Question

In the first lesson, does dim just stand for dimensional? And why do some of the derived dimensions seem different? IE dim kinetic energy= m*l^2*T^-2 instead of .5mv^2

Answer 1

I'm not exactly sure what you're referring to - the first 8.01 lecture on OCW doesn't incorporate the "dim" notation for dimension, it's just a colloquialism. In any case, the dimensionality of an expression - i.e. v*t - is the final unit of measurement that such an expression represents - i.e. meters/second. If you have an equation relating two different ways of calculating the kinetic energy of a particle, then, of course, the units that both sides leads to must be the same. If I were to tell you that x (position) = v*t, you would use dimensional analysis to check that indeed, x measures position in meters, and velocity times time gives you a number in units of meters.

In the case of kinetic energy, we define it to be 1/2 * m * v^2, meaning that its dimension is kilograms * meters^2 * seconds^(-2). To say that it's 0.5*m*v^2 is to just state the equation again, which isn't the goal of dimensional analysis.