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A good roller coaster causes dramatic swings in the rider's apparent 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:

as labeled in the picture below.

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Solution

We begin by using Point Particle Dynamics to relate the normal force from the seat to other quantities. This requires free body diagrams:

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

...

\begin{large}\[\sum F_{y} = N - mg = ma_{y}\]\end{large}

...

\begin{large}\[\sum F_{y} = N - mg = ma_{y}\]\end{large}

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\begin{large}\[\sum F_{x} = -N = ma_{x}\]\end{large}

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\begin{large}\[\sum F_{y} = - N - mg = ma_{y}\]\end{large}

...

\begin{large}\[\sum F_{x} = N = ma_{x}\]\end{large}

...

\begin{large}\[\sum F_{y} = N - mg = ma_{y}\]\end{large}

...

At each point of interest (A-F)

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the

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coaster

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is

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traveling

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a

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circular

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path.

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The

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Uniform

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Circular

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Motion

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model

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therefore

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implies

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the

...

presence

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of

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a

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radial

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component

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of

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acceleration

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given

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by:

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\begin{large}\[a_{y} = +\frac{v_{A}^{2}}{r_{1}}\]\end{large}

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\begin{large}\[a_{y} = +\frac{v_{B}^{2}}{r_{2}}\]\end{large}

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\begin{large}\[a_{x} = - \frac{v_{C}^{2}}{r_{2}}\]\end{large}

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\begin{large}\[a_{y}= - \frac{v_{D}^{2}}{r_{2}}\]\end{large}

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\begin{large}\[a_{x} = \frac{v_{E}^{2}}{r_{2}}\]\end{large}

...

\begin{large}\[a_{y}= -\frac{v_{F}^{2}}{r_{3}}\]\end{large}

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}|
||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

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the

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speed

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at

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point

...

A,

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we

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then

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write:

}\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:

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

We

...

can

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repeat

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this

...

procedure

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exactly

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for

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each

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of

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the

...

points

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if

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we

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can

...

define

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the

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height

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of

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the

...

person

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at

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each.

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From

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the

...

figures,

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we

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can

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see

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that

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these

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heights

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are:

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\begin{large}\[h_{A} = 0 \]\end{large}

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\begin{large}\[h_{B} = h_{2}\]\end{large}

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\begin{large}\[h_{C} = h_{2} + r_{2}\]\end{large}

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\begin{large}\[h_{D} = h_{2}+2r_{2}\]\end{large}

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\begin{large}\[h_{E} = h_{2} + r_{2}\]\end{large}

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\begin{large}\[h_{F} = h_{3}\]\end{large}

...

which means the speeds are given by:

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

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\begin{large}\[v_{B}^{2} = 2g(h_{1}-h_{2})\]\end{large}

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\begin{large}\[v_{C}^2 = 2g(h_{1} - h_{2} - r_{2})\]\end{large}

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\begin{large}\[v_{D}^{2} = 2g(h_{1} - h_{2}-2r_{2})\]\end{large}

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\begin{large}\[v_{E}^{2} = 2g(h_{1}-h_{2} - r_{2})\]\end{large}

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\begin{large}\[v_{F}^{2} = 2g(h_{1}- h_{3})\]\end{large}

...

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):

\begin{large}\[ N  = mg + m\left(\frac{2gh_{1}}{r_{1}}\right)\]\[= 7 mg = \mbox{4900 N}\]\end{large}

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\begin{large}\[N  = mg + m\left(\frac{2g(h_{1}-h_{2})}{r_{2}}\right)\]\[ = 7 mg = \mbox{4900 N}\]\end{large}

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\begin{large}\[N = m\left(\frac{2g(h_{1}-h_{2}-r_{2})}{r_{2}}\right)\]\[ = 4 mg = \mbox{2800 N}\]\end{large}

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\begin{large}\[N  = m\left(\frac{2g(h_{1}-h_{2}-2r_{2})}{r_{2}}\right)- mg\]\[ = mg = \mbox{700 N}\]\end{large}

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\begin{large}\[N  = mg - m\left(\frac{2g(h_{1}-h_{3})}{r_{3}}\right)\]\[ = \frac{2}{7}mg = \mbox{200 N}\] \end{large}

Follow Up Questions

1. 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, r₂ or r₃, should be larger in order to ensure that the normal force is always greater than zero during the ride?
2. Accelerations approaching 10_g_ can be dangerous. Suppose that we changed the parameters of our coaster to make h₂ = 0 and selected h₁ to give N = 0 at point D (the top of the loop). What is the minimum value for r₁ such that N < 8 mg?