Sunday, November 25, 2012

Mark Sanchez’s Butt Fumble



Dave Rappeport


One particularly miserable aspect of my thanksgiving was watching the Jets lose miserably at home to the Patriots. As a Jets fan, there was nothing more embarrassing than watching quarterback Mark Sanchez run into the backside of guard Brandon Moore, fumble the football, and then have it returned for a Patriots TD.
Although I am disgusted as a Jets fan, this situation intrigues me because of the unique physics involved. Sanchez’s kinetic energy is converted into rotational kinetic energy when his feet keep moving forward but Moore’s butt stops his upper body. Sanchez then hits the ground with such force that he fumbles the football.  In order to better understand the physics involved and partially vindicate Sanchez, I want to know how much force was exerted on Sanchez by the ground forcing him to fumble.

Assumptions
-Yards = meters
- Mark Sanchez is a long uniform rod rotating through its center

Mark Sanchez weighs 225 lbs (102 kg) and 6ft. 2 inches tall (1.88 m) and he travels a distance 7 meters in about a second before his upper body is abruptly stopped by Moore’s backside and his lower body keeps moving.

Vo = 7 m/s
Vf = 0 m/s
L = 1.88 m
ωo = 0 rad/s
ωf = ?

ΔKE = ΔKEt + ΔRKE
KEo + RKEo = KEf + RKEf

.5*M*Vo^2 + (.5)*((1/12)*M*1.88^2)* ωo^2 = .5*M*Vf^2 + (.5)*((1/12)*M*1.88^2)* ωf^2

.5 *Vo^2 + (.5)*((1/12)*1.88^2)* ωo^2 = .5*Vf^2 + (.5)*((1/12)*1.88^2)* ωf^2

24.5 + 0 = 0 + .147* ωf^2
ωf = 12.9 rad/s

In order to find the work done on Mark Sanchez by the ground, we must find the net torque the ground exerts on Sanchez to bring his rotational motion to a stop. Assuming that it takes about 0.2s for Sanchez to hit the ground and angular momentum is conserved…

L = Iω
Στ = ΔL/Δt = I*(Δω/Δt)

ωo  = 12.9 rad/s
ωf = 0 rad/s
M = 102 kg

Στ = ((1/12)*102*1.88^2)* (0-12.9)/0.2
Στ = -1940 Nm

τ  = R*F*sinΘ
Θ = 90°

τ = (1.88/2)*F*(1)
-1940 = .93*F

F = 2090 N

The ground exerts a force of 2090 N on Mark Sanchez forcing him to fumble the football. A .22 caliber bullet weighs 2.6 g and travels an average speed of 330 m/s. The force of a .22 caliber bullet hitting a body is given by the change in impulse over time. Assuming the change in time is .0005s for the bullet to stop…

F = Δp/Δt
F = (.0026)*(0-320)/ (.0005)
F = 1716 N

Yes, Mark Sanchez should not have run into Brandon Moore’s butt, but you can’t fault him for fumbling the football. I would fumble if I was shot. However, I’m not stupid enough to run into my own guard. 

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