Question

You have a summer job working for a company that designs amusement park rides. Their latest rollercoaster design involves a circular loop-the-loop with a radius of 12 meters. Neglecting friction, what's the minimum energy required for ensuring that the 1000 kg rollercoaster
One potential design for the rollercoaster involves having the car roll down from the top of a straight ramp (angled at 25 degrees above the horizontal) before it goes into the loop-the-loop. What is the minimum height required for this hill if there is a small friction coefficient of 0.15 for the full length of the hill?

Answer 1

@Gebooors

Answer 2

You didn't finish the first question - I'm assuming you want to know the minimum energy so that the roller coaster completes the loop.

The solution to this question is actually pretty simple, as long as you think in terms of conservation of energy. As the roller coaster goes up the loop, its kinetic energy is slowly converted into potential energy until it reaches the top of the loop. After this, the roller coaster starts coming back down and the potential energy is converted back into kinetic energy.

You can see, then, that we want to find the minimum energy, E(min), that the kinetic energy (KE) could be so that, when it is all converted to potential energy (PE), the roller coaster will be at the top of the loop. In other words:

\[E_\min=KE=PE_{loop}\]
We can't find KE because we don't know the speed of the roller coaster, but we can find PE using the information from the question. Keep in mind that the heigh of the roller coaster at the top of the loop will be 24 m, because it has a 12 m radius.

\[E_\min=mgh\]\[E_\min=1000 \times 9.8 \times 24\]\[E=235200\]
So the minimum energy required to complete the loop will be 235200 J.

Until now we've been ignoring friction. The second question now puts it back into play. Friction is doing work, W(f) against the motion of the roller coaster as it goes down the hill, so the hill will have to be a bit higher than if we were ignoring friction in order to add a bit more potential energy to account for the energy lost due to friction. This means we already know the height of the hill will be greater than the height of the loop, 24 m.

We can express this new situation as follows:

\[E_\min=PE_{hill}-W_f\]
The PE at the top of the hill is again mgh - this time we are solving for h, the height of the hill. Remember that W=Fd. The work done by friction is parallel to the ramp the roller coaster travels down, meaning the force is parallel as well. The force of friction is μF(N), which in this case is μmgcosθ. The distance the force is acting over is the length of the ramp, which is h/sinθ (you can get this from simple trig). Sooooo...let's plug this all back into our equation and solve for h!

\[E_\min=mgh- \mu mg \cos \theta \times {h \over \sin \theta}\]\[235200=1000 \times 9.8 \times h-0.15 \times 1000 \times 9.8 \cos 25 \times {h \over \sin 25}\]\[235200=9800h-3152h\]\[235200=6648h\]\[h=35.4\]
The minimum height of the hill needs to be 35.4 m in order for the roller coaster to clear the loop. This is exactly what we'd expect if we're considering friction - notice the height of the hill is greater than the height of the loop!

I know that's a lot to take in, so please let me know if you have any questions!