Question
An 840 Kg roller coaster car is launched from a giant spring of constant k=31 kN/m into a frictionless loop-the-loop track of radius 6.2 m, as shown below. What is the minimum amount that the spring must be compressed if the car is to stay on the track?
Answers
At the top of the loop mv^2/R=mg+N
However if we ant minimum speed put N to zero.
Thus to stay in the loop mv^2/R=mg
v at the top thus = sqrt(g/R)
To get to the top of the loop you are at a height of 2R, so the amount of energy lost getting there is 2mgR.
Thus the initial energy must be of magnitude 2mgR + mv^/2 so
kx = 2mgR + mg/2R
Divide by K and you're all set!
However if we ant minimum speed put N to zero.
Thus to stay in the loop mv^2/R=mg
v at the top thus = sqrt(g/R)
To get to the top of the loop you are at a height of 2R, so the amount of energy lost getting there is 2mgR.
Thus the initial energy must be of magnitude 2mgR + mv^/2 so
kx = 2mgR + mg/2R
Divide by K and you're all set!
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