# A rollercoaster car may be approximated by a block of mass m = 2.0 kg. The car which starts from...

## Question:

A rollercoaster car may be approximated by a block of mass m = 2.0 kg. The car which starts from rest is released at a height h above the ground and slides along a frictionless track. The car encounters a loop of radius R = 15 m. Find the initial height at which the car is released if the car's speed at the top of the loop is 15 m/s. Use the h to find the normal force (magnitude and direction) when the rollercoaster is on the top of the loop and the bottom of the loop respectively.

## Conservation of Energy

The principle of conservation of energy says that the energy in a system may only change from one form to another, but may never be created or destroyed. This means that if the total amount of energy in a system is 100 Joules at some point, then at any point later on the energy must still be 100 Joules, even if the energy has been converted into a different type of energy. A few of the most common types of energy that this principle is used with are gravitational potential energy and kinetic energy. Gravitational potential energy is equal to:

{eq}E_g = mgh {/eq}

Here m is the mass of the object, g is the gravitational acceleration (9.8 {eq}m/s^2 {/eq} near Earth's surface), and h is the height above some reference height. Kinetic energy is calculated as:

{eq}E_k = mv^2/2 {/eq}

Here m is again the mass of the object and v is the speed of the object.

To find how high our car was initially above the ground we will set up a conservation of energy equation. When our car is at the top of the hill, its only energy is gravitational potential energy due to it being at height h and not moving ({eq}E_k = 0 {/eq}). When our car is at the top of the loop it is at a height of 15m and has a velocity of 15 m/s, so it has both potential and kinetic energy. We will choose height h=0 to be the ground since it seems like the obvious choice. Equating these two cases gives:

{eq}mgh = mv^2/2+mgh_2 \:\rightarrow \: (2)(9.8)h = (2)(15)^2/2+(2)(9.8)(15) {/eq}

Solving for h gives:

{eq}h = 26.48 \: m {/eq}

## Finding the Normal Force

### Top of the Loop

To find the normal force at the top and bottom of the loop we need to look at the forces acting on the car at these points. For the top of the loop, we have the normal force due to the car touching the track, which always points away from the track (downwards) and we have the force of gravity which also points downwards. We also know that since our roller coaster car is undergoing circular motion that the net force on the car must be the centripetal force {eq}F_c = mv^2/R {/eq}, where R is the radius of the loop (15 m). Therefore we can say that the sum of our normal force ({eq}F_N {/eq}) and gravitational force (mg) must equal the centripetal force.

{eq}mv^2/R = mg + F_N {/eq}

{eq}(2)(15)^2/(15) = 2(9.8)+F_N \: \rightarrow \: F_N = 10.4 \: N {/eq}

So the normal force at the top of the loop is 10.4 Newtons.

### Bottom of the Loop

For the bottom of the loop, we again only have the normal force and gravitational force but this time the normal force acts oppositely to the gravitational force. Again the sum of these must be the centripetal force, which gives:

{eq}mv^2/R = F_N-mg {/eq}

Here mg is negative since it points in the opposite direction of the centripetal force. Before we can solve we need to find the cars velocity at the bottom of the loop. Once again we set up an energy equation between the cars starting point and the bottom of the loop.

At the bottom of the loop h=0, so the car only has kinetic energy. Thus:

{eq}mgh = mv^2/2 \: \rightarrow \: 2(9.8)(26.48) = v^2 {/eq}

This gives:

{eq}v = 22.78 \: m/s {/eq}

Now we can plug this into our force equation to solve for the normal force at the bottom.

{eq}2(22.78)^2/(15) = F_N - 2(9.8) \: \rightarrow \: F_N = 88.79 \: N {/eq}

Thus the normal force at the bottom of the loop is 89.79 Newtons. 