Skip to content

Torricelli’s Equation

I was reminded recently of a useful kinematics equation usually learned in high-school physics:

v_f^2 = v_i^2 + 2as

Strangely enough, it seems that not many people recognize this equation—apparently sometimes known as Torricelli’s equation—as conservation of mechanical energy in disguise. Even the Wikipedia articles mentioning it don’t make the connection.

Starting with conservation of mechanical energy with T and U denoting kinetic and potential energy, respectively:

T_f + U_f = T_i + U_i

using the relation of work to kinetic and potential energy:

W = \Delta T = -\Delta U

and rearranging, we get:

T_f = T_i - \Delta U = T_i + W

Taking the definition of work, converting it to an arc-length integral and applying the integral mean value theorem:

W = \int_{C} \vec F \cdot d\vec{s} = \int_{C} F_\parallel ds = \bar{F}_\parallel \Delta s

we then substitute:

T_f = T_i + \bar{F}_\parallel \Delta s

Applying the definitions of kinetic energy and force, assuming no change in mass:

\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + m \bar{a}_\parallel \Delta s

cancelling mass, rearranging, and renaming variables, we finally get:

v_f^2 = v_i^2 + 2as

The derivation implies that the equation is valid for arbitrary paths. Care must be taken, though, to define the variables suitably: v denotes, as usual, speed; a is the mean value of the tangential acceleration; and s is the arc length travelled.

Note that the only calculus needed was to derive the mean-value expression for work, which is usually given as the definition in high-school physics anyway; the conversion from the line integral definition was done to make explicit the applicability of the final equation for space paths.

Exercise for the reader: quickly derive the analogous equation for rotational kinematics:

\omega_f^2 = \omega_i^2 + 2 \alpha \Delta \theta

Post a Comment

Your email is never published nor shared. Required fields are marked *
*
*