Thursday, September 22, 2011
Give me the magic youtube video
Thermal energy of a caught baseball
I was intrigued by a question that was raised in the modeling workshop at SPU this past summer. Does FΔx for a baseball caught by a mitt give the change in kinetic energy of the baseball or the change in total energy of the baseball (including a rise in thermal energy)?
My model says that it depends on what Δx is. If Δx is the displacement of the center of mass of the ball then FΔxcm would equal the change in kinetic energy. If Δx is the displacement of the point of application of the force on the ball then FΔxforce would give the change in total energy of the baseball. Since the ball compresses a bit Δxcm will be slightly greater in magnitude than Δxforce. Both the change in kinetic energy of the ball and the change in the total energy of the ball would be negative but the magnitude of the change in total energy would be smaller. The difference would account for the rise in thermal energy. So maybe FΔxcm is -100J and FΔxforce is -95J this would mean the ball lost 100J of kinetic energy and gained 5J of thermal energy is produced in the ball. The remaining 95J of thermal shows up elsewhere.
Now I want to try out this thinking on two more scenarios; (A) a rigid metal ball that is stopped by a glob of clay and (B) a glob of clay that is stopped by a rigid metal wall. Let's assume that in both scenarios the change in kinetic energy is 100J, so FΔxcm is -100J in both scenarios.
In scenario (A) the ball is rigid so FΔxforce would be the same as FΔxcm , -100J. So the rigid ball doesn't get any of the thermal energy directly. The glob of clay gets the entire 100J.
In scenario (B) the wall is rigid so FΔxforce would be zero. In this case the glob of clay get's the entire 100J of thermal energy. The rigid wall doesn't get any.
I guess the moral of my story is that whatever squashes gets the thermal energy. In the case of the baseball, the baseball doesn't squash much so it doesn't get a very big share. The mitt, arm, shoulder squash a lot and get most of the thermal energy.
Wednesday, September 21, 2011
Walking uphill makes us sweat - some scientists agree
Cavagna GA, Saibene FP, Margaria R. 1963. External work in walking. J Appl Physiol 18:1–9.
One interesting piece of evidence that I came across was the following: Margaria (MARGARIA, R. Atti reale accad. naf. Lincei, Ser. VI. 7 : 5, 1938.) has shown that, in uphill walking (positive work), the efficiency, as expressed by the ratio of body lift to energy expenditure, tends to a maximum value of 0.25, while in downhill (negative work) walking or running it tends to 1.20.
I take this to mean that if I use 100J of chemical energy to walk uphill I could hope to gain as much as 25J of gravitation energy and generate 75J of thermal energy. On the other hand if I use 100J of chemical energy to walk downhill I would lose 120J of gravitational energy and generate 220J of thermal energy. In order to gain 100J of gravitational energy I would have to use at least 400J of chemical energy and generate 300J of thermal energy. So for the same change in height more thermal energy is generated on the way up, 300 vs. 220.
Sunday, September 18, 2011
Physics work and physical work
'Some physics isn't quite right here.
Last time I was on an inclined treadmill, my head (and attached torso) did NOT go "UP" as I hiked along. Most of me stayed at the same gravitational potential.
I did have to lift my legs more than on a flat treadmill but that, near as I can tell, was the only additional work I had to do. Not nearly as much work as climbing a hill.'