Question

How heavy do passengers feel as they enter the loop of a roller coaster that begins with a 90m hill followed by a loop 40m in diameter?

Answer 1

The equation for centripetal acceleration is:

\[a_r = \frac{v^2}{r}\]
velocity at the entrance to the loop will be given by\[ v=\sqrt{2gh}\]
"Heavyness" is basically "how many Gs" so \[\frac{g+a_r}{g}\]

Answer 2

Ok, so...
v=sqrt(2*9.8*90)
centripetal acceleration=v^2/20?
Weight= [9.8+(a sub r)]/9.8?

Answer 3

I think sa. 10 Gs sound high, but I also don't think this is a real-world roller coaster.

Answer 4

*I think so.

Answer 5

How would I calculate the gravity? I was just going off 9.8 m/s^2

Answer 6

Yeah, that is a good approximation. Doesn't matter as it cancels out anyway.

Answer 7

Didn't even think of that!

Answer 8

Ok, so working it out I got that the weight would be 90. Would that be correct?

Answer 9

I get \[\frac{1 G+9 Gs}{1}=10Gs\]
This has some of the derivations and whatnot:
http://www.real-world-physics-problems.com/roller-coaster-physics.html

Answer 10

How did you get 9G's?
\[a _{r}=\frac{ 2*90*G}{ 20}= \frac{ 180G}{ 20}=90G\]
Plugging in:
\[\frac{ G+90G}{ G}=91?\]

Answer 11

@mathmate

Answer 12

I must get some data straight.
A roller coaster usually goes up a hill (90m) and then drop down to a loop.
If we assume that's the case, then the passenger only falls through 40m (diameter of the loop, or the loop begins 50 m down?

Answer 13

So let us assume that "entering" the loop means it is at the lowest point where the potential energy is lowest.

Answer 14

And that it would be 90m below the hill.

Answer 15

Kinetic energy=(1/2)mv^2=mgh
so as Valpey puts it,
v=sqrt(2gh)=sqrt(180g).
acceleration at bottom is additive to g, so the acceleration (downwards) is
v^2/r +g
=180g/(40/2) + g
= (9+1)g
=10g
as @Valpey had it.

Answer 16

@mathsss you probably got one zero too many in 180g/20=9g

Answer 17

Wasn't thinking! Thank you guys!