How fast is this roller coaster?

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SUMMARY

The discussion focuses on calculating the speed and normal force acting on a 350kg roller coaster at point B of a frictionless loop-the-loop. The potential energy at point A is calculated as 85750 joules, while at point B it is 41160 joules, leading to a speed of 16 m/s at point B. The centripetal force is derived using the formula F = mv²/r, resulting in a value of 14933.3 N, although this value was initially questioned. The correct approach involves understanding the relationship between the normal force and gravitational force to find the net centripetal force.

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Bottomsouth
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Homework Statement


A 350kg roller coaster starts from rest at point A and slides down the frictionless loop-the-loop shown in the accompanying figure.

http://session.masteringphysics.com/problemAsset/1027487/5/1027487.jpg
1. How fast is this roller coaster moving at point B?
2. How hard does it press against the track at point B?

Homework Equations


1. it has potential energy at A = mgh =350*9.8*25= 85750 joules
at B it has PE = 350*9.8*12=41160 joules

the difference has been changed to kinetic energy = 1/2 mv^2

1/2 mv^2 =
1/2 (350)v^2 =85750 - 41160
=16 m/s this is good.

The Attempt at a Solution



2. F = mv^2/R = 350*16^2/6 = 14933.3

It says this isn't the answer.
 
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Bottomsouth said:

The Attempt at a Solution



2. F = mv^2/R = 350*16^2/6 = 14933.3

It says this isn't the answer.

There is your problem, the centripetal force,mv2/r is the resultant force. Normal reaction acts upwards,the weight acts downwards, the resultant of these two is the centripetal force (which you can calculate and you can also calculate the weight ,mg)
 
How do I find normal force?

n-mg = C_f?

n - 350(9.81) = C_f

what is n?
 
Bottomsouth said:
How do I find normal force?

n-mg = C_f?

n - 350(9.81) = C_f

what is n?

Cf is the centripetal force mv^2/r, which you can easily find since you have v at point B, mass,m and the radius r.
 

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