Sect. 2.4, Part III

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Transcript Sect. 2.4, Part III

classical
Special Cases of the 2nd Law: “Recipes”
• 1d discussion for now. Easily generalized to 3d.
• Newton’s 2nd Law equation for a particle has some
equivalent forms (F = total external force) which
might be useful in different cases:
F = ma = m(dv/dt) = m(d2x/dt2) (1)
Also, from the chain rule: dv/dt = (dv/dx)(dx/dt) = v(dv/dx)

F = mv(dv/dx)
• In general, F = F(x,v,t)
(2)
• Goal: Given F, & initial conditions, find v(t), x(t)
 F = ma can be a horrendous differential equation!
• It is easier in cases where F is a function of x, v, t separately.
2nd Law
F = ma = m(dv/dt) = m(d2x/dt2) = mv(dv/dx)
• General comments about these equivalent forms:
F = m(d2x/dt2):
2 time integrations to get x(t)
F = m(dv/dt):
1 time integration to get v(t)
F = mv(dv/dx):
Useful if F = F(v) or F=F(x)
F = ma = m(dv/dt) = m(d2x/dt2) = mv(dv/dx)
• Consider in 4 special cases:
1. F = constant
2. F = F(t), a function of time only
3. F = F(v), a function of velocity only
4. F = F(x), a function of position only
– We’ve already seen examples of 1., 2., & 3. Will
look at 4. in detail soon. Its helpful to (briefly)
summarize general procedures in the 4 cases.
Case 1
F = Constant

a = (d2x/dt2) = F/m = constant
• Obviously, get the familiar, 1d kinematics
equations for constant acceleration (Physics I!).
• Given, initial conditions:
t = 0, x = x0, v = v0
• Get:
v = v0 + at
x = x0 + v0t + ½at2
v2 = (v0)2 + 2a(x-x0)
Case 2
F = F(t) Time dependent forces
 m(dv/dt) = F(t) (1)
• Initial conditions: t = 0, x = x0, v = v0
• Integrate directly. General solution (integrals: limits on
t are 0  t, limits on v are v0  v, limits on x are x0  x):
(1)  dv = (F(t)/m) dt  ∫dv = ∫(F(t)/m)dt
 v(t) = v0 + ∫(F(t)/m)dt = (dx/dt) (2)
(2)  dx = v0dt + [∫(F(t)/m)dt]dt
 x(t) = x0 + v0t + ∫[∫(F(t)/m) dt]dt (3)
Case 3
F = F(v) Velocity dependent forces (like retarding forces).
• Method 1: Direct Integration
 m(dv/dt) = F(v)
(4)
• Initial conditions: t = 0, x = x0, v = v0
• General solution (integrals: limits on t are 0  t, limits on
v are v0  v, limits on x are x0  x):
(4)  dt = [m/F(v)]dv
 ∫dt = t = t(v) = m∫[1/F(v)] dv
(5)
(Gives t(v))
If possible, algebraically invert t(v) in (5) to get:
v(t) = (dx/dt) (6)
Then, integrate (6) to get x(t): (6)  dx = v(t)dt
 ∫dx = ∫v(t)dt
 x(t) = x0 + ∫v(t)dt (7)
Case 3
F = F(v) Velocity dependent forces (like retarding forces).
• Method 2:  m(dv/dt) = F(v) (4)
But, m(dv/dt) = mv(dv/dx).  (4) becomes:
mv(dv/dx) = F(v)
(8)
• Initial conditions: t = 0, x = x0, v = v0
• General solution (integrals: limits on t are 0  t, limits on
v are v0  v, limits on x are x0  x): Get v(t) from (4) as
before (previous page). Then, Integrate (8) to get x(v):
(8)  dx = [(mv)/F(v)]dv  ∫dx = m∫[v/F(v)] dv

x(v) = x0 + m∫[v/F(v)]dv
(9)
Finally, substitute v(t) into x(v):  x[v(t)]  x(t) (10)
Case 4
F = F(x): Position dependent forces.
 m(dv/dt) = F(x) = m(d2x/dt2) (11)
• Initial conditions: t = 0, x = x0, v = v0
• (11) is a 2nd order differential equation to solve for
x(t). Then compute v(t) = (dx/dt). Use standard
methods of differential equations. We’ll discuss this
in special cases as the course proceeds.
• Alternatively (& preferably!) use energy methods.
Introduce a potential energy function U(x) such that
F  - (dU/dx). Analyze the motion using this
function, as will be discussed in detail in the energy
discussion in the next section.
F = F(x): Outline using energy. More details next section.
• (11)

m(dv/dt) = F(x)
• Initial conditions: t = 0, x = x0, v = v0
• Integrate: (limits on v: v0  v, limits on x: x0  x):
 mv dv = F(x)dx  m∫v dv = ∫F(x)dx
 (½)mv2 – (½)m(v0)2 = ∫F(x)dx
(12)
• Introduce a potential energy function U(x) such that
F  - (dU/dx). Also define kinetic energy T  (½) mv2
(12)
or
and

T - T0 = -∫(dU/dx)dx
T - T0 = U(x0) - U(x)
T + U(x) = T0 + U(x0) = CONSTANT
Come back to this in the next section.
In General
• Newton’s 2nd Law: F = (dp/dt),
p = mv
• If we know F = F(t), integrate this to get (limits
on t are 0  t):
Δp  p - p0 = ∫F(t) dt
• Δp  Impulse produced by F
Δp  ∫F(t) dt  “Impulse-Momentum
Theorem”