The Physics of a Slap Shot

One thing that has always fascinated me is just how hard a hockey player can hit a slap shot. In baseball, exit velocities on hits can reach up to 120 miles per hour, Giancarlo Stanton of the Miami Marlins pushed past that threshold to 123.9. By comparison, hockey slap shot has topped out at 108.8 miles per hour, thanks to the massive build of Bruins defenseman Zdeno Chara. This makes sense though, a baseball arrives at the plate anywhere from 75 to 105 miles per hour, and a bat swing can have high velocities near that too. So in the collision of the bat hitting the ball, there is a huge Impulse. In hockey though, a slapshot is usually launched from a standstill, or off of a soft, controlled pass. So I wondered, where does all of the energy come from in a slapshot?

SmarterEveryDay, a YouTube channel that focuses on these types of real world questions, helped to enlighten me on where all of this energy comes from. When a hockey player fires a slapshot, they don't hit the puck cleanly. If they were to do so, then the only energy that would be acting on the puck would be that applied by the player. Now, some of these hockey players are big dudes that can swing fast, but there's a maximum human threshold. In order to put more power behind the puck, one would have to look at the stick. Today, sticks are designed to have a certain amount of bend to them. This bend not only keeps the stick from breaking, but adds a whole new type of energy to the swing. Instead of hitting the puck clean off the ice, a hockey player will instead hit the ice first, which bends the stick back as much as a few inches, and then as they follow through, the generated elastic potential from the bending stick is unleashed on the puck, lengthening the collision, and exerting more energy. It all happens in a few milliseconds, but that momentary drag on the ice makes all the difference.

Check out the video here! It's really cool to watch for hockey and physics fans!

https://www.youtube.com/watch?v=IsCdywftyok

Bernoulli Equation and Trains

After learning about Bernoulli Equation I recalled a situation that occurred early in my life. While waiting for the subway in NYC as mischievous 10 year old I always would stand near the edge of the platform before the train would come in. My mother, much to her credit, would always pull me back and berate me saying that the train could suck me in if I stood that close while the subway came by. As I know it all 10 year old I would laugh at this suggestion. Using Bernoulli's Equation however, it is possible to see if my mom was right.

p1+(1/2)p(v1^2)+pgy1=p2+(1/2)p(v2^2)+pgy2

Since we have no change in height and we are looking at the change in air pressure the equation we use becomes:

       Δp=(1/2)p(v1^2)-(1/2)p(v2^2)            p(air)= 1.225kg/m^3    v(air)= 500m/s                                                                                                      v(average subway)= 24m/s

      Δp= ((.5)( 1.225kg/m^3)(500m/s)^2) - ((.5)( 1.225kg/m^3)(524m/s)^2)=-15,052 N

Rearranging P=F/A to F=PA and assuming the average area of person is 1.7m^2 we get:

         F=(15,052N)(.85m^2)= 12,794N                *since the air pushes only on half a person's body the                                                                                  area used is multiplied by a factor of 0.5.

12,974 newtons would certainly be enough to move a small child though bear in mind that to fully experience this force an individual would need to be a the entrance to the tunnel and extremely close to the train. The force would dramatically decrease as the distance between an individual and a train increased. Even so it is not advisable to stand near a moving train, and I guess one should sometimes listen to their parents.
   

Earth's Rotation Is Slowing Down

According to a study, Earth's rotation's been slowing down by 1.8 milliseconds each century due to tidal friction. Thus, I wanted to calculate how long it would take for Earth to reach a rotational speed where a day would begin to last 25 hours, and how fast it is decelerating (in rad/s^2).

If the rotation slows by 1.8 milliseconds (1.8 x 10^-3 s) per century, the time it would take for the rotation to gain an hour (from 24 to 25 hours) would be quite long:

# of centuries = (3,600 s)(1.8 x 10^-3 s)
# of centuries = 2,000,000 centuries

This is essentially 200,000,000 years before Earth's rotation time increases by an hour. 

Next, I assumed that Earth's current rotation time is exactly 24 hours or 86,400 seconds. In order to calculate Earth's current rotational velocity, I used the following formula:

ω = ∆θ/∆t
ω = (2π)/(86,400 s)
ω = 7.2722 x 10^-5 rad/s

If each day lasted 25 hours (90,000 seconds), Earth's rotational velocity would be the following:

ω = ∆θ/∆t
ω = (2π)/(90,000 s)
ω = 6.9813 x 10^-5 rad/s

Using these numbers, I then calculated the rate at which the rotation is slowing down in rad/s^2:

200,000,000 years x 365 days/yr x 86400 s/day = 6.307 x 10^15 seconds

ωi = ωf + αt
7.2722 x 10^-5 rad/s = 6.9813 x 10^-5 rad/s + α(6.307 x 10^15 s)
α = 4.612 x 10^-22 rad/s^2

Sources
http://www.universetoday.com/26623/how-fast-does-the-earth-rotate/
https://www.sciencerecorder.com/news/2016/12/08/earths-rotation-slowing-1-8-milliseconds-century/

Falling Snow Pack

           

On my way to my last Latin class of the semester this morning, I observed a very large block of frozen snow that had broken off in the doorway roof of Lawrence Hall fall with a very heavy thud right besides me. For a moment, I reminisced about the past times when random falling chunks of frozen snow ended up on my neck, quickly melting into ice-cold water and flowing down my back underneath my layered clothes. I wondered if I had stood at that very spot when the ice fell around what force would I have been struck with how much work would have been done in the process. Force and work due to gravity being F=mg and W=mgh respectively, this was not too difficult to estimate. Assuming the height of the doorway roof to be around 3m, my height 1.8m, and the mass of the chunk of ice around 1.5 kg, I calculated as follows.

            (1.5 kg) * (9.8 m/s²) = 15 N

            (1.5 kg) * (9.8 m/s²) * (3-1.8m) = 18 J

            15 N being the force around holding a large book in the air, it did not seem that the falling block of ice would pose an immediate risk. However, it also occurred to me that a greater inconvenience would come from the piece of ice, if for some unfortunate circumstances it were to have lodged inside clothing and began to melt. Realistically assuming that the block of frozen snow would break on impact and perhaps only a fourth would end up in my neck, I could calculate the theoretical temperature change on my body temperature effected by the ice not accounting for the heat generated to counter the ice’s effect. The mass of me and the ice was estimated at 70 kg and 0.25 kg, respectively. The human specific heat was estimated at 3470 J/kg (http://www.engineeringtoolbox.com/human-body-specific-heat-d_393.html).

            Qi_ce=Q_body

mL_ice+mc_ice ΔT=mc_body ΔT

                  0.25kg*3.33*10⁵ J/kg+0.25Kg*4186 (J/kg) *t=70kg*3470J/kg*(273+36-t)

            t=307.3 K, 1.7 K decrease

Theoretically, if the body did not do heat regulation via homeostasis, the ice would decrease the body’s internal temperature by 1.7 C or K, which would be around 34.3 C, below the typical diagnosis for hypothermia.

Of course, as an endotherm mammal, homeostasis would ensure that my body generates heat to maintain body temperature at 36 C. However, if my body was not able to do so, that would result in hypothermia.

Physics behind airplane wings

    As I was booking plane tickets to go home for winter break, I wondered how airplanes can fly when the force of the engine only seems to push the plane forwards and not keep the plane up. As we learned in our physics class, there must be a force in a direction opposite of the force of gravity to keep the plane flying.

    When I was younger, I heard that the wings of planes are shaped in a way that creates lift. This is obviously not a satisfactory explanation, so I searched online to find out more.

Per sites online and our textbook, the wings have a curved top side, which makes the air move faster over the top. The bottom side of plane wings are flat and the air travelling by has a lower speed. According to Bernoulli’s Principle, the air on top will have a lower pressure because its velocity is high, and the air on the bottom will have a greater pressure. This pressure difference is what generates lift to keep giant metal birds in the air.

    However, Explainthatstuff.com claims that this explanation is not quite perfect. The article states that “as air flows over the curved upper surface, its natural inclination is to move in a straight line, but the curve of the wing pulls it around and back down. For this reason, the air is effectively stretched out into a bigger volume,” and this lowers the air pressure. The air that flows by the bottom of a wing experiences something opposite. The air molecules are squished into a smaller volume than normal, which means the pressure increases. And this “difference in air pressure between the upper and lower surfaces causes a big difference in air speed (not the other way around, as in the traditional theory of a wing”) A problem with the traditional explanation is that there is no reason that the air on top of a wing “has to travel a bigger distance in the same time.”

    This explanation makes a lot of sense to me. Either way, we know that the air flowing by the top of a wing has higher speed and lower pressure. The air flowing by the bottom of a wing has lower speed and high pressure. This pressure difference generates a force that keeps airplanes in the sky. 

Saving my coffee

After talking about coffee for so long in class on Wednesday, I realized that we were talking about an insulated mug - no wonder the coffee wasn't cooling down. My biggest problem these days as far as beverage consumption is concerned is not drinking my coffee fast enough, and since my microwave is broken, I can't even heat it up. Therefore, I have decided to determine how long I have to drink my coffee before it's at the point where it's no longer a pleasant temperature.

For the purposes of this blog, I'm going to assume that I made a cup of coffee and then proceeded to forget about it, which is unfortunately a common occurrence. Here are some basics:
-The average mug I own is 16 ounces, and I always fill that to the highest level I can while still leaving room for milk, which is about 12 ounces (354.8 grams).
-My Keurig brews coffee at a temperature of 192 ºF, or 88.8 ºC.
-My refrigerator is ~36ºF on a good day, so 2.2ºC
-For a 12 ounce cup of coffee, I probably add about 2 ounces of milk (61.5 grams) at the previously mentioned 2.2ºC.
-c(milk) = 3130 J/kg, c(water, also coffee) = 4186 J/kg
-The thermostat in my apartment is set to 68ºF (20 ºC) and no one is allowed to touch it
-I'd like my coffee to stay above 120ºF (48.8ºC)

So, here we go with some math. My coffee starts out at 88.8ºC, and then I add the milk, so let's find our starting temp:

My mug has an opening that is 7.32×10^-3 m2

Since it's getting late and I only have so much time to do this blog post, I'm going to pretend heat is only escaping through the opening in the top. This seems most like convection cus the air is interacting with the open mug, and some guy at MIT on the internet says the Kc of coffee is ~10 W/m2 x ºC, so I'm gonna use that.

So starting at 78.9ºC, in a 20ºC room, let's see how long it will take my coffee to reach 48.8ºC:

Now let's see how many Joules it will take to reach 48.8ºC, assuming the specific heat is not changed by the small amount of milk added:

So, 52453/4.31 is 12170 seconds, which is 202 minutes, aka 3.4 hours. Considering that the coffee is also losing heat through the mug itself, it must cool down quicker, but I still have more time than I thought I did.

Driving in Snow

Today as I was driving, I noticed that I drive slower with snow on the ground than when it is not on the ground. Through previous experiences, I knew the the road feels “slippery” when there is snow on the ground. I believe that this is because there is less friction with a snow-covered road. This would be caused by a difference in the coefficient of friction.

In fact, according “engineering toolbox”, the coefficient of friction of a dry road is 1, while the coefficient of friction of a wet road is 0.2. This means the force due to friction for a given object is 5 times greater on a dry road.

                                    F_{friction dry} / F_{friction wet} = (μ_kdry F_N) / (μ_kwet F_N)

F_{friction dry} / F_{friction wet} = (1 F_N) / (0.2 F_N)

F_{friction dry} / F_{friction wet} = (1) / (0.2)

F_{friction dry} / F_{friction wet} = 5

I thought it was thus useful to see how the breaking force is different between a wet and dry road. In order to do this, I modeled the car as a box with four forces acting on it: friction, applied (brake), normal, and gravity. For the purpose of this model, there will be no y-acceleration or velocity (F_normal = F_gravity). As such, I will only be examining the frictional and applied forces (which act opposite the direction of motion)

 

For my first example, I wanted to see how much more force would be needed to stop a 50 kg car moving 15 m/s in 10 m (v_i = 15 m/s and x = 10 m). Again, the only forces I will be exploring are the applied force of the brake and the frictional force. It is important to note the the normal force is equal and opposite to the force of gravity. As such, F_N is equal to m*g. Moreover, F_B will be used to denote the brake force; it is assumed that the brake force is uniformly applied over the stopping distance.

                                   

                                    Dry Road

                                    ΔKE=−ΔPE+W_NC

                                    ½ m (v_f² – v_i²) = -mg(h_f-h_i) + μ_k F_{N *} x + F_B * x

                                           ½ (50 kg) [(0 m/s)² – (15 m/s)²] = 0 - 1 (50 kg) (9.8 m/s) 10 m + F_B 10 m

                                           -5625 Nm = -4900 Nm + F_B  10 m

                                    -725 N m / 10 m = F_B

                                           F_B = -72.5 N

Wet Road

                                    ΔKE=−ΔPE+W_NC

                                    ½ m (v_f² – v_i²) = -mg(h_f-h_i) + μ_k F_{N *} x + F_B * x

                                           ½ (50 kg) [(0 m/s)² – (15 m/s)²] = 0 - .2 (50 kg) (9.8 m/s) 10 m + F_B 10 m

                                           -5625 Nm = -980 Nm + F_B  10 m

                                    -4645 N m / 10 m = F_B

                                           F_B = -464.5 N

With a wet road, the break force must be 392 N greater (still opposing the direction of motion). In other words, because the friction is much less on a wet road, the brake force much be greater to stop a given car with the the same initial speed in the same distance.

However, if the applied force from the brake is at most 100 N, the car would not be apple to stop in 10 m! As such, it is worth examining how much further the car would take to stop on a wet road (as compared to a dry road) if the same brake force was applied. For this example, the 50 kg car still initially had a velocity of 15 m/s. The applied brake force is assumed to be -100 N.

Dry Road

                                    ΔKE=−ΔPE+W_NC

                                    ½ m (v_f² – v_i²) = -mg(h_f-h_i) + μ_k F_{N *} x + F_B * x

                                           ½ (50 kg) [(0 m/s)² – (15 m/s)²] = 0 - 1 (50 kg) (9.8 m/s) x - 100 N x

                                           -5625 Nm = -490 N x - 100 N x

                                    -5625 Nm = -590 N x 

                                    x = 9.5 m

Wet Road

                                    ΔKE=−ΔPE+W_NC

                                    ½ m (v_f² – v_i²) = -mg(h_f-h_i) + μ_k F_{N *} x + F_B * x

                                           ½ (50 kg) [(0 m/s)² – (15 m/s)²] = 0 – 0.2 (50 kg) (9.8 m/s) x - 100 N x

                                           -5625 Nm = -98 N x - 100 N x

                                   -5625 Nm = -198 N x

                                    x = 28.4 m

                                   x_wet / x_dry = 28.4 m / 9.5 m = 2.98

For a given braking force, this particular car would take approximately 3 times the distance to slow down on a wet road as compared to a drive road. This explains why it is helpful to drive slow in snow, as it will take a much greater distance to slow down for a given velocity. A lower initially velocity would require a shorter distance as the change in kinetic energy would be lower—thus, the decreased friction can be offset by driving slower!