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Science > Physics |
| User: |
"" |
| Date: |
25 May 2007 01:41:05 AM |
| Object: |
Rollercoaster physics |
I know this is pretty basic stuff, but I'd like to here some input on
this.
The other day I was at an amusement park in Russia. The operator
wouldn't allow us to ride the rollercoaster unless it was completely
full with people. His reasoning was that the cars needed sufficient
weight in order to be able to gain enough speed to climb successive
hills. Does this make sense? I thought the speed of falling object is
equal, regardless of mass. But at the same time I see that there is
some mention that potential energy and kinetic energy are a function
of mass and height.
Does it make sense for the operator to fill the cars completely?
Thanks for your time.
Dave
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| User: "Greg Neill" |
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| Title: Re: Rollercoaster physics |
25 May 2007 09:07:58 AM |
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<daemon95131@gmail.com> wrote in message
news:1180075265.604157.253570@a35g2000prd.googlegroups.com...
I know this is pretty basic stuff, but I'd like to here some input on
this.
The other day I was at an amusement park in Russia. The operator
wouldn't allow us to ride the rollercoaster unless it was completely
full with people. His reasoning was that the cars needed sufficient
weight in order to be able to gain enough speed to climb successive
hills. Does this make sense? I thought the speed of falling object is
equal, regardless of mass. But at the same time I see that there is
some mention that potential energy and kinetic energy are a function
of mass and height.
Does it make sense for the operator to fill the cars completely?
The operator is giving a plausible excuse for the physics
illiterate (most of the public). The real reason is that
he wants to maximize profits. Running an empty (or nearly
empty) ride wastes power, ties up the equipment while other
paying customers may arrive, and wears out the equipment
without maximal return.
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| User: "Uncle Al" |
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| Title: Re: Rollercoaster physics |
25 May 2007 09:57:32 AM |
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wrote:
I know this is pretty basic stuff, but I'd like to here some input on
this.
The other day I was at an amusement park in Russia. The operator
wouldn't allow us to ride the rollercoaster unless it was completely
full with people. His reasoning was that the cars needed sufficient
weight in order to be able to gain enough speed to climb successive
hills. Does this make sense? I thought the speed of falling object is
equal, regardless of mass. But at the same time I see that there is
some mention that potential energy and kinetic energy are a function
of mass and height.
Does it make sense for the operator to fill the cars completely?
Thanks for your time.
Equivalence Principle. Other than a small term for air resistance,
the dynamics are independent of mass. He's maximizing his duty cycle.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
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| User: "b" |
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| Title: Re: Rollercoaster physics |
25 May 2007 04:03:46 AM |
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On May 25, 8:41 am, wrote:
I know this is pretty basic stuff, but I'd like to here some input on
this.
The other day I was at an amusement park in Russia. The operator
wouldn't allow us to ride the rollercoaster unless it was completely
full with people. His reasoning was that the cars needed sufficient
weight in order to be able to gain enough speed to climb successive
hills. Does this make sense? I thought the speed of falling object is
equal, regardless of mass. But at the same time I see that there is
some mention that potential energy and kinetic energy are a function
of mass and height.
The potential energy of the cars at the top of the first (highest)
hill is given by:
Ep = mgh1
Where m is the mass of the cars, g is the local acceleration due to
gravity and h1 is the height of the hill. When the cars are released,
the potential energy is converted to kinetic energy minus some losses
due to friction. As the cars climb the next hill, the kinetic energy
is again converted to potential energy minus friction. The potential
energy at the top of the second hill is:
Ep' = mgh2
Where h2 is the height of the second hill. The difference between the
potential energy at the top of each hill is:
deltaE = Ep - Ep' = mgh1 - mgh2 = mg(h1 - h2)
Clearly, if the cars are to climb the second hill deltaE must be
positive and greater than the work done by frictional forces.
Therefore,
deltaE > k * S(Fn)
Where k is the coeficient of kinetic friction and S(Fn) is the
integral of the normal force over the path taken by the car. We know
that the normal force is given by:
Fn = mg * cos(theta)
Where theta is the angle of the normal force with respect to the
vertical. Replacing in the equation above we find:
mg(h1 - h2) > mu * S(mg * cos(theta))
But since m and g are constant we have:
mg(h1 - h2) > kmg * S(cos(theta))
The mg term cancels out and we're left with:
(h1 - h2) > k * S(cos(theta))
This equation depends only on the relative heights of the two hills,
the coeficient of friction, and the path taken. Therefore, the mass of
the cars is irrelevant.
Does it make sense for the operator to fill the cars completely?
It makes sense from the point of view of his costs, but not for the
reasons given.
Regards,
Bart.
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| User: "b" |
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| Title: Re: Rollercoaster physics |
25 May 2007 04:06:36 AM |
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mg(h1 - h2) > mu * S(mg * cos(theta))
Sorry, mu should be k instead.
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