Paddling in the Deep Blue Fluid!

A calm Saturday morning before the start of a paddling race.

Ah yes, it’s that time of year again: paddling season. A time for boys to become men, a time to take the next stroke and paddle out into the uncertain blue sea. Although `Iolani Mens Paddling does not usually place well in the races, I and the rest of my crew enjoy it and continue to paddle out every day into the world’s most vast body of fluid: the ocean. The ocean is a fluid of uncertainty, you never know what you are going to encounter and you never  know how it is going to treat you. In this large fluid of uncertainty, there is one thing that’s for sure: we’ll all float on.

What is this concept of floating? Why is the canoe “buoyant“? In the most basic terms, the canoe floats because it is less dense than water. That is, the mass of the canoe over the canoe’s volume exists in a ratio such that it’s density is not greater than 1027 kg/m^3 (the density of sea water). More technically, the boat floats because there is an upward buoyant force pushing up on the boat that is equal to it’s weight. That is, Fb = ρVg = mg, where ρ is the density of the fluid, V is the volume of the object displaced in the fluid, g is gravity, and m is the mass of the object. Wait a minute, I thought that the floating canoe depended on it’s density, not it’s volume! How come buoyant force doesn’t factor in the object’s density? In short, it does. The V in the buoyant force equation can be rewritten as m/ρ_object = V. As long as the canoe is less dense than sea water, we’ll have a buoyant force equal to our weight, keeping us happily floating along.

That’s great that the canoe has a density greater than that of ocean water! That means that we won’t ever sink in races, right? Unfortunately, wrong. As previously mentioned, density depends on mass/volume. What would happen to the canoe, for example, if a rogue wave were to fill it with water while we were paddling (unfortunately, this has actually happened before, but we ended up alright)? Let’s think about this, as the canoe fills with water, it’s mass increases. What’s happening to it’s volume? Nothing. That’s the problem. As the mass of the canoe increases but its volume remains constant, its density increases. Once the canoe becomes more dense than ocean water, it will begin to sink. In paddling, we call this getting “swamped”. Once it begins to sink, buoyant force is no longer  Fb = ρVg = mg, it is  Fb = ρVg = mg – mg_apparent. Luckily, if we flip the canoe over and empty the water out, its density will again be less than that of ocean water. Hooray! We survived 🙂 

A photo of me and Terry Lam after surviving a life threatening paddle.

Phew, thank goodness for buoyant force.

Don’t Fall, Balance!

My Precalculus book providing a negative clockwise torque.

After coming home from a long school day, I carelessly threw my Precalculus and AP Economics textbooks on my desk. They landed in a peculiar fashion in which the AP Economics book was almost all the way off the edge of the table and the Precalculus book was on top of it. That’s strange, assuming the center of mass is located at the middle of the AP Economics book, shouldn’t it just fall off the table because most of it’s mass is over the edge? Of course not! The system, because of the Precalculus book on sitting on top of the AP Economics book, is in equilibrium. With knowledge of torque and equilibrium learned in AP Physics B, a situation like the two balancing textbooks is easy to explain.

Looking at the picture to the left, we choose the point below the second “c” on the AP Economics book to be the fulcrum. We can now analyze the forces acting on either side of the fulcrum to prove that the system is in equilibrium. To the left of the fulcrum, the (mass)(gravity) of the AP Economics book’s center of mass provides a positive counter-clockwise torque. To the right of the fulcrum, the (mass)(gravity) of the Precalculus book’s mass provides a negative clockwise torque. Knowing that the system is in equilibrium just by looking at it, we can conclude that the torque to the left of the fulcrum equals the torque to the right of the fulcrum. Because net torque in equilibrium is equal to zero, the system can be represented by the following equation: ΣΤ = 0 = (mg)(d2) – (mg)(d1). Recall that torque is equal to (force)(distance)(sin(theta)). In this scenario, theta is 90 degrees so it is simply equal to 1. If we were to rearrange the equation, we can easily see that the torques equal each other on both sides: (mg)(d1) = (mg)(d2).

Look what time it is! I think it’s time for me to do my math homework. If I remove the Precalculus book from the AP Economics book, what is going to happen? Well, removing the Precalculus book will remove the negative clockwise torque, leaving only the positive counter-clockwise torque of the AP Economics book’s center of mass. Because there is nothing to balance the AP Economics book’s torque after I remove the Precalculus book, the AP Economics book is bound to fall. Indeed, after I removed the Precalculus book from the AP Economics book, the latter came crashing to the ground off my desk. Some of my friends would probably be frightened by how I can relate such a simple thing like balancing books to Physics, but that’s just the way I do things.

Let’s Play Pool!

Pool balls at KMC in the Big Island.

NOTE:  To keep things simple, I neglected air resistance and friction in my discussion.

When Jordan and I were playing pool at Senior Camp this year, she said to me, “You should use this for your next physics blog! The collisions between the pool balls are nearly elastic.” At the time, I had no idea what an elastic collision was so I responded to her with a simple “Yeah, that’s a great idea!” Now, with a solid background in momentum and collisions, I would have whipped out my Casio Fx-CG10 as soon as Jordan reminded me about elastic collisions and I would have calculated the final velocities of the balls right there on the spot!

In pool, as Jordan pointed out, the collisions between the balls are nearly elastic. That is, both momentum AND kinetic energy are conserved. In pool, you start with two balls: your attack ball and your target ball. Once you set the attack ball into motion, one of your balls is moving while the other remains stationary. Soon enough, your attack ball hits your target ball and both balls scatter in different directions at different velocities! How exciting, right?!

A picture of me getting excited about elastic collisions at Senior Camp while Jordan watches me fail to shoot the pool ball in the pocket.

To find the new magnitude and direction of the balls’ velocities, we would need their angles of separation, which we could then use to componentize both balls in to their respective x- and y-velocity components. Assuming we have all of the necessary information and assuming that we are able to find the resultant vector of each ball’s velocities, we can observe conservation of momentum and kinetic energy in elastic collisions through the following relationships. Because both balls have the same mass and because ball 2 is initially stationary, the conservation of momentum equation (M1V1i + M2V2i = M1V1f + M2V2f) simplifies to V1i = V1f + V2f. Similarly, the conservation of kinetic energy equation (1/2M1V1^2i = 1/2M1V1^2f + 1/2M2V2^2f) simplifies to V1^2i = V1^2f + V2^2f.

Through the above relationships, we can see that both momentum and kinetic energy are conserved in pool (and all other examples of elastic collisions for that matter). Undoubtedly, Senior Camp would have been a much better experience if I had already known how to calculate the velocities of the pool balls. Oh well, at least I know what I’m going to do the next time I play pool.