The Iron Lotus

In the film,

http://www.youtube.com/watch?v=_7dr6PCeW4c (see from 0:44 - 1:45)

Is the physics shown in the clip accurate?

To get started, I assumed that the pair is skating on frictionless ice and that air resistance is negligible.

I found that Will Ferrel is 1.91 m tall and that Jon Heder is 1.85 m tall.

Then, I looked at the part of the move from the time Will Ferrell is released, to the time that Jon Heder catches him.

Using conservation of energy, I found the initial velocity at which Will Ferrell had to be released.

*Blades of Glory*, two figure skaters successfully complete an extremely difficult and dangerous move called "the iron lotus."http://www.youtube.com/watch?v=_7dr6PCeW4c (see from 0:44 - 1:45)

Is the physics shown in the clip accurate?

To get started, I assumed that the pair is skating on frictionless ice and that air resistance is negligible.

I found that Will Ferrel is 1.91 m tall and that Jon Heder is 1.85 m tall.

Then, I looked at the part of the move from the time Will Ferrell is released, to the time that Jon Heder catches him.

Using conservation of energy, I found the initial velocity at which Will Ferrell had to be released.

ΔKE = -ΔPE + W

_{NC}
I assumed non-conservative work is zero.

½ m (v

_{f}^{2}-v_{0}^{2}) = -mg(h_{f}-h_{0})
From these screen shots I found the initial and final heights:

Jon Herder is 1.85 m tall, and Will Ferrell is at about his shoulders, but Jon Herder is slightly bent. From the first picture, it can be inferred that Will Ferrell's initial height is 1.15 m.

Will Ferrell is 1.91 m tall. In the second picture it appears that he is about three times his height in the air, 3 x 1.91 m = 5.73 m. From the second picture, it can be inferred that Will Ferrell's final height is 5.73 m.

The masses cancel, and the final velocity of Will Ferrell is zero (he has stopped traveling up, but has not yet started falling back down). When values are plugged in, the equation becomes:

½(0-v

_{0}^{2}) = -(9.8 m/s^{2})(5.73 m-1.15 m)
And can be solved to find:

v

_{0}= 9.474 m/s**But is this initial velocity what is shown in the movie?**

Before Will Ferrell is released, Jon Heder is spinning him around in a circle. To find Will Ferrell's actual initial velocity, the linear velocity when he is released must be calculated.

I assumed that the last three rotations that Will Ferrell and Jon Herder make are at a constant angular velocity. Will Ferrell is released at the completion of the last rotation.

The clip shows that it takes 7 seconds for these last three rotations to take place.

Therefore, the angular velocity of the system can be calculated.

First the angular displacement is found:

ΔΘ= 3 revolutions x (2π radians / 1 revolution) = 18.8496
radians

From this value, and the change in time, the angular velocity can then be calculated.

ω= ΔΘ/ Δt

ω= 18.8496 radians/ 7s

ω= 2.69279 radians/s

Using the angular velocity, the linear velocity at the point when Will Ferrell is released can be calculated. I assumed the radius is Will Ferrell's height.

v= rω

v= (1.91 m)(2.69279 radians/s)

v= 5.1432 m/s

The linear velocity that Will Ferrell is released at is not nearly enough to get him to the height shown in the movie, and the height required for Jon Herder to complete his portion of the move.

**The physics shown in the clip is incompatible, and it seems that the Iron Lotus is indeed impossible, at least as it is shown in the movie.**

## No comments:

## Post a Comment