Roller Coaster Practice Problems

Energy Analysis and Freebody Diagrams:

In all of these general examples Fg remains the same as Fg=mg and neither m nor g has changed throughout the roller coaster. The cart enters from the left and exists at the right. Et is not a constant as the track is not frictionless. Eg changes depending on the distance between the cart and the ground (h=0) as seen below. Etot is the same throughout the rollercoaster loop because the energy in the system is not being affected by objects outside the system.The dot below represents the rollercoaster cart at different locations of the loop.

On the left side:

At the right and left side of the track the forces are identical in scale yet differ in direction of Fn on cart by track. 

Fn on cart by track still pushes the cart up with the same force as the cart pushes down on them regardless of the direction traveled.

At the bottom of the loop:

Since the bottom of the loop is on the ground, Eg=0.  

 At the top of the loop:

On the right side:

As the roller coaster cart exists the loop the cart is traveling slower than it was originally going. This is because Et has increased. 

Roller coaster problems tie in with both circular motion as well as the concept of energy. Energy is useful to find velocities at different points in the roller-coaster while circular motion is useful for finding forces and how you feel. Determining how you feel is very similar to the elevator physics we previously learned. 

What are "g's"?

g's are a ratio of how much normal force is exerted on you compared to your gravitational force. Mathematically:

Normally we experience 1 g because our F_n and F_g forces balance since we are typically moving uniformly vertically (i.e. at rest). In these situations we feel normal. However when we accelerate upwards the normal force on us by the ground must become larger then our weight in order to speed up or slow down. What if we were accelerating upwards so much that the size of the normal force was twice as big as our weight? In this case we would feel heavy and be experiencing 2 g's. 

Rules on g's

• If the # of g's = 1, you feel normal.
• If the # of g's is > 1 you feel heavier than normal
• If the # of g's < 1 you feel lighter than usual
• If the g's are negative this means the normal force on the person is actually pushing us down. We are not used to these types of forces are are not as able to withstand them.
• For example: a person experiencing -1 g has the same size normal force as usual, but this force is actually pushing the person down! 

Tip: g's are one of the few measurements in physics that does not have a unit. This is because g's are a ratio of forces and the Newton's cancel out. 

How many g's are safe?

If the # of g's is above 3.7 or below -1, the situation is not safe for a human. Although there is some leeway in terms of g's (i.e a human wouldn't be completely fine at 3.7 g's but die at 3.8 g's), being too far above or below the limit will cause a human to black out. However, people (usually pilots) can be trained to withstand higher amounts of g's than normal. The safe zone of g's is between -1 to 3.7. We are better able to handle normal forces which push us upwards since that is what our body is used to. 

Example 1:

The mass of the cart traveling on the roller coaster is 40 kg. Suppose the velocity of the cart at the bottom of the hill is 45 m/s. 

	A. Draw a free body diagram for the cart when it is at the bottom of the first hill.  When the cart is at the bottom of the first hill, Fn has to be the bigger force, which gives the thrill to the rider that they feel very heavy. They feel heavy because the Fn on cart by track is a much more powerful force than the Fg on cart by earth. Remember: we feel Fn as apparent weight, not Fg.
C. Write summation equations for the forces acting on the cart (remember the cart is going in a circle, therefore it is accelerating toward the center of the circle.     ∑Fx= Ff on cart= ma        ∑Fy= Fn on cart + Fg on cart = ma= mv2/ r	D. Calculate Fg and Fn for the cart at the bottom of the hill:   Fg=mg Fg= (40 kg) (10N/kg) Fg= 400 N     ∑Fy= Fn+Fg = ma= mv2/ r   Fn+ (-400N) = (40kg) (45 m/s)2 / 40m   Fn+ (-400N) = 2025 N   Fn= 2,425 N
E. Calculate the amount of g's.    g's = Fn/Fg= 2,425N / 400N g's= 6 g's	F. Will the rider feel safe or not?   No! 6 g's is way too much!

Example 2:

A 40 kg cart is moving over the top of a hill at 16 m/s. Find the number of g's on the rider, how the rider feels, along with determining if the roller coaster is safe or not.    Step 1: find the acceleration.    a=v2/r, which is 32 m/s2 up.   Step 2: find Fg and Fn   By Fg=mg, Fg must be 400 N down. By using Σf = ma, it can be deduced that Σf is 1280 N down (since the net force must point towards the center of the circle). Because Fg is 400 N down, Fn must be 880 N down so that we have enough net force to cause the cart to move in a circle while still moving so fast. By using the formula of Fn/ lFgI, we can find that the roller coaster has -2.2 g's, the rider feels heavy and the rider is not safe.

Tips for Solving Roller Coaster Questions:

1. Make sure that all of your “Fg on cart by earth” forces are the same magnitude. This is because F_g is dependent on the mass of the object, in this case the cart, and where you are, in this case the earth. Both of these remain the same, so the magnitude of F_g should also remain the same.
2. Make sure your summation equations are equal to something. This is especially important for the ∑F_y.
3. It is helpful to draw a circle above or below the cart. If done correctly, this can help you identify what direction the net force is pointing in (since it always points towards the center of the circle). 

Roller Coaster Project