A mass m = 86 kg slides on a frictionless track that has a drop,
followed by a loop-the-loop with radius R = 19.7 m and finally a
flat straight section at the same height as the center of the loop
(19.7 m off the ground). Since the mass would not make it around
the loop if released from the height of the top of the loop (do you
know why?) it must be released above the top of the loop-the-loop
height. (Assume the mass never leaves the smooth track at any point
on its path.)
1)What height above the ground must the mass begin to make it
around the loop-the-loop?
2)If the mass has just enough speed to make it around the loop
without leaving the track, what will its speed be at the bottom of
the loop?
3)If the mass has just enough speed to make it around the loop
without leaving the track, what is its speed at the final flat
level (19.7 m off the ground)?
4)Now a spring with spring constant k = 15200 N/m is used on the
final flat surface to stop the mass. How far does the spring
compress?
5)turns out the engineers designing the loop-the-loop didn’t really
know physics – when they made the ride, the first drop was only as
high as the top of the loop-the-loop. To account for the mistake,
they decided to give the mass an initial velocity right at the
beginning.How fast do they need to push the mass at the beginning
(now at a height equal to the top of the loop-the-loop) to get the
mass around the loop-the-loop without falling off the
track?
(1)
Let, height above the ground = h
From conservation of energy,
mgh = mg(2R) + (1/2)mv^2
where v = sqrt (Rg)
v = sqrt (19.7*9.8) = 13.89 m/s
h = 2R + (1/2)v^2 / g
h = 2*19.7 + (1/2)*(13.89)^2 / 9.8
h = 49.25 m
(2)
Let, speed at the bottom of the loop = vb
From conservation of energy,
mgh = (1/2)mvb^2
vb = sqrt (2*9.8*49.25)
vb = 31.06 m/s
(3)
Let, speed at the final flat level = vf
(1/2)mvf^2 = mg(H - R)
vf = sqrt (2*9.8*(49.25 - 19.7))
vf = 24.06 m/s
(4)
(1/2)*m*vf^2 = (1/2)kx^2
86*(24.06)^2 = 15200*x^2
x = 1.81 m
(5)
Here,
Centripetel force = gravitational force
mv^2 / R = mg
v = sqrt (Rg)
v = sqrt (19.7*9.8)
v = 13.89 m/s
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