Saturday, November 30, 2013

Driving in the north country

 I come from a small town that's just south of Canada, practically as far north as you can go in NY state.  This means that I still experience the same snow, ice, and overall poor weather at home that we experience at Colgate during the winter, only at home it's a little bit worse.  Driving in these wintery conditions is extremely dangerous.  A few days ago I was driving home and didn't realize that the road was covered in a layer of black ice.  I was driving slow, but when I applied my brakes to come to a stop at a red light, my tires locked and my car started skidding to a halt.  It was really disconcerting, but unfortunately happens a lot in upstate NY during the winter therefore I've just adjusted to not driving more than 10-15 mph in town when there's not sand or salt on the roads.  I knew I skidded for quite a ways the other day (fortunately I applied my brakes very early in advance so I still came to a stop about 5 m before the stop light), but I wanted to take a look at braking on the ice vs. the nice clear roads we have during the summer.

Velocity of my car: 10 mph (4.5 m/s)
Mass of my car and me (Toyota Rav4): 1150 kg
Coefficient of kinetic friction for tires on ice: .15
Coefficient of kinetic friction for tires on dry pavement: .80

To determine the distance I travelled while skidding:

  • Wnet = ΔKE = -ΔPE + Wnc

Since there is no change in potential energy here, there are only nonconservative forces at work

  • ΔKE= Wnc
  • Wnc= Force of friction * distance
  • F= (9.8 m/s^2)*(1150 kg)*.15 = 1691 N
  • .5*(1150 kg)*(4.5 m/s)^2 = 1691 N * distance
  • distance = 6.9 meters, or 23 feet
In comparison, if I had been driving on the dry pavement during the summer, the stopping distance would have been much less because of the more significant force of friction between the tires and the road

  • ΔKE= Wnc
  • Wnc= Force of friction * distance
  • F= (9.8 m/s^2)*(1150 kg)*.80 = 9016 N
  • .5*(1150 kg)*(4.5 m/s)^2 = 9016 N * distance
  • distance = 1.3 meters, or 4.3 feet
I also thought, what would be the difference in stopping distances if I had been driving my mother's car, which is equipped with studded snow tires? Looking online showed me that studded snow tires can increase a car's coefficient of friction with ice by a factor of up to .13.  This would make the new coefficient of friction equal to .28.

  • Force of friction= (9.8 m/s^2)*(1150 kg)*.28 = 3156 N
  • .5*(1150 kg)*(4.5 m/s)^2 = 3156 N * distance
  • distance = 3.7 meters, or 12 feet.
Studded snow tires would have reduced my stopping distance to 54% of what it originally was!  Definitely going to have my winter tires put on my car before making the drive back to Colgate tomorrow.

Just as a final thought, I was wondering what would have happened if there was a car in front of me that I had run into as a result of the poor braking capabilities on the ice.  If I had collided with another Rav4 at rest on the ice that didn't have snow tires (basically the same total mass as my car), going 4.5 m/s, how far would we both slide if the bumpers locked at it was a perfectly inelastic collision? If the car I hit was parked 1 meter behind another car, would we have collided into them as well?

  • M1V1 + M2V2 = (M1 + M2)(V)
  • (1150 kg)(4.5 m/s) + (1150 kg)(0 m/s) = 2(1150 kg)(V)
  • V= 2.25 m/s (resulting velocity after the collision)
  • ΔKE= Wnc
  • Force of friction= .15*(9.8 m/s^2)*(2300 kg) = 3381 N
  • .5*(2300 kg)*(2.25 m/s)^2 = 3381 N * distance
  • distance = 1.7 m or 5.6 feet 
- So yeah, we would have hit the car in front of us and perhaps caused a chain reaction of cars sliding into other cars (especially if I had been driving faster than 10 mph!)

Basically... winter driving is really hard. And we should all get studded snow tires on our cars! It helps prevent ending up in ditches as often as we would otherwise here in central and upstate NY.




This year, Thanksigiving and Hannukah share a night and the holiday has been named Thanksgivingkuh.  This won’t happen again for another 70,000 years so in its honor I am looking at the physics behind a dreidel.  The question I asked was how much rotational kinetic energy does a small plastic dreidel have when it is initially spun?

KErot = ½ Iω2

To calculate the moment of inertia of the dreidel I approximated it as a solid cylinder which is not entirely correct but I think it was the best option.  The mass of a plastic dreidel  is about 0.00907 kg according to a source online.  The diameter of the dreidel was approximately 1 inch which makes the radius is 0.0127 meters.

I= ½ mr2
I= ½ (0.00907kg)(0.0127m)2
I=7.314x10-7 kgm2

The initial angular velocity of the dreidel--it rotated approximately 3 rotations per second, therefore:

3 rotations x (2π radians / one rotation) = 18.85 radians/second

Therefore the total rotational kinetic energy of the dreidel when it was initially spun is:

KE = ½ (7.314x10-7)(18.85)2 = 1.299 x 10-4 J

Moose Power

By Kody Lyng

Over this break, I watched the newest Disney animation, Frozen. [SPOILER] In the film, a moose by the name of Sven must outrun a pack of wolves while pulling a sled that is carrying two humans. I would like to take a closer look Sven’s useful power output from the time he notices the wolves to the time he outruns them. A few assumptions must first be made:

Sven exerts a force of 1000 N at an angle 15 degrees above the horizontal
Sven starts at rest with a final velocity of 35mi/hr
Sven travelled 500 meters
The path travelled is completely horizontal

Power = Work/Time

Work = Distance*Force*cos(thetat)
W = (500m)*(500N)*cos(15) = 241,000 Joules

35miles/hour*1,610 m/1 mi*3600s/1 hr = 15.7 m/s

Distance = (vinital + vfinal)/2*time
time = (500m*2)/(15.7m/s) = 63.7 seconds

Power = 241,000 J/63.7 s = 3,800 Watts
746 Watts = 1 Horsepower
3,800 Watts = 5.1 Horsepower