# Mythbusters Mistake

1. May 5, 2010

### danielatha4

Mythbusters mistake:

They are currently testing that two cars impacting each other at 50mph each is similar to one hitting a solid wall at 100mph.

Regardless of this possible result, their small scale is bugging me.

The rig consists of a swinging arm, like a pendulum, and they are measuring the force exerted by the hammer coming down completely vertical and hitting a solid steel rod as an analogue for a solid wall. However, to compare what they say as “x” velocity and “2x” velocity they are starting the hammer at 45 degrees above the down/vertical to simulate “x” velocity and 90 degrees from the down/vertical to simulate the “2x” velocity.

Does anyone else realize that the change in height for the “x” velocity test will be sqrt(2)/2 and NOT 1/2 of the 90 degree, “2x” velocity, drop? Basic energy principle calculations will show that the 90 degree drop will not generate 2 times the velocity as the 45 degree drop. Thus, they should have used 90 degrees as “2x” velocity and 60 degrees as the “x” velocity.

Last edited: May 5, 2010
2. May 5, 2010

### danielatha4

Oh yeah, also that a downward swing pendulum will already be accelerating in the x direction due to circular motion. Taking away from the force exerted by the wall analogue.

3. May 5, 2010

### danielatha4

Ok sorry, I have to beat a dead horse here.

Not only is 45 degrees not half the height but it also won't yield what was thought to be half the velocity due to the energy principle relation.

v^2 is proportional to h

so in order to half the velocity we would have to find 1/4 h (if h=1 for V and we're trying to find (1/2)V)

does anyone know at what degrees would make the hammer 1/4 of the full height up?

4. May 5, 2010

### SpectraCat

Well, I also think their calculations are wrong, but not quite in the way that you are implying.

The perpendicular velocity of a pendulum at its aphelion is given by

$$\sqrt{k(1-cos\theta)}$$

where k is a constant and theta is the maximum angle. So for your case you need to set up the ratio:

$$\frac{v_{90}}{v_{45}}=\sqrt{\frac{1-cos(90)}{1-cos(45)}}}=\sqrt{\frac{2}{2-\sqrt{2}}}~=1.847$$

So it is definitely too far off for their purposes IMO, but not by as much as you are implying.

Last edited: May 5, 2010
5. May 6, 2010

### danielatha4

Right, I calculated that the "half velocity" angle to drop from should be about 41 degrees. Which works with the equations...

This has annoyed me.

6. May 6, 2010

### diazona

They often aren't particularly precise with the small-scale models, although you could argue that being precise isn't the point of small scale. (Usually, it isn't even the point of the show)

But it would be nice to mention that they're only using a rough calculation in a case like this.

7. May 6, 2010

### lurflurf

Watch more closely the 2x angle is labeled 90 degrees and the 1x angle is clearly labled 49 degrees which is close to Arcsin(.75)~48.59037789... as desired.

8. May 7, 2010

### diazona

Yeah, I just noticed this:
http://scienceblogs.com/dotphysics/2010/05/mythbusters_and_double_the_spe.php [Broken]
(I haven't seen the episode yet... clearly I need to get on that)

Last edited by a moderator: May 4, 2017
9. May 7, 2010

### danielatha4

Right! 41 from the bottom, 49 from the top. Touche Mythbusters. I'm just glad they didn't just put it at 45 like I initially thought they did.