# Normal Force at the Bottom of a Track

• sona1177
In summary, the problem involves finding the normal force exerted by the track on a roller coaster car at the bottom of a loop. The given information includes a speed of 25 m/s and a radius of curvature of 20 meters. The solution requires using the equation N - mg = mv^2/r and converting the given speed to g's. The correct answer is 4.2 mg, which can be obtained by substituting 31.25 m/sec^2 for v^2/r in the equation.
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?

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?

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 = ?).

Last edited:
PhanthomJay said:
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.

## What is the normal force at the bottom of a track?

The normal force at the bottom of a track is the force exerted by a surface on an object in contact with it, perpendicular to the surface. In the context of a track, it is the force that pushes the object in an upward direction to counteract the force of gravity pulling it down.

## How is the normal force at the bottom of a track calculated?

The normal force at the bottom of a track can be calculated using the formula: N = mg + ma, where N is the normal force, m is the mass of the object, g is the acceleration due to gravity, and a is the acceleration of the object on the track. This formula takes into account both the force of gravity and any additional acceleration that the object experiences on the track.

## What factors affect the normal force at the bottom of a track?

The normal force at the bottom of a track is affected by the mass of the object, the acceleration of the object, and the angle of the track. As the mass or acceleration of the object increases, the normal force will also increase. Additionally, a steeper angle of the track will result in a larger normal force.

## Why is the normal force at the bottom of a track important?

The normal force at the bottom of a track is important because it is responsible for keeping the object on the track and preventing it from falling or sliding off. It also plays a crucial role in determining the speed and acceleration of the object on the track.

## Can the normal force at the bottom of a track ever be negative?

No, the normal force at the bottom of a track can never be negative. It is always directed perpendicular to the surface and therefore cannot have a negative value. If the normal force were to become negative, it would mean that the object is being pushed into the surface, rather than being supported by it.

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