## Tuesday, October 30, 2012

### The Force of the Frankenstorm

We all know Hurricane Sandy rocked the East Coast over the past couple of days, and I was curious to check out some of the statistics for the Superstorm. I found that, along with huge amounts of rain and snowfall, gusts of wind reached as high as 94 mph (42 m/s) in places like Eaton's Neck, NY.

http://www.scientificamerican.com/article.cfm?id=the-stats-are-in-superstorm-sandy

This reminded me of a homework problem we'd done a little while back, with wind being strong enough to blow over a person. So, I wanted to look at situation that could definitely occur in a hurricane: wind blowing over an entire car.
I assumed that the wind is approaching the car from the side, so that the car is basically a rectangle. I found that the average length of a sedan is about 4.6 m, and the height is 1.5 m, and the mass is 1,500 kg. I also assumed the wind is striking the car at a rate of 66 kg/s per square meter and comes to rest. I wanted to: A. Find the approximate force of the wind on the car, and B.Compare this force to the typical maximum friction force between a car and the wet road.

A)
1. First find the area of the car on which the wind will be acting:
A = (l)(w) = (4.6 m)(1.5 m) = 6.9 m^2.

2. Use the area to find the rate of the wind in kg/s:
(66 kg/s/m^2)(6.9m^2) = 455.4 kg/s

3. Find the force of the wind:
We know that the force stopping the wind is exerted by the car, so the force on the car is equal in magnitude and opposite in direction to the force stopping the wind. So we can just look at magnitude to find the force of the wind:

Fwind on car = Fcar on wind = Δpwind/Δt = (mwind)(vwind)/Δt = (mwind/Δt)(vwind) = (455.4 kg/s)(42 m/s) = 19,127 N --> 19,000 N.

B)
Ffr = usmg = (0.40)(1500 kg)(9.8m/s^2) = 5880 N --> 5,900 N.

19,000 N is much greater than 5,900 N, meaning the force of the wind is strong enough to blow over a car.

Here's some evidence!

Part 2: If the wind got the car to travel 10.0 m/s and it was blown into a large building (and came to rest), A. What is the velocity of the car/building after the collision? B. How much energy is "lost" in the collision?

A) We would use conservation of momentum to find the final velocity of the car/building:

mcvci + mbvbi = (mc + mb)vf

But in this case, we can assume that the mass of the building is so much greater than the mass of the car that the velocity of the building will be unchanged in the collision, meaning the velocity of the car/building after the collision is 0 m/s (since the building starts at rest).

B) To find the energy "lost" in the collision, we look at the change in kinetic energy of the system:

ΔKE = KEafter - KEbefore

Since the velocity of the car/building after the collision is zero, the kinetic energy of the system after the collision is also zero. This means ALL of the initial kinetic energy is "lost". But we know this just means it is transformed into other types of energy, such as heat and sound!

Catherine Stecyk