Suppose a roller coaster starts out at 12 m above the ground and slides down and

Question

Suppose a roller coaster starts out at 12 m above the ground and slides down and around a loop that is 4 meters in radius. Ignoring friction what is the centripetal acceleration and downward normal force at the top of the loop.

28. The same roller coaster as above after it finished a ride collides horizontally with a spring whose spring constant is 10000 N/m. Suppose that the coaster is 1000 kg. How much does the spring compress. Does it compress more if the coaster collided with the spring vertically?

Explanation / Answer

here,

H = 12 m
R = 4 m

from conservation of energy we have :
potential Energy lost by coaster = Kietic Energy gained by coaster
mgh = 0.5mv^2
v = sqrt(2gh)
v = sqrt(2*9.8*12)
v = 15.336 m/s

Centripital acceleraion is given as :

Ac = v^2/R
Ac = 15.336^2/4
Ac = 58.798 m/s^2

The normal will be zero as it has very small value as it cannot be negatice
-N -mg = 0
n = -mg (not possible or will have very small value)

Now, From conservation of Energy we have :

Kinetic energy of coaster = potential energy stored in spring
0.5 * m * v^2 = 0.5 * k * x^2
x = sqrt(m * v^2 / k)
x = sqrt(1000 * 15.336^2 / 10000)
x = 4.849 m or 4.85 m

if coaster comes verically :
force = spring constant * compresion distance
mg = k*x
x = mg/k
x = (1000 * 9.8) / 10000
x = 0.98m

no compression will be less as compared to above case

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