### Newton's 2nd law, drag

#### Problem:

An object of mass m is launched from a stationary helicopter towards the Earth with the speed v0.   It experiences a force of air resistance F = -kv, where k is a positive constant.   The positive direction of all vector quantities is downward.
(a)  Draw a free body diagram showing the forces acting on the object.
(b)  What is the terminal speed of the object?
(c)  Find the speed of the object as a function of time, v0, m, g and k.
(d)  What is the direction of the acceleration of the object?

Solution:

• Concepts:
Newtonian mechanics
• Reasoning:
Fnet = ma is the equation of motion for the object
• Details of the calculation:
(a)  Since velocity is downward, air resistance is upward, in the opposite direction of gravity.

(b) When the object reaches terminal velocity, its acceleration is a = 0 and the sum of all forces acting on the object is 0.
0 = mg - kvt,  vt = mg/k.
(c) a = dv/dt
m(dv/dt) = mg - kv,  dv/(v - mg/k) = -(k/m)dt
∫ dv/(v - mg/k)  = -∫(k/m)dt
ln(v - mg/k) - ln(v0 - mg/k) = -(k/m)t
v = mg/k + (v0 - mg/k)e-(k/m)t
(d) a = dv/dt
a = (-k/m)(v0 - mg/k)e-(k/m)t
a = (g - v0k/m)e-(k/m)t
The acceleration can be positive, negative or zero.
If v0 < vt = mg/k, the acceleration is positive or downward
If v0 > vt = mg/k, the acceleration is negative or upward.
If v0 = vt = mg/k, the acceleration is zero.

#### Problem:

A particle of mass m falls subject to the pull of gravity, with acceleration g, and to the force of air resistance.  The particle is dropped from height z = z0.  The initial velocity is zero.  The force of air resistance may be modeled as linearly dependent on the speed, so that the height of the projectile satisfies

md2z/dt2 = -mg - bdz/dt.

(a)  Solve the equation of motion for the particle's height z(t).  Find the terminal velocity.
(b)  Take the b --> 0 limit and show that usual free-fall solution is obtained.

Solution:

• Concepts:
Newton's second law, integrating the equation of motion
• Reasoning:
The equation of motion is given.
• Details of the calculation:
(a)  Let v = dz/dt, then mdv/dt = -mg - bv.  Try a solution v = A + Be-kt.
v(t) = -(g/k) + (g/k)e-kt, where k = b/m.
As t approaches infinity, v approaches vtermial = -g/k.
z(t) = z0 + ∫0t(-(g/k) + (g/k)e-kt')dt' = z0  - (g/k)t + (g/k2)(1 - e-kt).
(b)  As b --> 0, k --> 0, e-kt = 1 - kt + k2t2/2 +... .
z(t) = z0  - (g/k)t + (g/k2)( kt - k2t2/2) = z0 - gt2/2.

#### Problem:

The forces acting on a sky-diver of mass m are the force of gravity and the drag force due to the air.  Assume the drag force is proportional to the square of the speed.  Find the diver's velocity as a function of time, and the diver's terminal velocity vf.  Assume vi = 0.

Hint:  ∫dx/(a2 - x'2) = (1/a) tanh-1(x/a)

Solution:

• Concepts:
Newton's second law
• Reasoning:
Let the diver's mass be m and the coefficient of proportionality between the drag force and v2 be mk.
mdv/dt = mg - mkv2.
• Details of the calculation:
Terminal velocity vf = V:  dv/dt = 0  --> V2 = g/k.
In terms of V, dv/dt = g(1 - v2/V2),    dv/(V2 - v2) = (g/V2)dt.
0vdv'/(V2 - v'2) = gt/V2.
Integrating and solving for v we get
v(t) = V tanh(gt/V).

#### Problem:

A boat with mass m is slowed by a drag force F(v).  Its velocity decreases according to the formula v(t) = c2(t - tf)2 for t ≤ tf, where c is a constant and tf is the time a which it stops.  Find the force F(v) as function of v.

Solution:

• Concepts:
Newton's second law
• Reasoning:
F = mdv/dt.
• Details of the calculation:
a(t) = dv/dt = 2c2(t - tf).
F(t) = ma(t) = 2mc2(t - tf).
t - tf = -√(v)/c  (Select the negative solution since t - tf is negative.)
F(v) = -2mc2√(v)/c = -2mc√(v).

#### Problem:

Assume a particle of mass m is subject to a damping and a driving force.  Its equation of motion is
dv/dt = -λv + (F/m)exp(iωt).
(The physics is represented by the real part of the equation.)
Find the general (real) solution for v(t).

Solution:

• Concepts:
Solving a differential equation
• Reasoning:
We find the solution to the inhomogeneous equation dv/dt + λv = (F/m)exp(iωt)
and add the solution of the homogeneous equation dv/dt + λv = 0.
• Details of the calculation:
(a)  inhomogeneous solution:
Try v = A exp(iωt),  iωA + λA = F/m.  A = (F/m)/(iω + λ).
homogeneous solution:
v = v0exp(-λt), v0 = arbitrary real constant.
general solution:
v = v0exp(-λt) - (F/m)exp(iωt)/(iω + λ).
Real part: v(t) = v0exp(-λt) + (Fω/m)sin(ωt)/(ω2 + λ2) - (Fλ/m)cos(ωt)/(ω2 + λ2).
v(t) = v0exp(-λt) + [(F/m)/(ω2 + λ2)](ω sin(ωt) - λ cos(ωt)).