Finding the range of speed on a banked curve without slipping

[totallyclone]

Homework Statement

A curve of radius 60 m is banked for a design speed of 100km/h. If the coefficient of static friction is 0.30, at what range of speeds can a car safely make the turn.

Homework Equations

Fun=ma
Ff=μsFn
Fc=mV²/r

The Attempt at a Solution

So, when it is "designed" for 100km/h that means there is no friction at 100km/h or 27.8m/s so this is my fbd for this speed. I've tried working it out and I manage to get okay looking answers but I'm not sure if I'm doing it right... :P

So, there is no acceleration in the Y-component:
ƩFy=ma
Fny-Fg=0
Fny=Fg
Fncosθ=mg 1st eqn.

For the X-component, there is acceleration which is the centripetal acceleration:
ƩFx=ma
Fnx=mV²/r
Fnsinθ=mV²/r 2nd eqn.

Now, to find the angle of the bank at that "design", so I isolated for θ:
2nd eqn./1st eqn.
Fnsinθ/Fncosθ=(mV²/r)(1/mg)
tanθ=V²/rg
θ=tan^-1(27.8²/60x9.8)
θ=52.7°

Now that I found the angle it is for 100km/h to make the turn in the bank without the need for friction, I'm trying to find the maximum speed for it in order not to slide up the bank so now I'll throw in friction and this is my fbd.

So I'll use the same concept with the Fun=ma just that I have Ff now:

ƩFy=0
mg=μsFnsinθ-Fncosθ eqn.1

ƩFx=ma
Ffx-Fnx=mv²/r
μsFncosθ-Fnsinθ=mv²/r eqn.2

eqn.2/eqn.1
(mv²/r)(1/mg)=(μsFncosθ-Fnsinθ)/(μsFnsinθ-Fncosθ)
V²/rg=(μscosθ-sinθ)/(μssinθ-cosθ)
V=√(rg(μscosθ-sinθ))/(μssinθ-cosθ)
V=31.3m/s
V=113km/h

So, that is the maximum speed for it in order not to slip up the bank. Now, I want to find the minimum.
I just flip some signs around in my fbd so:
ƩFy=0
Fg-Fny=0
mg=Fncosθ eqn.1

ƩFx=ma
Fnx-Ffx=mv²/r
Fnsinθ-μsFncosθ=mv²/r eqn.2

eqn.2/eqn.1
(mv²/r)(1/mg)=(Fnsinθ-μsFncosθ)/Fncosθ
V²/rg=(sinθ-μscosθ)/cosθ
V=24.4m/s
V=87.8km/s

∴Range of speed in order to make the turn safely is from 87.8km/h to 113km/h.

I appreciate if there's any corrections or pointers that I missed or anything! :)

 

[SammyS]

Homework Statement

A curve of radius 60 m is banked for a design speed of 100km/h. If the coefficient of static friction is 0.30, at what range of speeds can a car safely make the turn.

Homework Equations

Fun=ma
Ff=μsFn
Fc=mV²/r

The Attempt at a Solution

So, when it is "designed" for 100km/h that means there is no friction at 100km/h or 27.8m/s so this is my fbd for this speed. I've tried working it out and I manage to get okay looking answers but I'm not sure if I'm doing it right... :P
[ IMG]http://24.media.tumblr.com/2bb5bea0ef717a347f78a1a9eee3a18e/tumblr_mk8mi4yIz71qe908uo1_500.jpg[/PLAIN]
So, there is no acceleration in the Y-component:
ƩFy=ma
Fny-Fg=0
Fny=Fg
Fncosθ=mg 1st eqn.

For the X-component, there is acceleration which is the centripetal acceleration:
ƩFx=ma
Fnx=mV²/r
Fnsinθ=mV²/r 2nd eqn.

Now, to find the angle of the bank at that "design", so I isolated for θ:
2nd eqn./1st eqn.
Fnsinθ/Fncosθ=(mV²/r)(1/mg)
tanθ=V²/rg

That answer looks good !

θ=tan^-1(27.8²/60x9.8)
θ=52.7°

Now that I found the angle it is for 100km/h to make the turn in the bank without the need for friction, I'm trying to find the maximum speed for it in order not to slide up the bank so now I'll throw in friction and this is my fbd.

So I'll use the same concept with the Fun=ma just that I have Ff now:

ƩFy=0
mg=μsFnsinθ-Fncosθ eqn.1

That should be \displaystyle \ \ mg=F_n\cos(\theta)-\mu_sF_n\sin(\theta)\ .

ƩFx=ma
Ffx-Fnx=mv²/r
μsFncosθ-Fnsinθ=mv²/r eqn.2

That should be \displaystyle \ \ \mu_sF_n\cos(\theta)+F_n\sin(\theta)=m\frac{v^2}{r}\ .

If we're looking at maximum speed, then the direction of friction is down the incline & toward the center of the curve.

eqn.2/eqn.1
(mv²/r)(1/mg)=(μsFncosθ-Fnsinθ)/(μsFnsinθ-Fncosθ)
V²/rg=(μscosθ-sinθ)/(μssinθ-cosθ)
V=√(rg(μscosθ-sinθ))/(μssinθ-cosθ)
V=31.3m/s
V=113km/h

So, that is the maximum speed for it in order not to slip up the bank. Now, I want to find the minimum.
I just flip some signs around in my fbd so:
ƩFy=0
Fg-Fny=0
mg=Fncosθ eqn.1

ƩFx=ma
Fnx-Ffx=mv²/r
Fnsinθ-μsFncosθ=mv²/r eqn.2

eqn.2/eqn.1
(mv²/r)(1/mg)=(Fnsinθ-μsFncosθ)/Fncosθ
V²/rg=(sinθ-μscosθ)/cosθ
V=24.4m/s
V=87.8km/s

∴Range of speed in order to make the turn safely is from 87.8km/h to 113km/h.

I appreciate if there's any corrections or pointers that I missed or anything! :)

For the minimum speed, just reverse the signs on the force of friction.

 

[totallyclone]

That answer looks good !


That should be \displaystyle \ \ mg=F_n\cos(\theta)-\mu_sF_n\sin(\theta)\ .


That should be \displaystyle \ \ \mu_sF_n\cos(\theta)+F_n\sin(\theta)=m\frac{v^2}{r}\ .

If we're looking at maximum speed, then the direction of friction is down the incline & toward the center of the curve.



For the minimum speed, just reverse the signs on the force of friction.

Alright, so I've corrected my mistakes for finding maximum speed and I've worked it out and it makes so much sense. I thought the Ff is pointing away from the incline because isn't Ff what's keeping you from sliding down the bank? But anyways, I corrected and worked it out for maximum speed:

ƩFy=0
Fg-Fny+Ffy=0
Fg=Fny-Ffy
mg=Fncos-μsFnsinθ eqn.1

ƩFx=ma
Ffx+Fnx=mv²/r
μsFncosθ+Fnsinθ=mv²/r eqn.2

eqn.2/eqn.1
(mv²/r)/(1/mg)=(μsFncosθ+Fnsinθ)/(Fncos-μsFnsinθ)
v²/rg=(μscosθ+sinθ)/(cos-μssinθ)
v=√[rg(μsFncosθ+Fnsinθ)]/(Fncos-μsFnsinθ)
v=39.6m/s
v=143km/h ∴maximum speed

This makes more sense than my previous answer (113km/h) because since it's designed for 100km/h because a 13km/h allowance would be kind of... little... to be able to make that curve.

Now, for the minimum. You said to change the directions of the Ff. Would Ff now point away from the center? So:
ƩFy=0
Fg-Fny-Ffy=0
Fg=Fny+Ffy
mg=Fncosθ+μsFnsinθ eqn.1

I'm unsure how I would put it for ƩFx=ma... would it be
Fnx-Ffx=mv²/r
OR
Ffx-Fnx=mv²/r
?

 

[SammyS]

If you're taking the curve very fast, the the car will tend to go up the banking. Friction will help prevent that by pointing down the banking toward the center of the curve. It helps the normal force produce centripetal force (toward the center of the circle).

If you're taking the curve very slowly, the the car will tend to want to slide down the banking. Friction will help prevent that by pointing up the banking which happens to be away from the center of the curve. It opposes the normal force as far as producing centripetal force .

For the minimum speed: Use the same equations as for max. speed but change the signs on the frictional force ... in both equations.

 

[totallyclone]

If you're taking the curve very fast, the the car will tend to go up the banking. Friction will help prevent that by pointing down the banking toward the center of the curve. It helps the normal force produce centripetal force (toward the center of the circle).

If you're taking the curve very slowly, the the car will tend to want to slide down the banking. Friction will help prevent that by pointing up the banking which happens to be away from the center of the curve. It opposes the normal force as far as producing centripetal force .

For the minimum speed: Use the same equations as for max. speed but change the signs on the frictional force ... in both equations.

That makes sense! So for the minimum speed, the car tends to slide down the bank so that means Fnx would be greater than Ffx but Ffx still prevents it from sliding therefore it is:
ƩFx=ma
Fnx-Ffx=mv²/r
Fnsinθ-μsFncosθ=mv²/r eqn.2

eqn.2/eqn.1
(mv²/r)/(1/mg)=(Fnsinθ-μsFncosθ)/(Fncosθ+μsFnsinθ)
V=√[rg(sinθ-μscosθ)/(cosθ+μssinθ)]
V=20.7m/s
V=74.5km/h

So that tends to be my minimum value.

This makes much more sense. And plus, I tried the other way and I turned out getting an error ∴ it does not exist so yeah.