[Examples from Energy]
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:
h1 |
h2 |
h3 |
r1 |
r2 |
r3 |
|---|---|---|---|---|---|
75 m |
15 m |
70 m |
25 m |
20 m |
14 m |
as labeled in the picture below.
Solution
System:
Interactions:
Models:
Approach:
Diagrammatic Representation
We begin by using Point Particle Dynamics to relate the normal force from the seat to other quantities. This requires free body diagrams:
|
|
|
|
|
|
Point A |
Point B |
Point C |
Point D |
Point E |
Point F |
|---|
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.
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?
Dynamics
These free body diagrams give the following form for the relevant component Newton's 2nd Law at the points of interest:
Unknown macro: {latex} \begin Unknown macro: {large} [\sum F_ Unknown macro: {y}
= N - mg = ma_ ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [\sum F_ Unknown macro: {y}
= N - mg = ma_ ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [\sum F_ Unknown macro: {x}
= -N = ma_ ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [\sum F_ Unknown macro: {y}
= - N - mg = ma_ ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [\sum F_ Unknown macro: {x}
= N = ma_ ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [\sum F_ Unknown macro: {y}
= N - mg = ma_ ]\end |
Point A |
Point B |
Point C |
Point D |
Point E |
Point F |
|---|
We have only included the equation for the direction that includes the normal force.
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:
Unknown macro: {latex} \begin Unknown macro: {large} [a_ Unknown macro: {y}
= +\frac{v_ Unknown macro: {A}
^{2}}{r_{1}}]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [a_ Unknown macro: {y}
= +\frac{v_ Unknown macro: {B}
^{2}}{r_{2}}]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [a_ Unknown macro: {x}
= - \frac{v_ Unknown macro: {C}
^{2}}{r_{2}}]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [a_ Unknown macro: {y}
= - \frac{v_ Unknown macro: {D}
^{2}}{r_{2}}]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [a_ Unknown macro: {x}
= \frac{v_ Unknown macro: {E}
^{2}}{r_{2}}]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [a_ Unknown macro: {y}
= -\frac{v_ Unknown macro: {F}
^{2}}{r_{3}}]\end |
Point A |
Point B |
Point C |
Point D |
Point E |
Point F |
|---|
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.
Energy
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):
\begin
[ E_
= mgh_
]\end
To find the speed at point A, we then write:
\begin
[ E_
= K_
+ U_
= \frac
mv_
^
+ mgh_
= \frac
mv_
^
= E_
= mgh_
]\end
which implies:
\begin
[ v_
^
= 2gh_
]\end
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:
Unknown macro: {latex} \begin Unknown macro: {large} [h_ Unknown macro: {A}
= 0 ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [h_ Unknown macro: {B}
= h_ Unknown macro: {2}
]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [h_ Unknown macro: {C}
= h_ Unknown macro: {2}
+ r_ ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [h_ Unknown macro: {D}
= h_ Unknown macro: {2}
+2r_ ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [h_ Unknown macro: {E}
= h_ Unknown macro: {2}
+ r_ ]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [h_ Unknown macro: {F}
= h_ Unknown macro: {3}
]\end |
Point A |
Point B |
Point C |
Point D |
Point E |
Point F |
|---|
which means the speeds are given by:
Unknown macro: {latex} \begin Unknown macro: {large} [v_ Unknown macro: {A}
^ Unknown macro: {2}
= 2gh_ Unknown macro: {1}
]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [v_ Unknown macro: {B}
^ Unknown macro: {2}
= 2g(h_ Unknown macro: {1}
-h_ )]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [v_ Unknown macro: {C}
^2 = 2g(h_ Unknown macro: {1}
- h_ Unknown macro: {2}
)]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [v_ Unknown macro: {D}
^ Unknown macro: {2}
= 2g(h_ Unknown macro: {1}
- h_ -2r_ Unknown macro: {2}
)]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [v_ Unknown macro: {E}
^ Unknown macro: {2}
= 2g(h_ Unknown macro: {1}
-h_ - r_ Unknown macro: {2}
)]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [v_ Unknown macro: {F}
^ Unknown macro: {2}
= 2g(h_ Unknown macro: {1}
- h_ Unknown macro: {3}
)]\end |
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):
Unknown macro: {latex} \begin Unknown macro: {large} [ N = mg + m\left(\frac{2gh_{1}}{r_{1}}\right)][= 7 mg = \mbox Unknown macro: {4900 N}
]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [N = mg + m\left(\frac{2g(h_ Unknown macro: {1}
-h_ Unknown macro: {2}
)}{r_{2}}\right)][ = 7 mg = \mbox Unknown macro: {4900 N}
]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [N = m\left(\frac{2g(h_ Unknown macro: {1}
-h_ Unknown macro: {2}
-r_ )}{r_{2}}\right)][ = 4 mg = \mbox Unknown macro: {2800 N}
]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [N = m\left(\frac{2g(h_ Unknown macro: {1}
-h_ Unknown macro: {2}
-2r_ )}{r_{2}}\right)- mg][ = mg = \mbox Unknown macro: {700 N}
]\end |
Unknown macro: {latex} \begin Unknown macro: {large} [N = mg - m\left(\frac{2g(h_ Unknown macro: {1}
-h_ Unknown macro: {3}
)}{r_{3}}\right)][ = \frac Unknown macro: {2}
Unknown macro: {7}
mg = \mbox Unknown macro: {200 N}
] \end |
Point A |
Point B |
Point C or Point E |
Point D |
Point F |
|---|
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.
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, r2 or r3, 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 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?