TOPIC 3
Projectile motion
If a particle moves in a vertical plane without being constrained to a specific curve, it has two degrees
of freedom. If it’s moving under local gravity (as well as possibly other forces) we call it a projectile.
Because we are now working in the full freedom of more than one dimension, Newton’s second law is a
vector differential equation of the form
F
d2 r
= .
dt2
m
Let’s review, mathematically, what we already know about differential equations and see what carries
across to the vector case.
3.1
Vector differential equations
There are two broad categories of technique we can use:
[A] Methods from scalar calculus which work in this context if we’re careful;
[B] Completely new methods which have no equivalent in scalar calculus.
In general, the methods from category [A] are systematic: you already know how they work, you just
have to be careful in applying them. The methods from category [B] are a bit more ad hoc: I will show
you various things you can try, but there’s no guarantee they will help in every case.
The basic principle here is this:
Any method from scalar calculus is worth a try as long as it doesn’t involve illegal vector operations.
3.1.1
Scalar methods that don’t carry across
We start with the most obvious technique that doesn’t carry over, and show you where the illegal step
would be if you did this with vectors.
Separation of variables
To solve the equation
dv
= F (v)G(t)
dt
we would divide both sides by F (v) and integrate w.r.t. t:
Z
Z
1
dv = G(t) dt.
F (v)
Given a similar equation for vectors, though:
dv
= g(t)(v · v)v
dt
we can happily divide by v · v, as it’s a scalar; but we can’t divide by v as it is a vector, so we cannot
use separation of variables.
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3.1.2
26
Methods from scalar calculus that do carry across
Let’s look at a few examples, and see how they generalise to vectors.
Plain integration
When we integrate a vector, any constant of integration must also be a vector. Otherwise, integration
works as you expect.
Example
dv
integrates directly to give
v = bt2 + c.
= 2bt
dt
The integrating factor method
Suppose we have the equation
dv
+ 3t2 v = t2 b
dt
If the vector v(t) were instead a scalar function v(t), this would be a standard integrating factor example.
We would multiply by the factor
Z
I(t) = exp
3t2 dt = exp [t3 ]
and then integrate. The method works because dI/dt = 3t2 I(t) so the new LHS is the derivative of
I(t)v(t).
We can still multiply by I(t): that’s not an illegal operation; and we still have
dv
d
[I(t)v(t)] = I(t)
+ 3t2 I(t)v(t)
dt
dt
so we write
d
[I(t)v] = t2 I(t)b = bt2 exp [t3 ].
dt
Then we can integrate (remember that the constant of integration must be a vector):
exp [t3 ]v(t) =
1
exp [t3 ]b + c
3
v(t) =
1
b + c exp [−t3 ].
3
The “CF and PI” method
This method applies to linear differential equations of any order. Because the principle of linear superposition applies just as well for vectors as for scalars, it carries through to vector differential equations.
We first find the general solution to the corresponding homogeneous equation (equation with no “forcing
term”): we call this the Complementary Function. Then we add to that any one solution to the whole
equation: the Particular Integral. That provides the most general solution to the whole equation.
Example 7: Constant coefficients example
Look at this equation, with initial conditions:
d2 x
dx
=g
+k
dt2
dt
At t = 0, ẋ = U and x = 0.
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It is a linear differential equation with constant coefficients. For the CF, we solve the equivalent
homogeneous equation:
d2 x
dx
+k
=0
2
dt
dt
To solve a scalar equation like this, we would use the trial function C exp [λt] and look for values
of λ that can work with C 6= 0. For vectors, we do the same, but use a vector constant:
xtrial = c exp [λt].
Substituting in gives
λ2 c exp [λt] + kλc exp [λt] = 0
leading us to the characteristic equation
λ2 + kλ = 0
λ = 0, −k
The general solution to the homogeneous equation is
xCF = Ae−kt + B.
Helen’s Handy Hint #4
Never apply your initial conditions to the CF;
wait until you have the full general solution.
Then we need a particular solution: we usually try something that looks like the forcing term on
the RHS. In this case we have a constant vector on the RHS, but there is a constant vector in the
CF so we know that won’t work. Instead we adde another factor of t:
Substituting gives
xtrial = ct.
kc = g.
The general solution to the full equation is
x = xCF + xPI = A + B exp [−kt] +
1
gt.
k
Now we need to apply the initial conditions. At t = 0, x = 0 so
0=A+B
A = −B.
x = B(1 − e−kt ) +
ẋ = Bke−kt +
At t = 0, ẋ = U so
U = Bk +
Our final solution is
x=
1
g
k
1
gt.
k
1
g.
k
Bk = U −
1
g.
k
1
1
1
U − 2 g (1 − e−kt ) + gt.
k
k
k
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3.1.3
28
Techniques specific to vectors
Now we move on to techniques that are entirely vectorial. Broadly speaking, these can be described as
Use the dot product or the cross product to multiply your equation by something helpful
It’s usually the dot product that helps, but it’s worth bearing the cross product in mind too.
Example 8: Dot product
dv
= g(t)(v · v)v
dt
This is the example that failed under separation of variables above. The presence of the dot product
suggests that another dot product might help. . . so we dot the whole equation with v.
v·
dv
= g(t)(v · v)(v · v).
dt
We need one more moment of inspiration: notice that
dv
dv
d
dv
(v · v) = v ·
+
· v = 2v ·
dt
dt
dt
dt
so our equation becomes
d
(v · v) = g(t)(v · v)2 ,
dt
which is a scalar equation (and an easy one if you solve for v · v). Once you know v · v you can put
it back into the original equation and solve for v with an integrating factor as before. You need to
check at the end that your values for v and v · v are consistent.
1
2
Example 9: Dot product: Energy Equation
Suppose our force is conservative, F = −∇V . Then the governing equation:
mr̈ = F = −∇V
responds well to a dot product with velocity.
mṙ · r̈ =
ṙ · [−∇V ] = −
d
dt
2
1
2 m|ṙ|
∂V dx ∂V dy ∂V dz
+
+
∂x dt
∂y dt
∂z dt
=−
dV
dt
so we get an energy equation (which is a first-order differential equation):
2
1
+ V (r) = E.
2 m|ṙ|
Example 10: Energy Equation with a Constraint Force
Now suppose in addition to the conservative force, we have a reaction force whose only purpose is
to impose a geometrical constraint. This could be the force keeping our particle on a wire (which
lies in the plane perpendicular to the wire at any given point) or the reaction force stopping a
particle from “falling through” a plane (which is normal to the plane).
Either way, this constraint force R is perpendicular to the curve or surface within which the particle
travels.
Now our governing equation is
mr̈ = F = −∇V + R
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and again, it responds well to a dot product with velocity:
mṙ · r̈ = ṙ · [−∇V ] + ṙ · R
Because R is perpendicular to the surface (or curve) and the particle’s velocity is within the surface
(or along the curve), we have ṙ · R = 0 so
mṙ · r̈ = ṙ · [−∇V ] .
This is exactly the equation we had in the last example, so the derivation of the energy equation
goes through unchanged:
2
1
+ V (r) = E.
2 m|ṙ|
Example 11: Cross product: Angular momentum
In the case of a central force, F = α(r)r, a cross product with the position vector allows us to
make progress:
mr × r̈ = 0
mr̈ = αr
Now we know
d
(mr × ṙ) = mṙ × ṙ + mr × r̈
dt
so we have shown angular momentum L is a constant, where
L = mr × ṙ.
3.2
Particle projected in a vertical plane moving under gravity
Figure 7: Particle projected in vertical plane under gravity
See figure 7. The particle, of mass m, is projected from position r 0 at time t = 0 with velocity U .
Take i and j as unit vectors in the x and y directions. Then the position of the particle at time t is
r(t) = x(t)i + y(t)j. Let v = dr/dt be its velocity. Then Newton’s law gives
F = ma = mv̇,
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where F = mg(−j) is the total force acting on the particle. Thus
v̇ = −gj,
(11)
This is a vector differential equation, which we can solve in the usual way. Integrating, we have
Z
Z
v̇ dt = −gj dt
so v = C − gtj. For C we use the initial conditions: at t = 0, v = U so C = U and v = U − gtj. The
position r is obtained by integrating again:
Z
Z
1
r = v dt = (U − gtj) dt = U t − gt2 j + C ′ .
2
Using the initial condition gives C ′ = r 0 and so
1
r = r0 + U t − gt2 j.
2
Expression in terms of speed and angle
If the particle is projected from the origin (r 0 = 0) with speed U at an angle α to the horizontal then
U = U cos αi + U sin αj and hence r = U t cos αi + U t sin αj − (gt2 /2)j. Equating components of i, j
gives
x =
U t cos α
y
U t sin α − 21 gt2 .
=
So the particle moves with constant speed in the horizontal (there is no horizontal component of the
force) and under constant acceleration in the vertical (under the constant gravitational force).
Horizontal Range
To find the horizontal range of the particle, we need to find T when y(T ) = 0. This is T such that
0 = y = U sin αT − gT 2 /2. Thus either T = 0 or T = 2U sin α/g. T = 0 is where the particle starts, so
we need T = 2U sin α/g and hence the range
xmax = x(T ) =
U2
sin(2α).
g
Cartesian equation for the path of the particle (α 6= π/2)
For the path of the particle, we must eliminate t to find y as a function of x. Thus we have t = x/(U cos α)
(as long as α 6= π/2) and hence
y = U t sin α − 21 gt2 = U sin α(x/U cos α) − 21 g(x/U cos α)2 = x tan α − (gx2 /2U 2 ) sec2 α.
So
y = x tan α −
gx2
(1 + tan2 α)
2U 2
π
.
for α 6=
2
(12)
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y
α
x
Figure 8: Parabola of safety (the dotted line)
Safety Zone
Suppose that an anti-aircraft gun is placed at the origin and it fires shells at a fixed speed U , but at
any angle α. If an aeroplane wishes to fly over the gun, what is the shape of the area of space that it
must avoid? If the position of the aeroplane is (X, Y ), then we would like to find condition(s) that the
shell cannot reach the point (X, Y ) no matter what angle α is used. Thus if X 6= 0, from the previous
paragraph we want (X, Y ) such that equation (12) at (X, Y ) :
Y = X tan α −
gX 2
(1 + tan2 α)
2U 2
has no real solutions in α, or equivalently tan α. (If X = 0, from the earlier paragraph, we need
Y > U 2 /2g.) Write our equation as a quadratic in tan α: we get
gX 2
gX 2
2
= 0.
tan
α
−
X
tan
α
+
Y
+
2U 2
2U 2
This has no real solutions in tan α if
X2 < 4
gX 2
2U 2
Y +
gX 2
2U 2
.
Hence the plane is safe if Y > U 2 /2g − gX 2/2U 2 when X 6= 0. This formula also gives the right solution
when X = 0. Thus the aeroplane is safe if it lies above the parabola
Y =
3.3
U2
gX 2
−
.
2g
2U 2
Particle projected in a vertical plane moving under gravity and air resistance
We have a particle projected from the origin at time t = 0 with velocity U . Suppose we are told that the
air resistance on the particle is kv per unit mass. Then the total force on the particle is F = mg+mk(−v).
Thus from Newton’s 2nd law we obtain
r̈ = g − k ṙ.
This is exactly the equation we solved (for x) as an example of CF-and-PI; and the solution was
1
1
1
1
1
U − 2 g (1 − e−kt ) + gt = U (1 − e−kt ) + 2 g(e−kt − 1 + kt).
r=
k
k
k
k
k
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Behaviour for small k
Physically, we expect the limit of no air resistance to give us back the solution we had before for a pure
projectile. But looking at our expression for r(t), it looks at first glance as if it’s badly behaved for small
k. Let’s investigate it properly.
We start from the general expression:
r=
1
1
U (1 − e−kt ) + 2 g(e−kt − 1 + kt).
k
k
Now we can make a Taylor expansion of each e−kt term (note, this is only really valid when kt is small:
for large enough time, even tiny air resistance will have a significant effect):
e−kt = 1 − kt + 21 k 2 t2 − 61 k 3 t3 + O(k 4 )
and substitute in:
r(t) = U
(1 − 1 + kt − 21 k 2 t2 + O(k 3 ))
(1 − 1 + kt − 21 k 2 t2 + 61 k 3 t3 + O(k 4 ) − kt)
−g
k
k2
r(t) = U
(kt − 21 k 2 t2 + O(k 3 ))
(− 1 k 2 t2 + 61 k 3 t3 + O(k 4 ))
−g 2
k
k2
r(t) = U (t − 12 kt2 + O(k 2 )) − g(− 21 t2 + 16 kt3 + O(k 2 ))
and finally:
r(t) = U t + 12 t2 g − k
1 2
2t U
+ 61 t3 g + O(k 2 ).
This gives the right answer if we set k = 0 and also tells us the first effect of k when it is very small.
Behaviour for large time
Note that as t → ∞ we have v(t) → (1/k)g, which is the terminal velocity. Eventually (after infinite
time) the particle moves vertically at the terminal velocity: (ẋ(t) → 0, ẏ(t) → −g/k). This is the velocity
at which the air resistance balances the weight so there is no acceleration.
For a more complex force law (say, air resistance proportional to speed squared) we may not be able to
solve the whole system; but we can always find the terminal velocity as it is the velocity at which the
forces are balanced, so acceleration is zero.