Circular Motion Practice

Question

A 600kg roller coaster car is at the top of a loop (it is upside down) that has a radius of 6m. If the normal force on the roller coaster car is -12,000N, how fast is the car moving?

Answer 1

Hi Lilyana,

The position of the car at the top of the loop is a very important detail in this problem. When the car is at the top of the loop, both the normal force from the loop and the force due to gravity point in the same direction (towards the center of the loop). Let's also take the downwards direction (from top of the loop towards the center) to be positive so that we don't have to deal with negative values.

Let's see what consequence this has on the force summation we need to write using Newton's second law:

∑F = ma

What kind of acceleration (a) do we have here? Centripetal acceleration, which is = v²/R. So this becomes:

∑F = mv² / R

What is ∑F? The forces acting on the car are the normal force (F_{due to loop}) and the force of gravity (mg). They both point in the same direction and can therefore be added directly without worrying about vector components. Both point downwards in our coordinate system, which we defined as being a positive direction.

∑F = F_{due to loop} + mg = mv² / R

12,000 N + (600 kg * 9.8 m/s²) = mv² / R

Rearrange and solve for v