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Excerpt
hiddentrue

Two examples of drawing free body diagrams for objects navigating a banked curve.

Composition Setup
 Deck of Cards
idbigdeck
 Wiki Markup
h3. Part A

!velodrome.jpg!
Velodromes are indoor facilities for bicycle racing (picture by Keith Finlay, courtesy of Wikimedia Commons). Olympic velodromes are usually ovals 250 m in circumference with turns of radius 25 m. The peak banking in the turns is about 42°. Assuming a racer goes through the turn in such a velodrome at the optimal speed so that no friction is required to complete the turn, how fast is the racer moving?

System:  {color:maroon}The rider will be treated as a [point particle].  The rider is subject to external influence from the earth (gravity) and from the track (normal force).  We assume friction is not present, since we are told to determine the speed at which friction is unnecessary to complete the turn.{color}

Model:  [Point Particle Dynamics].

Approach:  
{color:maroon}Since we are assuming no friction is needed, we have the free body diagram and coordinate system shown above.  The corresponding equations of Newton's Second Law are:
{note}Notice that we have used a true vertical y-axis and true horizontal x-axis.  The reason will become clear in a moment.{note}

{latex}
 Card
labelPart A
 Image Added 
 Photo by Keith Finlay, courtesy of Wikimedia Commons 
Part A
Velodromes are indoor facilities for bicycle racing. Olympic velodromes are usually ovals 250 m in circumference with turns of radius 25 m. The peak banking in the turns is about 42°. Assuming a racer goes through the turn in such a velodrome at the optimal speed so that no friction is required to complete the turn, how fast is the racer moving?
 Toggle Cloak
idsysa
 System: 
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idsysa
The rider will be treated as a .
 Toggle Cloak
idinta
 Interactions: 
 Cloak
idinta
The rider is subject to external influence from the earth (gravity) and from the track (normal force). We assume friction is not present, since we are told to determine the speed at which friction is unnecessary to complete the turn.
 Toggle Cloak
idmoda
 Model: 
 Cloak
idmoda
  and . 
 Toggle Cloak
idappa
 Approach: 
 Cloak
idappa
 Toggle Cloak
iddiaga
 Diagrammatic Representation
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iddiaga
Image Added
 Since we are assuming no friction is needed, we have the free body diagram and coordinate system shown above.
 Cloak
diaga
diaga
 Toggle Cloak
idmatha
 Mathematical Representation
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idmatha
The corresponding equations of Newton's Second Law are: 
 Note
Notice that we have used a true vertical y-axis and true horizontal x-axis. The reason will become clear in a moment. 
 Latex
\begin{large}\[\sum F_{x} = N \sin\theta
 = \frac{mv^{2}}{r} 
\]\[\sum F_{y} = N \cos\theta - mg = 0 \] \end{large}
{latex}

Where _r_ is the radius of the turn, _v_ is the speed of the racer and _m_ is the racer's mass (including bike and gear).
{note}The situation here is very different than that of an object sliding down an inclined plane. In the case of an object moving _along_ the plane, the acceleration will have both _x_ and _y_ components.  For an object moving along a banked curve, the object will not be moving up or down and so _a_~y~ must be zero.  The x-component of the acceleration will of course not be zero because the object is following the curve.  This difference between motion along a banked curve and motion down an inclined plane is the reason we use tilted coordinates for the incline and traditional coordinates for the curve.{note}
{warning}Because _a_~y~ = 0 here, it is _not_ appropriate to assume _N_ = _mg_ cosθ. That equation follows from assuming the object accelerates along the incline.{warning} 

From the y-component equation, we find:

{latex}
The vertical forces cancel out, but the resultant force in the x-direction is not zero
 Latex
\begin{large}\[\ F_{res} = \sum F_{x} = N \sin\theta\]\end{large}
Image Added
This force is directed horizontally inwards toward the center of the curve, causing the bicycle to move in a horizontal circle. We therefore set this resultant Force equal to the force needed to keep the bicycle of mass m moving in a circle of radius r
 Latex
\begin{large}\[N \sin\theta = \frac{mv^{2}}{2} \]\end{large}
 Note
The situation here is very different than that of an object sliding down an inclined plane. In the case of an object moving along the plane, the acceleration will have both x and y components if our untilted coordinates are used. For an object moving along a banked curve, the object will not be moving up or down and so ay must be zero. The x-component of the acceleration will of course not be zero because the object is following the curve. This difference between motion along a banked curve and motion down an inclined plane is the reason we use tilted coordinates for the incline and traditional coordinates for the curve. 
 Warning
It is not appropriate to assume N = mg cosθ. Note that we have not tilted our coordinates to align the x-axis with the ramp. Thus, the y-direction is not perpendicular to the ramp. 
From the y-component equation, we find:
 Latex
\begin{large} \[ N = \frac{mg}{\cos\theta} \]\end{large}
{latex}

{note}Note both the similarity to the standard inclined plane formula and the important difference.{note}

Substituting into the x-component equation then gives:{color}

{latex}
 Note
Note both the similarity to the standard inclined plane formula and the important difference. 
Substituting into the x-component equation then gives:
 Latex
\begin{large} \[ v = \sqrt{gr\tan\theta} = 15 \: {\rm m/s} = 33 \:{\rm mph} \]\end{large
}{latex
}


{
 Tip
}
Is 
 
 this 
 
 speed 
 
 reasonable 
 
 for 
 
 a 
 
 bike 
 
 race? 
{tip}

{info}Note that our result is independent of the mass of the rider.  This is important, since otherwise it would be impractical to construct banked curves.  Different people would require different bankings!{info}

h3. Part B

The Indianapolis Motor Speedway (site of the Indianapolis 500 race) has four turns that are each 0.25 miles long and banked at 9.2°.  Supposing that each turn is 1/4 of a perfect circle, what is the minimum coefficient of friction necessary to keep an IndyCar traveling through the turn at 150 mph from skidding out of the turn?

System:  The car plus contents will be treated as a [point particle].  The car will be subject to external influences from the earth (gravity) and from the track (normal force _and_ friction).  

Model: [Point Particle Dynamics]

Approach:

The free body diagram from Part A is modified to include friction, as shown above.

{note}In this problem, it is not clear _a priori_ which way friction should point.  Under certain conditons friction will point up the incline while in others it will point down (can you describe the conditions for each case?).  This ambiguity should not prevent you from solving.  Simply guess a direction.  If you guess wrong, your answer will come out negative which will indicate the correct direction.{note}

The resulting form of Newton's Second Law is:

{latex}\begin
 Info
Note that our result is independent of the mass of the rider. This is important, since otherwise it would be impractical to construct banked curves. Different people would require different bankings! 
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matha
matha
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 Card
labelPart B
 Image Added 
 Photo courtesy Wikimedia Commons, uploaded by user The359. 
Part B
The Indianapolis Motor Speedway (site of the Indianapolis 500 race) has four turns that are each 0.25 miles long and banked at 9.2°. Supposing that each turn is 1/4 of a perfect circle, what is the minimum coefficient of friction necessary to keep an IndyCar traveling through the turn at 150 mph from skidding out of the turn?
 Toggle Cloak
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 System: 
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The car plus contents will be treated as a .
 Toggle Cloak
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 Interactions: 
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The car will be subject to external influences from the earth (gravity) and from the track (normal force and friction).
 Toggle Cloak
idmodb
 Model: 
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  and .
 Toggle Cloak
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 Approach:
 Cloak
idappb
 Toggle Cloak
iddiagb
 Diagrammatic Representation
 Cloak
iddiagb
Image Added
 The free body diagram from Part A is modified to include friction, as shown above. 
 Note
In this problem, it is not clear a priori which way friction should point. Under certain conditons friction will point up the incline while in others it will point down (can you describe the conditions for each case?). This ambiguity should not prevent you from solving. Simply guess a direction. If you guess wrong, your answer will come out negative which will indicate the correct direction. 
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diagb
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idmathb
 Mathematical Representation
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The resulting form of Newton's Second Law is:
 Latex

\begin
{large} \[ \sum F_{x} = F_{f} \cos\theta + N \sin\theta = \frac{mv^{2}}{r} \]\[ \sum F_{y} = N \cos\theta - F_{f} \sin\theta - mg = 0\] \end{large}
{latex}




We 
 
 have 
 
 three 
 
 unknowns 
 
 (
_
N
_
, 
 _
F
_~f~
f, 
 
 and 
 _
m
_
) 
 
 but 
 
 only 
 
 two 
 
 equations. 
  
 To 
 
 solve, 
 
 we 
 
 must 
 
 develop 
 
 another 
 
 constraint. 
  
 To 
 
 do 
 
 so, 
 
 we 
 
 must 
 
 notice 
 
 a 
 
 key 
 
 phrase 
 
 in 
 
 the 
 
 problem 
 
 statement. 
  
 We 
 
 are 
 
 asked 
 
 to 
 
 find 
 
 the 
 _
minimum
_ 
 coefficient 
 
 of 
 
 friction. 
  
 The 
 
 minimum 
 
 coefficient 
 
 will 
 
 be 
 
 the 
 
 value 
 
 such 
 
 that 
 
 the 
 
 static 
 
 friction 
 
 force 
 
 is 
 
 maximized, 
 
 satisfying:


{latex}
 Latex

\begin
{large} \[F_{f} = \mu_{s} N \] \end{large}
 {latex}

{note}Remember that the point of contact of tires with the road surface is static (unless the car is in a skid) so static friction applies here.  That is the reason that there is a _minimum_ coefficient.{note}
{warning}Whenever static friction applies, it is important to justify using the equation _F_~f~ = &mu;_N_, since it is also possible that _F_~f~ < &mu;_N_.{warning}

With this substitution, we have:

{latex}\begin


 Note
Remember that the point of contact of tires with the road surface is static (unless the car is in a skid) so static friction applies here. That is the reason that there is a minimum coefficient. 
 Warning
Whenever static friction applies, it is important to justify using the equation Ff = μN, since it is also possible that Ff < μN. 
With this substitution, we have:
 Latex

\begin
{large} \[ \sum F_{x} = \mu_{s}N \cos\theta + N \sin\theta = \frac{mv^{2}}{r} \]\[ \sum F_{y} = N \cos\theta - \mu_{s} N \sin\theta - mg = 0\] \end{large
}{latex
}



Proceeding 
 
 as 
 
 in 
 
 Part 
 
 A:


{latex} 
 Latex

\begin
{large} \[ N = \frac{mg}{\cos\theta - \mu_{s} \sin\theta} \]\end{large
}{latex
}



which 
 
 is 
 
 then 
 
 substituted 
 
 into 
 
 the 
 
 y-component 
 
 equation 
 
 to 
 
 give:


{latex}
 Latex

\begin
{large}\[\mu_{s} \ge \frac{v^{2}\cos\theta - gr \sin\theta}{v^{2} \sin\theta + gr \cos\theta}\]\end{large
}{latex
}



The 
 
 substitutions 
 
 here 
 
 require 
 
 some 
 
 thought. 
  
 The 
 
 turns 
 
 are 
 
 quarter-circle 
 
 segments 
 
 with 
 
 a 
 
 length 
 
 of 
 
 a 
 
 quarter 
 
 mile 
 
 each. 
  
 Thus, 
 
 they 
 
 have 
 
 the 
 
 same 
 
 radius 
 
 of 
 
 curvature 
 
 as 
 
 a 
 
 full 
 
 circle 
 
 with 
 
 circumference 
 
 1 
 
 mile. 
  
 The 
 
 corresponding 
 
 radius 
 
 is 
 
 256 
 
 meters. 
  
 Then, 
 
 converting 
 
 the 
 
 speed 
 
 to 
 
 m/s 
 
 and 
 
 using 
 
 the 
 
 formula 
 
 derived 
 
 above 
 
 gives:


{latex}
 Latex

\begin
{large}\[\mu_{s} \ge 
0
1.
89
3\]\end{large
}{latex
}


{
 Tip
}
Given 
 
 our 
 
 answer 
 
 to 
 
 Part 
 
 A, 
 
 we 
 
 expect 
 that &mu;~s~ can be as small as _zero_ if _v_^2^ = _gr_ tan&theta;.  Is that fact reflected in our formula?{tip}
{info}Our answer is slightly in error.  The normal force exerted on the car is actually increased beyond our assumed value by the presence of airfoils at the front and back of the car that generate _negative_ lift (a downward force).  This force is specifically intended to increase the maximum friction force available from the tires.  As a result, the friction coefficient can likely be lower than our answer and still keep the car on the track.{info}


that μs can be as small as zero if v2 = gr tanθ. Is that fact reflected in our formula? 
 Info
Our answer is slightly in error. The normal force exerted on the car is actually increased beyond our assumed value by the presence of airfoils at the front and back of the car that generate negative lift (a downward force, often simply called "downforce"). This force is specifically intended to increase the maximum friction force available from the tires. As a result, the friction coefficient can likely be lower than our answer and still keep the car on the track. How large (as a fraction of the car's weight) would the downforce have to be keep the minimum acceptable coefficient of friction below 1.0? Assume the downforce presses straight into the surface of the track. (IndyCars can generate more than three times their weight in downforce when traveling near top speed.) 
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