Thursday, November 16, 2017

Physics in Dropping a Phone

One morning, I unfortuently dropped my phone from my bed. I have a top bunk in a bunk bed, which is about 6 ft, or 1.83 meters, above the floor. Assuming an initial velocity of zero and neglecting air resistance, I calculate the velocity of which my phone hit the ground to be 5.99 m/s.
Vf2 = Vi2 + 2ad
Vf2 = 0 + 2 (9.8 m/s) (1.83 m)
Vfground= 35.9 m2/s2 
Vf= 5.99 m/s

Furthermore, my phone is an iPhone 5S which, according to Apple, has a mass of 112 grams. Setting my dorm room floor at a height of zero, my phone had 2.00 J of energy being converted from potential to kinetic energy throughout its fall. 

Total Energy = K.E + P.E
Total Energy = .5mv2 + mgh
Total Energy = .5 (.112 kg) (5.9 m/s)2   
Total Energy = 2.00 J

Luckily, my screen did not shatter when my phone dropped. There are many factors that play a part into whether a phone will shatter when dropped. The impulse equation, Ft=p is useful in determining this. The higher the force, the more likely it will shatter. Thus the height from which the phone was dropped and the surface it lands on will effect the likelihood of it shattering. The higher the phone is dropped, the larger the momentum and force are. The surface the phone lands on also plays a role in the likelihood of the phone cracking because it effects the time of the collusion. A larger time will occur when landing on something like a pillow and will exert a smaller force. A smaller time which occurs from landing on a surface like tile will exert a larger force. Another equation that determines whether the phone will shatter is P= F/A. The larger the area that ultimately hits the ground, the less pressure exerted on the phone and the less likely it is to shatter.

Why does an hourglass run with a constant speed?

     Physics News

     My friend has an hourglass and I enjoy staring at it. It occurred to me that the hourglass seemed to be going at a constant speed.

     I did research online, and found out that hourglasses do go at a relatively constant speed. ("Why Hour Glasses Tick" Physical Review Letters, Volume 71, Number 9, 30 August 1993 https://journals.aps.org/prl/pdf/10.1103/PhysRevLett.71.1363)


     However, this is a little puzzling to me. Sand in an hourglass can be thought of as liquid. We know from our class that pressure at the bottom of a container is affected by the height of the liquid, and has nothing to do with the shape of the container. So as the height of the sand decreases, shouldn't the speed of the hourglass decrease because the pressure at the bottom hole decreases?
    
     It turns out that this has something to do with the special shape of the hourglass. The hour glass is designed in a certain shape, so that the friction between the sides and sand would cancel out certain pressure that is directed to the bottom hole. Pressure at the bottom hole stays relatively constant, resulting in the hourglass going at a constant speed.

Bungy Jumping

While studying abroad in Sydney, Australia, I had the opportunity to spend a week in Queenstown, New Zealand. Queenstown, known as the adventure capital of the world, is home to the Kawarau Bridge, the location of the world’s first commercial bungy jumping operation. I had never considered bungy jumping before my week in Queenstown, but once I got there, the locals convinced me I had to while in New Zealand. Exactly one year ago today, I jumped off of the Kawarau Bridge.

The Kawarau Bridge is 43 meters above the river below. It is an option for some to be dunked in the river at the bottom of the jump – full body, just your head, or even just your hair and fingertips. Given that I was very afraid of water as a kid, and to this day fairly apprehensive about swimming and being around water, I was absolutely not interested in hitting the water. Therefore, I stood on the edge of the bridge for a full two minutes unable to jump, having convinced myself I was going to be dunked in – even though I said I didn’t want to. The nice men working the bungy thought this was hilarious – I wasn’t scared to jump off a bridge, but I was scared of going head first into the river below. They convinced me that it wasn’t even an option for me if I wanted to, since I didn’t weigh enough to make it that far down.

At the time, I didn’t understand the physics behind bungee jumping and therefore was unsure about hitting the water. How that I understand the physics, I understand that given the spring constant of the bungy cord used (I tried to find this online and couldn’t) and given the potential energy I had on the bridge – mgh = 50kg*9.8m/s2*43m = 21070 J – I could not have reached the spring potential energy necessary for the change in length needed in order to reach the water, given the conservation of energy.


Therefore, I now understand why I had nothing to be worried about!

Sunday, November 12, 2017

Why Throw A Spiral?

We thought about in the homework: does a rotating ball travel faster or slower than a non-rotating ball? If we look at the kinetic energy of the ball (which has a translational and rotation component), they we see that any rotational movement would only take away from the overall kinetic energy of the ball. This would therefore keep the ball from traveling as far or as fast as if the ball was not rotating at all. Then why do football players throw their balls with spirals? 

We have to take into account air resistance and how this affects the angular momentum of the ball. If we picture a ball tumbling through the air, its shape would continuously be changing and therefore so would the force of air resistance on the ball. This change in shape and force of air resistance makes it more difficult for the thrower and receiver to predict where the trajectory of the ball (and therefore where it will end up). However, if we have no external torques (such as the ball tilting), and the ball is spinning continuously, our angular momentum is conserved. 


Therefore, the putting spin on the football does not make it go farther or faster (quite the opposite), it merely makes the thrower more confident in placing the ball where they want it to go. 

Physics in Ballet

I have been a dancer for my entire life, however I have never, until this point, stopped to think about the physics behind some of the most fundamental dance moves. Physics is not only able to explain the precise technique needed to achieve certain movements, but also helps explain why a specific body type tends to be seen in professional dancers (mostly ballet). We can easily examine the physics of turns – how to get moving and stay moving, as well as change speed. A common type of turn in ballet is a fouette turn, the progression of which is shown below. 


In order to get turning (position 1), the dancer applies a force to the floor, causing friction to push against the foot in the opposite direction, creating torque. This external torque is responsible for starting the turning motion. The motion of a fouette turn is to move the leg and arms in and out at a specific time (position 2 to position 3), decreasing and increasing speed, respectively. When the turn is first initiated, the dancer pulls their legs and arms close to their body. The purpose of this is to decrease moment of inertia and which would lead to a high rotational velocity. While the dancer is turning, no external torque is acting so rotational momentum is conserved. Moving the legs and arms outward (position 2) slows the dancer down for a moment, because inertia is temporarily increased. Pulling the arms and leg back in (position 3) decreases inertia again and speeds up the turn. This sequence is repeated as many times as desired. Since inertia is proportional to mass of an object, decreasing mass is an efficient way of increasing rotational speed. This is (one of the reasons) why professional ballet dancers all have a very slender figure. Another, perhaps healthier, way to decrease inertia is to work on holding the leg and arms tight to the body, decreasing the radius component of inertia. Often times dancers use a combination of both of these methods in order to obtain the highest rotational speed in turns.

Saturday, November 11, 2017

You + Hula Hoop = Physics!

Who doesn't love hula hooping?
Me.
So in the hopes of improving my skills, let's study it!
Keeping the hula hoop on a person's waist and maintaining its circular movement requires forces acting upon the hoop. The friction between the hula hooper and the hoop keeps the hoop positioned on the person's waist; this friction also causes the hoop to slow down as it brushes against clothing. The hula hooper's hips exert an upwards force on the hoop, which opposes gravity and also serves to maintain its position. The person also produces torque from the hips, which is an outward turning force required to maintain the centripetal force of the hoop; without this, the hoop would fall to the ground.

When a person steps inside of a hula hoop, he or she becomes the hoop's axis of rotation (denoted by the hoop's radial symmetry). The individual must perform work by moving his or her hips to maintain the hoop's momentum created by the initial spinning of the hoop. According to angular momentum, it is less challenging to keep a heavier hula hoop moving than it is to maintain a light hoop's movement, as long as the hoops' radii are constant. This is because the hoop's angular momentum is directly related to the hoop's mass. When we consider the hula hooper to be a point mass, then angular momentum can be quantified by:
                                                    L=rpsinθ                                        where p=mv       
If we consider the hula hooper a continuous object, the principle still holds:
                                                    L=Iω                                              where I=mr and ω= Δθ/Δt
So those hula hoops with the sand in them? Turns out the extra weight of the sand is intended to help challenged hula hoopers like myself keep the hoop moving! If we increase the radius of the hula hoop, we would need to spin it faster (to achieve the same change in angle over time) in order to maintain the same momentum as a hula hoop with a smaller radius. This can also be seen in the angular velocity formula, in which the radius is inversely proportional to velocity:
                                                     v=rω 
Moral of the story? If you struggle with hula hooping like I do, buy a weighted hoop with a small radius so you have a chance at hula hooping success.

Sources
https://www.scientificamerican.com/article/bring-science-home-hula-hoop-physics/
https://people.howstuffworks.com/hula-hoop1.htm
https://www.youtube.com/watch?v=ZGK770HE7os