Explain these coasters physics wise
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Explain these coasters physics wise

[From: ] [author: ] [Date: 11-12-29] [Hit: ]
......

the physics person can figure out exactly how fast a coaster will be going if he knows the difference in height from one hill to the next.. assuming low friction from the wheels

say the initial hill is 300 m high

and say the cart has a mass of 1000kg and the persons inside are figured into the 1000 kg

raising the cart to 300 m means it gets

Pe = mgh = 1000*10*300
Pe = 300*10000
Pe = 3*10^6 J

if the cart is just let go when it reaches the top of the hill its INITIAL SPEED IS 0
SO that is what we now use in the calculations

now if we were designing the coaster when it fell down the first hill it would reach a point where the track starts to level out

say this is 10m for the curve of the track from the point it starts to level to the point it is level

that means that the g's you would feel are equal to the change in speed down

now if the first drop were vertical
then the track flattened out at 200 m point

we know the speed the coaster will have at 200 m when falling straight down

Ke = 0.5mv^2

we know Ke gained will be the same as Pe lost

lost Pe = 1000*10*200 = 2*10^6 J

this means Ke will be 2*10^6 J

2*10^6 = 0.5*1*10^3 * v^2

we solve for v^2

v^2 = 2*Ke/m
v^2 = 2*2*10^6/10^3
v^2 = 4*10^3

v = 63.25 m/s

this means it must go from 63.25 m/s to 0 m/s down if this happens in only 10 m

from physics we know

v^2 - u^2 = 2ad
solving for a

a = (v^2 - u^2)/2d

this means that the accelleration you feel as you slow down at the bottom will be

a = (-4*10^3)/2*10
a = -2*10^2 m/s^2

a = -200 m/s^2

if you know physics I used the average value of 10 m/s^s for gravity

200 m/s^2/10 m/s^2 = 20 g's

if a roller coaster were designed to slow drop 200 m and to change direction in the last 10 meters to horizontal you would feel the same as 20g's like a FIGHTER JET falling straight down and trying to level out in 10 m would kill the fighter and would probably hurt or kill you if you tried
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