Question

in 8.01 when t=C*sqrt(h/g) how did he find C? and how is it proportional to sqrt(h)? this equation confuses me

Answer 1

Where was this equation? In a problems set, lecture, or equation sheet? If possible please link this location.

Answer 2

lecture http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/units-and-dimensional-analysis/ in the first lecture my bad

Answer 3

around 25:50

Answer 4

This isn't too complicated, but its easy to miss how this works. In the original problem the professor had \[t \alpha h^\alpha*m^\beta*g^\gamma\] In this equation the first alpha \[\alpha\] means is proportional to.
Using \[\beta = 0, \alpha = \frac{1}{2}, \gamma = \frac {-1}{2}\]
We can plug in the values and get:
\[t \alpha \sqrt\frac {h}{g}\] or t is proportional to the square root of h/g. Now for the part that is confusing you...
Instead of using alpha to denote proportionality we can use an equal sign and a constant!
so replacing alpha for a equals sign and a constant denoted by the letter C we get:
\[t=C\sqrt\frac {h}{g}\]

Answer 5

thanks!

Answer 6

In the first lecture, the aforementioned formula was "derived" using dimensional analysis. Point is dimensional analysis is a matter of juggling with the units of quantities. When you equate two quantities their units must obviously be equal. Derivations based on dimensional analysis purely rely on this fact. So you cannot calculate exact quantities( i.e. magnitudes) using this as multiplying or dividing dimensionless quantities wont change the units on either side of the equation. So you cannot precisely derive an equation using dimensional analysis. The professor did not derive the value of the constant as 1. The equation actually is derivable precisely using the fundamental laws of physics in which case it is found to be 1. The professor merely stated this result for the time being