Why is 2.5 times the radius the minimum height needed to do a loop?

In summary: Semi circle is a figure 8.I found it slightly easier, to grok the relationships between energies and velocities at various points, by removing a superfluous 'r' from the height ; "semicircle" meaning "arch" - an upside-down cross-section of a half-pipe, if you insist.In summary, the height can be determined by conservation of energy (ignoring all friction). The mechanical energy when the car is at rest, equals the mechanical energy when the car is in the middle of the loop (at the top of the loop):E_{0} = E_{loop}mgh_0 = \frac{1}{2}m
  • #1
IsakVern
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0
TL;DR Summary
In the classic "loop the loop" problem where a model car rolls from rest down a ramp from a height h0 to perform a "loop the loop" stunt around a circular track of radius R in the vertical plane, why is the minimal starting height required in order to perform a loop without leaving the track exactly 5/2 of the radius R?
The height can be determined by conservation of energy (ignoring all friction). The mechanical energy when the car is at rest, equals the mechanical energy when the car is in the middle of the loop (at the top of the loop):
\begin{equation}
E_{0} = E_{loop}
\\
mgh_0 = \frac{1}{2}mv^2+mgh_{loop}
\\
gh_0 = \frac{1}{2}v^2+2gR
\end{equation}

In the middle of the loop (at the top of the circular track), the only force acting on the car is gravity, which equals the centripetal force:
\begin{equation}
F = mg = \frac{mv^2}{R}
\\
\rightarrow v^2=gR
\end{equation}

Inserting this into the first equation, we get:

\begin{equation}
gh_0 = \frac{1}{2}gR+2gR
\\
h_0= \frac{5}{2}R
\end{equation}

Note that the height doesn't depend on the magnitude of the gravitational force at all, so no matter what planet you perform this experiment on, the minimal height needed in order for the car to loop the loop is always 2.5 times the radius of the loop. Is this just a algebraic coincidence/necessity, or is there some neat explanation for why this is?
 
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  • #2
Do you know any dimensional analysis? As with your other thread about escape velocity and circular orbital velocity, it's inevitable once you've eliminated ##v## that ##g## has to cancel out because it's the only thing with a time dimension. Either it cancels out or the units can't match.
 
  • #3
Ibix said:
Do you know any dimensional analysis? As with your other thread about escape velocity and circular orbital velocity, it's inevitable once you've eliminated ##v## that ##g## has to cancel out because it's the only thing with a time dimension. Either it cancels out or the units can't match.

I realize that, but I'm wondering whether there's a reason why there is a factor of 5/2, as opposed to any other number. Is there anything special about 5/2?
 
  • #4
What would happen in the case that the starting height equals the diameter of the loop (2 x the radius R)?
 
  • #5
IsakVern said:
I realize that, but I'm wondering whether there's a reason why there is a factor of 5/2, as opposed to any other number.
Well, it's that number because that's the way the equations work out.
IsakVern said:
Is there anything special about 5/2?
I'm not sure what you mean by "special". Is there an example of an equation you can quote where there is a number you would describe as "special"?
 
  • #6
I believe that the answer to your question is that you need some amount of extra energy to be able to sustain certain amount of forward speed at the top of the loop in order to generate the centrifugal force that counteracts the weight.
The combination of the required initial height to reach and stop at the top (2R) and the minimum forward velocity to overshoot it (V proportional to sqrt R from v^2=gR) gives you the 2.5 factor.
 
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  • #7
Lnewqban said:
The combination of the required initial height to reach and stop at the top (2R) and the minimum forward velocity to overshoot it (V proportional to sqrt R from v^2=gR) gives you the 2.5 factor.

If one starts at R, then you could achieve an "orbit" , where at apogee, having no forward horizontal momentum, gravity pulls you straight down to the ground. Of course, in reality, the car would just reverse direction and travel back down the curved "ramp".
 
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  • #8
There is a similar 'coincidence' that applies to fairground rides. If you are sitting in a swing with a rigid support and stationary at the top of the circle (strapped in of course). The swing starts to move and you fall under just gravity and you end up going past the lowest point pulling 3g, whatever height the swing.
 
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  • #9
IsakVern said:
or is there some neat explanation for why this is?
The release height must have enough mgh to get the car to the top of the loop and also to provide enough KE for the car to be going fast enough to balance the weight force at the top. The centripetal force involves v2 and so does the KE. There is no other factor involved so you would expect the release height to have the same constant of proportionality with the radius. The 'neatness' is just to say that there is no reason for the constant of proportionality to vary.
But why not let the maths do the talking? It's the appropriate language for such things.
 
  • #10
IsakVern said:
Is there anything special about 5/2?

Or, ##\frac{3}{2}r## on a semi-circle.

Or, ##\frac{9}{2}r## on a figure '8'.
 
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  • #11
hmmm27 said:
Or, ##\frac{3}{2}r## on a semi-circle.

Or, ##\frac{9}{2}r## on a figure '8'.
Can you tell me the requirements for those two moves? What is a "semi circle" in this context? Not a half pipe?
 
  • #12
sophiecentaur said:
Can you tell me the requirements for those two moves? What is a "semi circle" in this context? Not a half pipe?

I found it slightly easier, to grok the relationships between energies and velocities at various points, by removing a superfluous 'r' from the height ; "semicircle" meaning "arch" - an upside-down cross-section of a half-pipe, if you insist.

Not sure what you mean by "requirements" - putting it all in context of a fairground rollercoaster ?... okay, the semicircle would have 45deg trampolines at each end to bounce the car between horizontal and vertical (not sure it would be a popular ride) The figure 8 - in order to keep the vehicle from flying off at various points - would need either a twist in the track (to maintain its consistent "down" to the riders), or hand off points in the middle between the circles (which would result in the car pulling negative G's during parts of its journey).
 
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  • #13
hmmm27 said:
"semicircle" meaning "arch" - an upside-down cross-section of a half-pipe, if you insist.
OK thanks. The word arch gets things the right way up.
I do have a slight problem with your extension of this topic because of the KE needed at the top of the loop is the same, however many times the car does a loop. So why would a figure of eight need any more energy than is needed for the first loop? The car exits the first loop with the same Kinetic energy that was available from the initial mgh. OR . . . . is it just a matter of the way the situation is being described (my badly chosen word "requirements")
 
  • #14
sophiecentaur said:
I do have a slight problem with your extension of this topic because of the KE needed at the top of the loop is the same, however many times the car does a loop.
The OP seemed to have difficulty understanding 5/4, so I presented 3/2 as an alternative. The bottom half of the loop is pretty boring, anyways : energy/speed in = energy/speed out ; no chance of falling off. With the exception of your 3g observation/calculation.
So why would a figure of eight need any more energy than is needed for the first loop? The car exits the first loop with the same Kinetic energy that was available from the initial mgh. OR . . . . is it just a matter of the way the situation is being described
It doesn't need any more extra energy : it just illustrates that the extra bit of "oomph" needed is always 1/2r's worth of a drop.
 
  • #15
hmmm27 said:
It doesn't need any more extra energy : it just illustrates that the extra bit of "oomph" needed is always 1/2r's worth of a drop.
"oomph" means Energy to me so what are you actually saying if energy is conserved? Certainly any extra will depend on efficiency and not be a fixed amount.
So going up and over as many times as you like will not require any extra mgh when there are no losses. This is why I was questioning your statement.
 
  • #16
sophiecentaur said:
"oomph" means Energy to me so what are you actually saying if energy is conserved? Certainly any extra will depend on efficiency and not be a fixed amount.
So going up and over as many times as you like will not require any extra mgh when there are no losses. This is why I was questioning your statement.

"Extra" as in "We have to raise it up more than the height of the loop (which would be all that would be required if it was just a bouncing ball dropped straight down and coming back up, again).

I don't see how "going up and over as many times as you like" is relevant to anything except motion sickness : it's a circle : if you can get through it once, you can continue doing so. When you're finished you can scoot back up to 3/2r above the center, or 5/2r above the ground.
 
  • #17
Doing the math, it appears as it is some sort of conspiracy, that the ##u^2## we get from conservation of energy gets replaced by ##gR## which comes by applying Newton's 2nd law and the expression for centripetal acceleration. So conservation of energy, gravity ##mg##, Newton's 2nd law and the expression for centripetal acceleration do a conspiracy and we end up with a final result ##\frac{5}{2}R## .
 
  • #18
Delta2 said:
some sort of conspiracy
Your "conspiracy" is all to do with common forms in mathematical expressions. I still cannot believe how people are suspicious when given mathematical explanations, preferring arm waving and verbal constructions.
Surprisingly, though, basic Arithmetic is acceptable when dealing with money. Try asking how much money a shopkeeper wants for a load of bread and imagine what your reaction would be if he replied "whatever you have in your hand at this moment". How much money would you put on the counter?
You have to go back to 'the dark ages' and beyond to find general innumeracy. Perhaps in a few more decades / centuries most people will accept Algebra, Calculus and Logic as just part of our language.
 
  • #19
I just used a bad word "conspiracy" it doesn't mean that I don't know math or that I am living in the middle ages. I just tried to put myself in the position of "average Joe" and think how it might look to him the fact that we get 5/2R. I might be the average Joe as well so I don't need a lot of effort to put myself in his position lol.
 
  • #20
My comments are aimed at the "Average Joe". If you appreciate and use Maths then they are not aimed at you. AND, if you are happy with the mathematical derivation of 5R/2 then why try too hard to 'translate' that derivation into the limited language of AJ?

I know we all like to 'feel' the way Science works but I would suggest that a lot of that feeling is actually based on the Maths we know and actually take for granted. Familiar things always seem perfectly reasonable arguments for us.
 
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FAQ: Why is 2.5 times the radius the minimum height needed to do a loop?

Why is 2.5 times the radius the minimum height needed to do a loop?

The minimum height needed to do a loop is 2.5 times the radius because it is the point at which the centripetal force is equal to the force of gravity, allowing the object to maintain enough speed to complete the loop without falling.

How is the minimum height for a loop calculated?

The minimum height for a loop is calculated using the equation h = 2.5r, where h is the height and r is the radius of the loop. This equation takes into account the necessary centripetal force and the force of gravity to maintain the object's speed throughout the loop.

Can a loop be done at a height lower than 2.5 times the radius?

No, a loop cannot be done at a height lower than 2.5 times the radius because the object would not have enough speed to maintain the loop and would fall due to the force of gravity.

Is 2.5 times the radius the same for all objects?

No, the minimum height for a loop may vary for different objects depending on their mass and speed. However, the ratio of 2.5 times the radius is a general rule of thumb for most objects.

What happens if the minimum height for a loop is not met?

If the minimum height for a loop is not met, the object will not have enough speed to maintain the loop and will fall due to the force of gravity. This can result in the object crashing or not completing the loop successfully.

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