Friday, November 30, 2012

Physics of Tom Petty’s Free Fall

 By Jessica Mondon

In the song “Free Fallin’” Tom Petty repeats the words “I’m free falling” a lot. I
decided to look at the physics of this free fall. I want to see how far he would free
fall if he fell for the whole time that he said he was free falling. Then I want to see if
he produces enough power with this fall to light the Christmas tree at Rockefeller
Center. He says that he’s free falling for two minutes and 28 seconds (148 s) in the
song. I am assuming he starts at rest and since he is free falling, the only force acting
on him is the force of gravity. Also, since he is free falling the acceleration is just the
acceleration due to gravity, or 9.8 m/s2. Using kinematics we can find the distance
(Δy) that he falls:
Δy = vot + ½ at2
Δy= ½ (-9.8 m/s2) (148 s)2
Δy = -107330 m

I looked up Tom Petty’s weight, and he is approximately 75 kg. This means the force
with which he is falling is:
F = mg
F = (75 kg) (9.8 m/s2)
F = 735 N

I can then calculate the work done by gravity, which is the total work for the system
because gravity is the only force. We now know the force of gravity, and the distance
he free falls so:
W = FdcosΘ
W= (735 N) (-107330 m) cos(270)
W = 7.9 x 107 J

Now that we found work, and we know time we can calculate the power with which
he falls:
P = W/t
P = (7.9 x 107 J)/(148 s)
P = 533022 W

Now that I found the power with which he falls I need to find the power it takes to
light the tree in Rockefeller Center. I found that there are roughly 30,000 bulbs on
the tree, and each bulb is 7.5 W, so the power needed to light the tree is 225,000 W.
The power with which Tom Petty fell is over twice this, so he fell with enough power
to light the tree in Rockefeller Center twice.

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