Normal Force at the Bottom of a Track

[sona1177]

Homework Statement

If the speed of a roller coaster at the bottom of the loop is 25 m/s, what is the normal force exerted by the car by the track in terms of the car's weight mg?

Homework Equations

N - mg =mv^2/r

The Attempt at a Solution

N=mv^2/r + mg
N=m(v^2/r+ g)
N=m (31.25 + g)

My book says the answer is 4.2 mg. What did I do wrong?

 

[PhanthomJay]

Apparently the radius of cuvature is 20 meters at the bottom of the loop, which you did not state. If so, your answer is correct, but they asled for the answer in terms of mg. Convert 31.25 m/sec^2 to how many 'g's , where g = ?).

 

[sona1177]

Apparently the radius of cuvature is 2 meters at the bottom of the loop, which you did not state. If so, your answer is correct, but they asled for the answer in terms of mg. Convert 31.25 m/sec^2 to how many 'g's , where g = ?).

The Problem never says that the radius of curvature is 2 meters at the bottom. How did you find that? The problem Only says that the roller coaster includes a vertical circular loop of radius 20.0 meters, which I already took into account into the problem. However, I do apologize for not putting that in the question. 31.25 is 3.19 g's. so that makes m (3.19g + g)=4.2mg. Thank you so much! Now I understand.