Free Fall with Air Resistance Question 1 Obtain the equation of motion for a particle falling vertically under the influence of 1 gravity when frictional forces obtainable from a dissipation function ππ£ 2 are present. 2 Integrate the equation to obtain the velocity as a function of time and show that the maximum possible velocity for a fall from rest is π£ = ππ π . SOLUTION Let us consider the point of release of the object under free-fall is the origin of the coordinate system. The vertical direction is defined to the π§-axis, such that the object moves along the negative π§-axis as it falls. Our modelling suggests that the case is one-dimensional. With this, we can say that the position of the object can be defined by π§(π‘) and the velocity of the object at any given time is defined by π§Μ (π‘). The potential energy of the system is, πΈπ = −πππ§ The kinetic energy of the falling object is, πΈπΎ = With this, we can find the Lagrangian πΏ. 1 ππ§Μ 2 2 1 πΏ = ππ§Μ 2 + πππ§ 2 1 Goldstein, Poole & Safko, Classical Mechanics (3rd edition), Chapter 1, Exercise 23 Lagrange’s formalism of equation of motion is given by, π ππΏ ππΏ ππ ( )− + =0 ππ‘ ππ§Μ ππ§ ππ§Μ Now, we find each term separately, for the sake of simplicity. ππΏ = ππ§Μ ππ§Μ π ππΏ ( ) = ππ§Μ ππ‘ ππ§Μ ππΏ = ππ ππ§ 1 The dissipation function π is given to be π = ππ£ 2 . 2 ππ = ππ§Μ ππ§Μ Now, we can find the Lagrangian equation of motion for the system. ππ§Μ − ππ + ππ§Μ = 0 ππ§Μ + ππ§Μ = ππ The equation above is the Lagrangian equation of motion for the system. We can find the velocity function of the object if we solve the differential equation in π§Μ (instead of π§). π π π§Μ + π§Μ = π π ππ‘ π We introduce a constant π such that π2 = . So, we can write, π π π§Μ + π2 π§Μ = π ππ‘ This is a first order ordinary differential equation. The integrating factor is π ∫ π π π2π‘ . Therefore, ππ 2π‘ π 2 2 π§Μ + π2 π π π‘ π§Μ = ππ π π‘ ππ‘ π π2π‘ 2 (π π§Μ ) = ππ π π‘ ππ‘ 2 2 π π π‘ π§Μ = π ∫ π π π‘ ππ‘ 2 π π π‘ π§Μ = π§Μ = π§Μ = π π2π‘ π +πΆ π2 πΆ π + π2π‘ 2 π π ππ‘ ππ + πΆπ − π π Here, πΆ is the integration constant. To find it, we need to consider the initial condition. As the object was released from a point, π§Μ (0) = 0. So, we can write, π§Μ (0) = π×0 ππ + πΆπ − π π ππ +πΆ =0 π πΆ=− ππ π Putting this value for the integration constant in the solution, π§Μ = π§Μ = ππ ππ −ππ‘ − π π π π ππ‘ ππ (1 − π − π ) π This is the particular solution for the velocity of the falling object. 2 ππ‘ = Now, we can find the terminal velocity of the object from this solution. Terminal velocity is reached when π‘ → ∞. So, π§Μ π = ππ ππ (1 − 0) = π π So, the maximum possible velocity for a fall from rest is ππ π .
