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h2. Part A

!bigcoaster1.png!

A good roller coaster causes dramatic swings in the rider's [apparent weight|Apparently I've Lost Weight].  Calculate the magnitude of the force exerted by their seat on a rider with a (regular) weight of 700 N at the points labeled A-F in the picture above, assuming the coaster starts from rest at the top of the first hill and that the coaster is frictionless.  (Recall that this seat force will give an idea of how "heavy" the rider feels at each point in the ride.)  For Part A, use the values:

||_h_~1~ ||_h_~2~||_h_~3~||_r_~1~||_r_~2~||_r_~3~||
|75 m|15 m|70 m|25 m|20 m|14 m|

as labeled in the picture below.

!bigcoaster2.png!

h4. Solution

*System:*  Rider as [point particle].

*Interactions:*  The [conservative|conservative force] influence from the earth (gravity) will be assigned a [potential energy] for this problem.  The seat's influence (normal force) is the only [non-conservative] force, but since the rider travels with velocity always parallel to the track, the normal force is always perpendicular to the path and does zero [work], and so we will neglect it when using the [Mechanical Energy and Non-Conservative Work] model.  The normal force will be relevant for the dynamics of the car.

*Models:*  [Uniform Circular Motion], [Point Particle Dynamics] and [Mechanical Energy and Non-Conservative Work].

*Approach:*  We begin by using Point Particle Dynamics to relate the normal force from the seat to other quantities.  This requires [free body diagrams|free body diagram]:

|!bigcoasterfbdab.png!|!bigcoasterfbdab.png!|!bigcoasterfbdc.png!|!bigcoasterfbdd.png!|!bigcoasterfbde.png!|!bigcoasterfbdf.png!|
||Point A||Point B||Point C||Point D||Point E||Point F||

{note}As usual, we have made a guess about the relative sizes of the normal force and gravity.  The accuracy of this guess is not vital to solving the problem, only the direction of the forces matter (parallel to gravity, perpendicular to gravity, etc.).  If the relative sizes do not make sense to you now, consider them again _after_ solving Part A.{note}
{info}Technically, roller coaster seats can also exert forward (from the seat back) and backward (from the restraining straps or bars) forces on the rider.  In the limit of a very short coaster (say one car) with very little friction/air resistance these forces are basically zero because gravity will accelerate the person and the coaster at the same rates.  What would the seat force on a rider in the front or back car of a very long (many cars) roller coaster look like?{info}

These free body diagrams give the following form for the relevant component [Newton's 2nd Law|Newton's Second Law] at the points of interest:

|{latex}\begin{large}\[\sum F_{y} = N - mg = ma_{y}\]\end{large}{latex}|{latex}\begin{large}\[\sum F_{y} = N - mg = ma_{y}\]\end{large}{latex}|{latex}\begin{large}\[\sum F_{x} = -N = ma_{x}\]\end{large}{latex}|{latex}\begin{large}\[\sum F_{y} = - N - mg = ma_{y}\]\end{large}{latex}|{latex}\begin{large}\[\sum F_{x} = N = ma_{x}\]\end{large}{latex}|{latex}\begin{large}\[\sum F_{y} = N - mg = ma_{y}\]\end{large}{latex}|
||Point A||Point B||Point C||Point D||Point E||Point F||

{note}We have only included the equation for the direction that includes the normal force.{note}

At each point of interest (A-F) the coaster is traveling a circular path.  The [Uniform Circular Motion] model therefore implies the presence of a radial component of acceleration given by:

|{latex}\begin{large}\[a_{y} = +\frac{v_{A}^{2}}{r_{1}}\]\end{large}{latex}|{latex}\begin{large}\[a_{y} = +\frac{v_{B}^{2}}{r_{2}}\]\end{large}{latex}|{latex}\begin{large}\[a_{x} = - \frac{v_{C}^{2}}{r_{2}}\]\end{large}{latex}|{latex}\begin{large}\[a_{y}= - \frac{v_{D}^{2}}{r_{2}}\]\end{large}{latex}|{latex}\begin{large}\[a_{x} = \frac{v_{E}^{2}}{r_{2}}\]\end{large}{latex}|{latex}\begin{large}\[a_{y}= -\frac{v_{F}^{2}}{r_{3}}\]\end{large}{latex}|
||Point A||Point B||Point C||Point D||Point E||Point F||

{note}It is very important to correctly identify the sign of the acceleration.  The acceleration must be directed toward the center of the circular path in each case.{note}

We can now solve our equations for the normal force if we can find the speed of the person at each of the points of interest.  We can accomplish this by using the fact that the [mechanical energy] of the person is constant in this problem (since the normal force does no work).  

Since the person is at rest atop the first hill, the person's total mechanical energy is (taking h=0 to be at the location of the dotted line in the figure above):

{latex}\begin{large}\[ E_{tot} = mgh_{1}\]\end{large}{latex}

To find the speed at point A, we then write:

{latex}\begin{large}\[ E_{A} = K_{A} + U_{A} = \frac{1}{2}mv_{A}^{2} + mgh_{A} = \frac{1}{2}mv_{A}^{2} = E_{tot} = mgh_{1} \]\end{large}{latex}

which implies:

{latex}\begin{large}\[ v_{A}^{2} = 2gh_{1} \]\end{large}{latex}

We can repeat this procedure exactly for each of the points if we can define the height of the person at each.  From the figures, we can see that these heights are:

|{latex}\begin{large}\[h_{A} = 0 \]\end{large}{latex}|{latex}\begin{large}\[h_{B} = h_{2}\]\end{large}{latex}|{latex}\begin{large}\[h_{C} = h_{2} + r_{2}\]\end{large}{latex}|{latex}\begin{large}\[h_{D} = h_{2}+2r_{2}\]\end{large}{latex}|{latex}\begin{large}\[h_{E} = h_{2} + r_{2}\]\end{large}{latex}|{latex}\begin{large}\[h_{F} = h_{3}\]\end{large}{latex}|
||Point A||Point B||Point C||Point D||Point E||Point F||

which means the speeds are given by:

|{latex}\begin{large}\[v_{A}^{2} = 2gh_{1} \]\end{large}{latex}|{latex}\begin{large}\[v_{B}^{2} = 2g(h_{1}-h_{2})\]\end{large}{latex}|{latex}\begin{large}\[v_{C}^2 = 2g(h_{1} - h_{2} - r_{2})\]\end{large}{latex}|{latex}\begin{large}\[v_{D}^{2} = 2g(h_{1} - h_{2}-2r_{2})\]\end{large}{latex}|{latex}\begin{large}\[v_{E}^{2} = 2g(h_{1}-h_{2} - r_{2})\]\end{large}{latex}|{latex}\begin{large}\[v_{F}^{2} = 2g(h_{1}- h_{3})\]\end{large}{latex}|
||Point A||Point B||Point C||Point D||Point E||Point F||

Substituting into the expressions for the radial acceleration and then substituting _them_ into the equations from Newton's 2nd Law gives the normal force at each point (we combine points C and E since they give the same value):

|{latex}\begin{large}\[ N  = mg + m\left(\frac{2gh_{1}}{r_{1}}\right)\]\[= 7 mg = \mbox{4900 N}\]\end{large}{latex}|{latex}\begin{large}\[N  = mg + m\left(\frac{2g(h_{1}-h_{2})}{r_{2}}\right)\]\[ = 7 mg = \mbox{4900 N}\]\end{large}{latex}|{latex}\begin{large}\[N = m\left(\frac{2g(h_{1}-h_{2}-r_{2})}{r_{2}}\right)\]\[ = 4 mg = \mbox{2800 N}\]\end{large}{latex}|{latex}\begin{large}\[N  = m\left(\frac{2g(h_{1}-h_{2}-2r_{2})}{r_{2}}\right)- mg\]\[ = mg = \mbox{700 N}\]\end{large}{latex}|{latex}\begin{large}\[N  = mg - m\left(\frac{2g(h_{1}-h_{3})}{r_{3}}\right)\]\[ = \frac{2}{7}mg = \mbox{200 N}\] \end{large}{latex}|
||Point A||Point B||Point C or Point E||Point D||Point F||

{info}The difference of 6 _mg_ between the normal force at the top of the loop-the-loop and the bottom is characteristic of a circular loop-the-loop without frictional losses.{info}

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h2. Part B -- Follow Up Questions

# If our equations result in a negative normal force, it means that the person will tend to fly out of their seat.  Even though passengers are belted in, amusement parks take care to avoid this situation since the restraints are meant to be a safety backup rather than a primary component of the ride.  Suppose that our coaster parameters were changed so that points D and F were at the same height.  Which radius, _r_~2~ or _r_~3~, should be larger in order to ensure that the normal force is always greater than zero during the ride?  
# Accelerations approaching 10_g_ can be dangerous.  Suppose that we changed the parameters of our coaster to make _h_~2~ = 0 and selected _h_~1~ to give _N_ = 0 at point D (the top of the loop).  What is the minimum value for _r_~1~ such that N < 8 _mg_?