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The rider will be treated as a .

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.

and .

Notice that we have used a true vertical y-axis and true horizontal x-axis. The reason will become clear in a moment.

h3. Part A
!velodrome.jpg|width=600!
Velodromes are indoor facilities for bicycle racing (picture by Keith Finlay, courtesy of [Wikimedia Commons|http://commons.wikimedia.org]). 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:id=sysa} *System:* {cloak:id=sysa}The rider will be treated as a [point particle].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}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.{cloak}

{toggle-cloak:id=moda} *Model:*  {cloak:id=moda}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:*

{cloak:id=appa}

{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}
{cloak:id=diaga}
!bankFBDnofric.jpg!  
Since we are assuming no friction is needed, we have the free body diagram and coordinate system shown above.  

{cloak:diaga}
{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}
{cloak:id=matha}
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}\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 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 _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}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.{warning} 

From the y-component equation, we find:

{latex}

\begin{large}\[\ F_{res} = \sum F_{x} = N \sin\theta\]\end{large}

\begin{large}\[N \sin\theta = \frac{mv^{2}}{2} \]\end{large}

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 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.

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.

\begin{large} \[ N = \frac{mg}{\cos\theta} \]\end{large}{latex}

{note}Note both the similarity

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

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

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}

{

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!

The car plus contents will be treated as a .

The car will be subject to external influences from the earth (gravity) and from the track (normal force and friction).

and .

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.

\begin

h3. Part B

!indy500.jpg|width=75%!


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?  (Photo courtesy [Wikimedia Commons|http://commons.wikimedia.org], uploaded by user [The359|http://en.wikipedia.org/wiki/User:The359].)

{toggle-cloak:id=sysb} *System:* {cloak:id=sysb} The car plus contents will be treated as a [point particle]. {cloak}

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}The car will be subject to external influences from the earth (gravity) and from the track (normal force _and_ friction).{cloak}

{toggle-cloak:id=modb} *Model:* {cloak:id=modb}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appb} *Approach:*

{cloak:id=appb}

{toggle-cloak:id=diagb} {color:red} *Diagrammatic Representation* {color}
{cloak:id=diagb}
!bankFBDwfric.jpg!
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}

{cloak:diagb}

{toggle-cloak:id=mathb} {color:red} *Mathematical Representation* {color}

{cloak:id=mathb}

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}

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

{

Whenever static friction applies, it is important to justify using the equation F_f = μN, since it is also possible that F_f < μN.

\begin
{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{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}

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

{latex}\begin
{large}\[\mu_{s} \ge 1.3\]\end{large}{latex}

{

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.)