Show all your work. Write out your work in terms of variables given and then sol
ID: 1996783 • Letter: S
Question
Show all your work. Write out your work in terms of variables given and then solve with numbers at the end.
You have been tasked with designing a roller coaster ride. Since this is a preliminary design, you may treat the roller coaster car as a point-mass. Assume the roller coaster car rolls on the track with no friction. The track begins with a horizontal section which is h = 20 meters above the ground. The ride begins by the car being pushed off by a horizontal spring which is compressed x = 2 meters. The car leaves the spring and then heads down a ramp to another horizontal section of the track which is at ground level. Next it has to go through a vertical loop. The diameter of the loop is h = 20 meters. After leaving the loop, the car travels along horizontally at ground level and then goes up a ramp to another horizontal section of track which is 10 meters above the ground. Finally, the car is brought to rest by a sticky section of track. This sticky section of track may be modeled as a coefficient of kinetic friction that increases linearly from zero to a maximum value of mu_max = 0.5 in a distance of L = 50 meters. Here's where your design comes in. Let the mass of the car and the riders be m = 500 kg.
a) Calculate the spring stiffness necessary to launch the car so that it barely makes it around the vertical loop while remaining in contact with the track.
b) Calculate the spring stiffness necessary so that the car has twice the speed at the top of the vertical loop that you found was necessary in part (a).
c) Calculate the length of sticky track needed to bring the car to rest at the end of the ride for each of the spring stiffnesses found above.
Explanation / Answer
Consider the roller coaster as point mass or a ball. In order to arrive the speeds required to make the roller coaster complete one cycle is determined by the 20m diameter vertical loop.
Minimum velocity at the top of the vertical loop to make roller coaster to remain in contact is dictated the centrifugal force must be equal to the ‘g’ the acceleration due to gravity.
Let ve the velocity at the top of loop of 20m Diameter (R=10m) the centrifugal force =mv2/R=mg
For ease of calculations we take g=10m
Therefore ve= Rg = (10 x 10) = 10m/sec.
Let Vc be the velocity at the entrance to the loop. Vc has to be higher than Ve as the roller coaster climbs up it must lose part of its Kinetic Energy to potential energy.
Thus we have mVc2/2 = mVe2/2 +mgh (here h=20m, ve= 10m/sec)
Vc2 = 100+2x10x20=500 therefore Vc=22.32 m/s
Now let us consider velocity Vb required in the roller coaster before it sgoes down the 20m ramp to have
Vc = 22.36 m/s.
Again summing up sum of energies at point b with that e.
At point b, PE = mgh (h=20m), KE = 0.5Vb2
At point e, PE = 0 (ground level), KE 0.5Vc2
Thus we have m.gh+mVb2/2=Vc2/2
Vb = (Vc2-gh) = (500-10x20)= 17.32 m/s
To have Vb=17.32 m/s, the PE imparted by the spring must be kx2/2=mVb2/2
We are given x=2m, m=500kg, therefore k = (500 x17.322/22)= 37500 N/m
Part b:
IF we desire the max velocity at Vc=2 x Vc of part a= 22.36*2= 44.72 m/s
Corresponding Vb= use the same equation as above, Vb= (Vc2-gh) = (44.722-10x20)= (2000-200)= 42.42 m/s
Corresponding Spring stiffness required = (500 x42.422/22)= (500 x1800/22)= 225000 N/m
Part C:
By symmetry the velocity of the roller coaster at the exit of the vertical curve is same as the entry point velocity. First calculate the velocity Vf after roller coaster climbs ramp of 10m as it loses part of KE to PE.
Vf = (Vc2-gh) = (500-10 x10)=20,/sec
This KE with Vf=20m/s (KE = 500 x 202/2=100000 joules) loses thru friction varying from zero at the beginning and at the end after travelling a distance d.
All KE available in the car must be dissipated friction. In the first 50m, avg =025. Thus the enrgy dissipate is avg mgd = 0.25 x 500 x 10 x 50 = 62500 joules. he remaining (100000-62500 = 37500) has to be dissipated at =0.5. and distance it travels = 37500/(0.5 x500x10) = 15m. Thus in the first case the car comes to stop in about 50+15 = 65m.
IF Vc was 44.72m/sec, then Vf=(2000-10 x10)=43.588 m/s
Thus the car has KE= 500x1900/2= 475000 joules.
In first 50m it dissipates 62500 joules. The ramining (475000-62500) =412500 with =0.5. the distance it travels = 412500/(0.5 x500x10) = 165m
Thus in the second case the car comes to stop in about 50+165 = 215m.