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h3. Part A
{excerpt:hidden=true}Two examples of drawing free body diagrams for objects navigating a banked curve.{excerpt}
!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:

!bankFBDnofric.png!  
{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}\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} \[ N = \frac{mg}{\cos\theta} \]\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}\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}

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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, uploaded by user The359.)

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:
!bankFBDwfric.png!
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{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~, 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}\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{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} \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}\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}\begin{large}\[\mu_{s} \ge 1.3\]\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, 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.){info}