According to Newton’s Second Law, the net force on an object is proportional to
ID: 3005841 • Letter: A
Question
According to Newton’s Second Law, the net force on an object is proportional to its acceleration. That is, F = ma. Furthermore, acceleration is dened as the rate of change of velocity with time, a = dv dt . It can be shown that air resistance acts opposite the direction of motion, and for larger velocities obeys a velocity-squared rule. That is, |Fair| = kv2 , where k is a constant. You may assume that the force due to gravity is Fg = mg where g = 9.81m/s2 is acceleration due to gravity. a) (*) Formulate a differential equation in velocity v describing the motion of an object falling vertically under the forces of gravity and air resistance. b) (**) Using the techniques you have learned, solve the resulting ODE. Plot your solution if v(0) = 0, m = 1, and k = 1. c) (***) Determine the terminal velocity (the velocity at which the forces of gravity and air resistance perfectly balance), and linearise the ODE around this point. Solve the resulting ODE and plot against your solution to part (b). How do your solutions differ? Why wouldn’t it have made any sense to linearise around the initial velocity?
Explanation / Answer
a> according to Newton's Law of mothion
force in the direction - force opposite to the direction of motion = F
=> Fg - Fair = ma
mg - kv^2 = mdv/dt
=> dv/dt = g - kv^2/m
dv/dt = 9.81 - kv^2/m ----------> this is the required differential equation
b> dv/dt = 9.8 - kv^2/m
dv/(9.81 - kv^2) = dt/m
-dv/(kv^2 - 9.81) = dt/m
integrationboth sides
=> .3194tanh^-1[.3194sqrt(k)*v] = t/m +c
gven at t=0 , v(t)=0 , m=1 and k = 1
=> .3194tanh^-1[.3194sqrt(1)*0] = 0/1 +c
c = 0
=> .3194tanh^-1[.3194sqrt(k)*v] = t/m
.3194sqrt(k)*v = tanh[t/.3194m]
=> v = 1/.3194sqrt(k)*tanh[t/.3194m]
k=1 and m= 1
therefore , v = 1/.3194*tanh[t/.3194]
c> terminal velocity
Fg = Fair
mg = kv^2
v = +- sqrt[mg/k]
v = +- sqrt[9.81m/k]
m =1 and k = 1
=> v = +- sqrt(9.81)