Question

What is the minimum speed required for a roller coaster to stay on the track at the top of a loop with a radius of 5 meters?

Answer 1

To determine the minimum speed required for a roller coaster to stay on the track at the top of a vertical loop, we need to consider the forces acting on the coaster at that point. At the top of the loop, the gravitational force provides the necessary centripetal force to keep the coaster on the track.

Let's denote:
- \( m \) as the mass of the roller coaster (this will later cancel out),
- \( g \) as the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( r \) as the radius of the loop (which is \( 5 \, \text{m} \)),
- \( v \) as the speed of the roller coaster at the top of the loop.

At the top of the loop, the following relationship must hold:

1. The centripetal force required to keep the roller coaster in circular motion is given by:
\[
F_c = \frac{m v^2}{r}
\]

2. The gravitational force acting on the roller coaster is:
\[
F_g = mg
\]

For the roller coaster to just stay on the track, the centripetal force required must be equal to the gravitational force weight acting downwards. Thus, we have:
\[
\frac{m v^2}{r} = mg
\]

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)), resulting in:
\[
\frac{v^2}{r} = g
\]

Now, we can solve for the speed \( v \):
\[
v^2 = g \cdot r
\]
\[
v = \sqrt{g \cdot r}
\]

Substituting \( g = 9.81 \, \text{m/s}^2 \) and \( r = 5 \, \text{m} \):
\[
v = \sqrt{9.81 \times 5}
\]
\[
v = \sqrt{49.05}
\]
\[
v \approx 7.0 \, \text{m/s}
\]

Therefore, the minimum speed required for the roller coaster to stay on the track at the top of the loop is approximately **7.0 m/s**.