Thursday, November 30, 2017

Juggling Clubs

     I juggle, it’s a very serious hobby and I’ll always be grateful for the time I took to learn. I can’t juggle many thing, mainly balls and rings. I have been trying to juggle something new this semester: clubs. Clubs are like bowling pins with a rounded bottom and a skinnier neck.
     How can one learn how to juggle such an object? Well the answer is one club at a time. I’ll start with the physics of trying to throw one of these things up in the air and catching it. The starting position of juggling a single is grasping the handle a little less than halfway with your thumb on top and your forefinger adjusted a little further up from the rest of the fingers
         Now to start with the physics. You need to apply a force with your thumb downward on the handle of the club.  The club will rotate about in a counter clockwise direction (toward you). The goal is for the club to make one full rotation before it lands into your hand again. You'll want to move your hand out the way to allow the club to rotate fully then place it back in the original position before it ends its rotation in order to catch it nicely. If done correctly the club should rotate in place, not moving in the y direction at all. To accomplish this, you need to apply a torque that will overcome the club's moment of inertia and anticipate when the club will cover 2𝜋 radians so that you know when it's time to put your hand back to catch it.

     We start with calculating the moment of inertia and we’ll need to use the parallel axis theorem because we are going to be rotating this club about a point that’s not it’s center of mass which is located near the bulge at the top of the object. We could say that our club is a long uniform rod with a point mass at the end. The equation for the total moment of inertia for the object would be the sum of the moments of inertia of the long uniform rod rotating about its center of mass and the point mass located at the end of the long uniform rod. The equation would be I= Icm + Ipoint mass. For the long uniform rod the moment of inertia would be I=1/12ML2.  Let's say the mass of the rod portion of the club is 100g. Let us say that the length of the rod portion of the club is around 40 cm. Now if we plug that into this equation for the moment of inertia of just the long uniform rod we would get I= 1/12 (.100kg) (.45m)2=1.7 x 10^-3 kgm2. The equation for the moment of inertia of the point mass would be I=Mh2, where h is the distance the point mass is from the axis of rotation of the uniform rod. For that we would get I=(.200kg) (.25m)2=1.3 x10-2 kgm2. The sum of these two moments of inertia would be I= (1.3 x10-2 kgm2) +(1.7 x 10^-3 kgm2) =1.5 x 10^-2 kgm2. So, 1.5 x 10^-2 kgm2 is the resistance to the rotation of the club that we must to overcome with the torque we apply with our thumb.
     Before we determine the torque, we need to apply with our thumb, let's first think about the angular momentum that we want the club to have. We basically go from having an angular velocity of 0 rad/s to about 2𝜋 radians per 0.5s or 12.6 rad/s. This change in angular velocity occurs within half a second so I'd say that your angular acceleration would be 25.2 rad/s2. The equation of the total torque equals the product of the total moment of inertia and the angular acceleration. With this equation, we can find the torque we need to apply with for thumb. So, 𝜏= (1.5 x 10^-2 kgm2) (25.2 rad/sec)=0.378Nm. It turns out that you don’t need to apply too much torque to juggle a single club which makes sense since you are applying the force with your thumb. However, this is just the beginning we haven't even gotten into the combination of rotational and translational motion as well as adding in two other clubs into the mix but, for starting out, that's enough for now.

Righting Reflex in Cats

Cats possess the righting reflex which allows them to always land on their feet after a fall, even if they are dropped with their back towards the ground. At first, this ability may seem like it violates the Law of Conservation of Angular Momentum; the cat starts with zero angular momentum and no external force acts on it, so it must turn its self over while maintaining zero angular momentum. To achieve this, a cat will bend its back, creating two different axes of rotation on its body, one on its front half and one on its back half. Next, the cat will tuck its front legs towards itself, lowering the moment of inertia of the front half, and rotate so its front half is facing the ground. Simultaneously, the cat will stretch its hind legs out, increasing the moment of inertia of its back half while it rotates in the opposite direction as the front half did to maintain zero angular momentum. Due to the back half's large inertia, the angular velocity of the back half will not be as big as the front half's and it will not rotate as much. Once the front half of the cat is facing the ground, the cat will rotate its hind legs to face the ground. To do this, the cat will straighten the front legs, increasing the moment of inertia and lowering the angular velocity. The hind legs are brought into the body, lowering the moment of inertia and increasing angular velocity. As a result, the rear end of body rotates to face the ground while the front half barely rotates due to its large inertia.

Iaωa + Ibωb = L = 0   

How I Shattered My Computer Screen

Last week when I was home over Thanksgiving break, I accidentally broke my computer screen.  My laptop was lying open my couch, but it was not visible because I had previously thrown a shirt on top of it when changing.  After I changed, without thinking I jumped onto the couch and landed directly on the screen of the computer causing it to shatter.  Even though I was very distraught after breaking my laptop, I immediately began to consider the physics of what had just happened.  First, I calculated the force exerted on the laptop by my body, using my mass (about 70 kg) and my acceleration due to gravity (my body is in free-fall after the highest point of the jump.
F = ma = (70 kg)(9.8 m/s^2) = 686 N

I was also curious about the amount of energy that my body had just before landing on the computer, so I could know the energy that was transferred into the computer causing it to break.  To do so, I first calculated my final velocity (just before hitting the couch) estimating that the highest point of my jump was about 0.5m above the couch.   
Vf^2 = Vo^2 + 2ad = 0 + 2(9.8 m/s^2)(0.5 m)
Vf = 3.13 m/s

Once final velocity was calculated to be 3.13m/s, the total energy transferred into the computer screen can be calculated using: Total E = KE + PE.  Calling the top of the couch h=0, the potential energy at the collision is 0, so the only energy is kinetic energy.
KE = 1/2mv^2 
KE = (.5)(70 kg)(3.13 m/s)^2 = 343 J

My laptop is a 12-inch Macbook weighing only 2.30 pounds, or 0.92 kg. After calculating the force acting on the laptop from my fall to be 686 N as well as the energy transferred to be 343 J, it is clear why my screen shattered the way it did.  These values for force and energy are both very large in comparison to the very small size of the laptop.  Weighing only 0.92 kg and being only a few millimeters thick, my Macbook’s screen stood no chance of surviving the force of my entire body falling directly onto the screen.