T want it to slide down thus
μ ≥ tan θ

lol

 

[3unitz]

iii) to stop slipping down plane the centripetal force and friction has to balance or be greater then the weight component acting down the plane

mg sin@ < (mv^2/r)cos@ + uN

mg sin@ - (mv^2/r)cos@ < uN

[mg sin@ - (mv^2/r)cos@] / N < u

since (mv^2/r)cos@ cant be negative, the LHS will be greatest when v=0 (i.e. object not moving)

to stop sliding down the plane when this occurs: mg sin@ / N < u

and we can deduce mg/N = 1/cos@ from some triangles, so

tan@ < u

hence so long as friction constant is greater than tan@, no sliding down the plane will occur no matter what velocity.

 

[vds700]

Thanks Trebla and 3unitz for your help.

my brain is dead now so ill try to understand it tomorrow when i can think clearer

 

[Trebla]

i dont think this is true? =S

The question says that the particle does NOT move up or down the bank and also that -μN ≤ F ≤ μN. Hence the system is 'stable' within that range of F.
So, if F < -μN, then the friction force is down the slope hence the particle will slip up and if F > μN, then the friction force is up the slope hence the particle will slip down.
(Also, if this wasn't true the statement in the last question will be false)

 

[3unitz]

iii) The particle only slides down when F > μN, if -μN < F < μN, then it will not slide.
So assume the particle does slide down, then
F/N > μ (N > 0)
but mv²/r = Nsin θ - Fcos θ > 0
=> F/N < tan θ
So μ < F/N < tan θ
=> μ < tan θ
But we actually DON'T want it to slide down thus
μ ≥ tan θ

lol

so the particle doesnt slide down when: F < μN
F/N < μ
tan θ < μ

rather then going through everything else

 

[vds700]

The question says that the particle does NOT move up or down the bank and also that -μN ≤ F ≤ μN. Hence the system is 'stable' within that range of F.
So, if F < -μN, then the friction force is down the slope hence the particle will slip up and if F > μN, then the friction force is up the slope hence the particle will slip down.
(Also, if this wasn't true the statement in the last question will be false)

why is it that the particle slips in the opposite direction to which way the friction force, F, is directed?

 

[3unitz]

why is it that the particle slips in the opposite direction to which way the friction force, F, is directed?

the friction force acts to oppose the slipping, hence opposite direction. friction forces are always opposite to motion.

 

[vds700]

the friction force acts to oppose the slipping, hence opposite direction. friction forces are always opposite to motion.

ah ok. Thanks man

 

[Trebla]

so the particle doesnt slide down when: F < μN
F/N < μ
tan θ < μ

rather then going through everything else

The problem with that way is that you can't immediately conclude to the last line.
So if F/N ≤ μ (1)
You want to use: F/N < tan θ (2)
but you can't tell for sure (or at least without some explanation) if tan θ is less than or greater than μ.
From (1) to (2), you can also have tan θ ≥ μ for (2) to be true, as well as tan θ ≤ μ.