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A popular circus stunt is for a motorcycle rider to ride inside a bowl shaped depression called a "Wall of Death." The rider goes higher and higher up the wall until they are actually horizontal. I wonder if a human could do the same.

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2 Answers 2

The speed you need (explained http://physics.mut.ac.th/PhysicsMagic/wall.htm)

$$ V^2 > R g/u $$

$v$ = the velocity (m/s)
$R$ = Radius of the pit (m)
$g$ = acceleration of gravity (9.8m/s^2)
$u$ = coeff of static friction

So if u is 1 (the maximum possible), and you are an olympic athlete that can run 100m in 10s
Then you apparently could with a radius of about 10m

$$ R = uV^2/g = 10^2/9.8 = 10m $$

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How do you get that the maximum possible $\mu$ is $1$? –  Mark Eichenlaub Jul 29 '11 at 7:54
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If $\mu$ were always less than one, a conventional dragster would be unable to accelerate at more than 1 g. –  mmc Jul 29 '11 at 11:58
    
I believed only static friction exceeded 1. The dragster site proves me wrong. –  Michael Luciuk Jul 29 '11 at 12:20
    
what difference does it make that a runner is not in contact with the wall for a significant portion of the time while running? The frictional force would note be exerted on the runner during this time. –  luksen Jul 29 '11 at 13:26
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@Ben - well obviously it's only valid for a smooth spherical non-relativistic athlete running in a vacuum –  Martin Beckett Jul 30 '11 at 2:36

Short answer: No.

Longer answer: I found it easiest to approach this by asking the question in a different way ---> If a runner were to run around in circles on an inverted conical ramp, how fast would he have to run to remain perdendicular to the ramp's surface? The answer depends on three things: radius for the running path; slope of the ramp (B) and height/mass distribution of the runner (shorter runners do better, since more of their mass is at a larger $v^2/R$).

Assumptions: 1.6m (5'3") tall runner, with a center of gravity (CoG) 0.8m from feet, on a 2m radius (approx 13 foot diameter) path. Since he is tilted in from vertical, his CoG traces a path which has a diameter somewhat less than that: for now, estimate B = 40 degrees tilt, which give a path for the CoG of radius (2 - 0.8 sin(40))m = about 1.5m.

$a_d = 9.8 m/s^2$ is acceleration (due to gravity) straight down. $a_L = V^2/R$ is acceleration laterally outward, due to centrifugal force. $a_t$ is total acceleration, which will be perpendicular to ramp surface.

$a_d = a_tcos(B)$

$a_L = a_tsin(B)$

Therefore, we have $a_L = a_d tan(B)$ ----> $B = tan^{-1}(V^2/(R a_d)$

Remembering that the R in that equation is the R for CoG, and that the velocity of that CoG will be slower than speed of the runner's feet, we have...

$B = tan^{-1}((6.7 (m/s)(1.2/2))^2/(1.5(m) 9.8 (m/s^2))$ = about 47.8 degrees from horizontal. Compare to the original guess of 40 degrees. I'm not making a second iteration to correct path length: instead guess about 45 degrees. This assumes that the runner is running a 4 minute mile, which is quite a feat to pull off on a conical ramp with path of radius 2 m, with your head (experiencing less centripetal force than CoG) constantly trying to fall in towards the center while your feet (experiencing more) trying to slide out from under you in the opposite direction.

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Your calculation is incorrect. In the nonrotating frame, there is no centrifugal force, the acceleration is toward the center of the circle, not perpendicular to the surface, and it's inward, not outward. In the rotating frame, the acceleration is zero. You might want to start by drawing a free-body diagram, then applying Newton's second law. –  Ben Crowell Jul 29 '11 at 19:40

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