Unfortunately it's that time of year in Central New York that the summer playthings really need to be put away. After coming home for Thanksgiving Break I found time for one more short ride on my motorcycle and it was there that I started thinking about physics. We all know that turning quickly on a bike or in a car or even ice skating causes a shift in weight and the feeling of being pulled out of the turn so I took it upon myself to figure out what was going on.

I started with the angular kinematics of a smooth turn and used rough values for my velocity, a turning radius of 6 m, and a 90 degree (1.57 radians) turn.

v0 = 5 m/s

w0 = 0.83 rad/s

vf = 7 m/s

wf = 1.17 rad/s

∆Theta = 1.57 radians

Ideally one wants a constant angular acceleration and these values result in a value of 0.22 rad/s^2. However, the real crux of the matter comes in the conceptual understanding of torque.

Any object traveling in a circular path must necessarily have a force acting on it in the direction of the center of the circle because, as we know from Newton's first law, any object wants to continue in straight line motion. If the friction force from the tires acts in the direction of the center of the turning circle then the center of our mass is "left behind" due to the nature of objects proceeding in said straight line. Because of this we experience the "phantom force" known as centrifugal force which is the sensation of being pulled out of any circular path. Torque comes into play when we think about where these forces are acting.

(This is easier to see and understand with a diagram but imagine the bike and rider as a lever)

Because the axis of rotation is where the wheels meet the ground the friction force has no torque. If the center of mass is directly above the axis of rotation (no lean) then the force of gravity results in zero torque because the angle is 180 degrees. This leaves the phantom centrifugal force which acts away from the center of the circle at whatever distance above the axis of rotation that the center of mass is at. By not leaning into the turn a biker experiences a net torque away from the center of the circle which results in a crash.

So what happens when we lean? When we lean we alter the angle that the centrifugal and, more importantly, the gravitational forces act on the center of mass in relation to the axis of rotation. A biker riding away from us (i.e. into the board) making a left turn would experience a lessened clockwise torque and introduce a counterclockwise torque thereby canceling each other out and allowing the rider to proceed through the turn without incident.

Now comes the part that proves professional racing riders are master physicists. If, upon exiting the turn, a rider is still leaning to one side they will inevitably fall over because the centrifugal torque disappears. As a result when learning to ride a motorcycle one is taught to accelerate out of turns to increase the centrifugal torque and bring the bike upright. The trick is that if you don't speed up enough you're not upright when the turn ends and if you speed up too much you will be upright before the turn ends and end up crashing. The ideal scenario is to have the perfect acceleration that results in the rider being upright the moment the turn ends and they can continue in the straight line that Newton's first law describes.

# Physics 111: Fundamental Physics I

## Tuesday, November 21, 2017

## Sunday, November 19, 2017

### Debunking the myth that women can’t do pull ups

A few weeks ago at lift, our coach told us that pull ups are
much harder for girls than guys because our center of mass is lower. Of course
I wanted to prove her wrong, so I attempted to look at the pull up from a
physics perspective.

The work that a person needs to do for a pull up does not
depend on center of mass at all.

W= m*g*d

It depends only on the mass of the person and the distance,
which in this case would be the length of their arms (from shoulder to top of
knuckles). This means that pull ups would theoretically be harder for a person
with greater mass or longer arms. Because men are typically built heavier than
women, this puts women at an advantage. Arm length is generally proportional to
height, and since men are typically taller than women, this gives women the
advantage again. Even though I am heavily generalizing here, there is
definitely no way that someone can conclude that women are at a huge
disadvantage based on height and arm length.

I then decided to see how big of a difference a small change
in either mass or arm length would make in the amount of work that would need
to be done for a pull up.

W=m*g*d

W= (70kg)(9.8m/s

^{2})(0.48m)
W= 329.28 J

Changing arm length:

W=(70kg)(9.8m/s

^{2})( 0.55m)
W=377.3 J

Changing mass:

W=(90kg)(9.8m/s

^{2})(0.48m)
423.36 J

So ultimately, I was able to prove that a lower center of
mass will not affect the difficulty of pull ups, and so women should theoretically
be able to do them with the same ease as men (with the same muscle mass).

Sources:

https://blogs.scientificamerican.com/guest-blog/the-mechanics-of-the-pull-up-and-why-women-can-absolutely-do-them/

## 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.

Vf

^{2}= Vi^{2}+ 2ad
Vf

^{2}= 0 + 2 (9.8 m/s^{2 }) (1.83 m)
Vf

^{2 }ground= 35.9 m^{2}/s^{2}^{ }
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 = .5mv

^{2}+ 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/s

^{2}*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.

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