Human loop speed

Rhett Allain’s post about a human running around a loop has really got me (and him!) thinking (click through to see the video). I wondered if there was a more sophisticated way to do the calculation for the minimum speed needed. While Rhett tweeted out an approach based on integrating the “fake” forces involved, I wanted to see if I could do it more generally for a body with any moment of inertia.

My approach was to figure out the speed that an object with a moment of inertia (as measured about its center of mass) would have to hit the loop with so that it would have enough speed at the top to not lose contact with the surface.

Just like Rhett, I found it easier to think of the angular frequency, \omega, instead of the speed, at least at first. After playing around with this for a while, I’ve convinced myself that the angular speed necessary at the top is independent of the moment of inertia and is, in fact, the same as what you get with Rhett’s initial calculation:

\omega_\text{top}\geq\sqrt{\frac{g}{r_\text{cm}}}

How can you determine that? Well the centripetal force at the top has to be at least as big as the gravitational weight force, so that the normal force of the loop is at least zero (floors push, they can’t pull).

So if the rotation speed at the top is independent of the moment of inertia, is that the whole story? No, for two reasons: 1) what does it mean to say “how fast do you have to run?” and 2) do you slow down from the bottom up to the top due to the increase of potential energy?

First number 1: from the angular speed above, you can certainly figure out the linear speed that the center of mass has. However, that’s not where your feet are, and that’s probably a better place to measure your speed.

Now 2: if you were to do this with an ice loop with skates, you could just get a lot of speed at the bottom and coast through the ramp. That’s basically what I assumed for the rest of this post. If you’re running, I’m not sure if you’d be able to keep your speed up while running around, I guess I assume you’d likely slow down in a similar fashion.

So, given the rotational speed at the top, can we figure out the speed you’d have to enter the loop so that you’ll have that rotation at the top? Sure! All I did was figure out the total energy at the top and assume you have that same (total) energy at the bottom. If you have less potential energy at the bottom, you’d have to move faster. The nice thing is that the speed of your center of mass at the bottom is the same as your feet (assuming you’re approaching the loop on flat land).

E_\text{total}^\text{top}=\frac{1}{2} I_\text{cm} \omega_\text{top}^2+\frac{1}{2} M r_\text{cm}^2 \omega_\text{top}^2+2 M g r_\text{cm}

where the first term is the rotational kinetic energy around the center of mass, the second term is the translational kinetic energy of the body, and the third term is the additional gravitational potential energy compared with when the body enters the loop.

To find the speed at the bottom, we plug in the expression for \omega_\text{top} and set that energy equal to the translational kinetic energy at the bottom 1/2 M v^2, and solve:

v_\text{bottom}=\sqrt{\left(r_\text{cm}^2+\frac{I_\text{cm}}{M}\right)\frac{g}{r_\text{cm}}+4 g r_\text{cm}}

Note how you get the well known v=\sqrt{5 g r} if the moment of inertia about the center of mass is zero (which would mean that the r becomes the radius of the loop).

But that’s not the whole story, since I’d rather express it in terms of these variables:

Schematic showing an extended body going around a loop

Schematic showing an extended body going around a loop

(sorry for the crappy drawing, I was in a hurry :) Note that now r_\text{cm}=R-h/2. With that we get:

v=\sqrt{\left(R^2-Rh+\frac{5 h^2}{12}\right)\frac{g}{R-\frac{h}{2}}+4 g \left(R-\frac{h}{2}\right)}

Ugly, right? But still interesting. Here’s a plot of the speed at the bottom for an R=1.5m loop for people ranging from 1 to 2 meters in height (note that I’m modeling a human as a rectangular bar with I_\text{cm}=\frac{1}{12} M h^2):

minimum running speed versus person height for R=1.5m

minimum running speed versus person height for R=1.5m

So what’s the upshot? Raise your hands when trying to run around the loop and you won’t have to run as fast.

Thoughts? Here’s some starters for you:

  • I’m not convinced that the moment of inertia doesn’t affect the angular speed at the top. Prove it!
  • I tried this after reading this post and now I’m in the hospital. What’s the name of your lawyer?
  • Whatever Rhett says is law. You haven’t contradicted him have you?
  • Don’t you have a real job?
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About Andy "SuperFly" Rundquist

Associate professor of physics at Hamline.
This entry was posted in fun, physics, twitter. Bookmark the permalink.

3 Responses to Human loop speed

  1. bretbenesh says:

    I have nothing insightful to say. I just enjoyed this.
    Bret

    P.S. Thanks for the retweet today.

  2. rhettallain says:

    I like that part that says whatever I say is law. I should let my kids read this and that way they will have to listen to me.

    But I’m still not convinced. If I understand this, you are just looking at energy – right? My assumption was that the person had already made it to the highest point of the loop and then I just looked at forces. But now I am even doubting my own calculation (which, btw, I did on a plane so there could be errors).

    I wonder if I could make this into a numerical model? Maybe the person is 4 masses connected by springs. Then I could let this spring person move around a loop. Maybe I will try that.

    • Andy "SuperFly" Rundquist says:

      Here’s what I did: I formulated the problem as a constrained Lagrangian one where the variables were the r and \theta of the center of mass. The constraint was that the r was fixed. I then took a look at a plot of the constraint force (the \lambda in the Lagrangian) and found the initial speed when the constraint didn’t change sign at the top. I then realized that I could do it analytically and got the results above.

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