Download MATH1302, Week 1 Vector dynamics and simple planar

MATH1302, Week 1
Vector dynamics and simple planar motion
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Vector dynamics
We will be considering particles moving in 1,2 and 3 dimensions. In 2 and 3 dimensions vectors
are used to describe particle position, velocity and acceleration.
Let a particle move along a curve; we can describe the position of the particle as a function of
t as the position vector r (relative to a suitable origin O). Then we define
Velocity:
Acceleration:
dr
dt
1
= lim (r(t + τ ) − r(t)).
τ →0 τ
dv
d2 r
a(t) =
=
.
dt
dt
v(t) =
These are vector quantities: they have magnitude and direction. The magnitude |v| of the
velocity is called the speed. By definition it is non-negative. Thus a particle thrown vertically
upwards with velocity U and speed U = |U | (see figure 2) returns to the thrower with speed
U (prove this!), but its velocity has then changed to −U . Notice that the set of basis vectors
Figure 1: Definition of velocity
used to describe r does not come into the above definition. Note also that the velocity is
independent of the choice of origin. When r is written in terms of the standard basis i, j, k we
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Figure 2: When a ball thrown upwards returns, its speed is the same,
but its velocity has changed sign.
have r = xi + yj + zj and hence
v(t) =
dr
dt
1
= lim (r(t + τ ) − r(t))
τ →0 τ
1
= lim (x(t + τ ) − x(t))i
τ →0 τ
1
1
+ lim (y(t + τ ) − y(t))j + lim (z(t + τ ) − z(t))k
τ →0 τ
τ →0 τ
= ẋi + ẏj + żk
The speed |v| =
p
ẋ2 + ẏ 2 + ż 2 .
Note: The lectures will start with solving some mechanics problems in vector notation. The
problems can be solved without vector notation, by resolving force and accelerations vertically,
horizontally, parallel to a plane, etc. The idea is to achieve some familiarity with vector mechanics before moving on to more difficult problems where the motion is on a curved plane, wire
or surface, and where the basis vectors we work with change with position.
Example 1: Simple harmonic motion
A particle mass m moves in one dimension under a linear restoring force. Thus r = xi and
v = ẋi, a = ẍi. The force F = mkx(−i) and so ma = F (see later) gives mẍi = mkx(−i) i.e.
ẍ = −kx.
2
√
This has general solution x = A cos( kt + δ) where A, δ are constants. Hence
√
r = A cos( kt + δ)i
√
√
v = −A k sin( kt + δ)i
√
a = −Ak cos( kt + δ)i.
Note that the speed is NOT ẋ but |ẋ|!
Example 2: Ball sliding down an inclined plane
Figure 3: Ball rolling on inclined plane in standard coordinates (left)
and in coordinates that use distance d travelled down the plane (right).
First using i, j (left figure 3) we have r = xi + yj for the position vector of the particle. The
velocity of the particle is thus v = ẋi + ẏj.
On the other hand, in terms of the basis vectors e1 , e2 (see right figure 3) we have
r = h + de1 ,
where h is the vector stretching from O to the top of the plane. Thus the velocity is
d
d
˙ 1.
r = (h + de1 ) = de
dt
dt
It is easy to show that the two velocities are equal: We note that e1 , e2 are just i, j rotated by
angle α, so that
e1 = cos αi − sin αj, e2 = sin αi + cos αj.
Moreover, cos α = x/d so that d˙ cos α = ẋ. Hence
˙ 1=
de
But tan α =
ẋ
ẋ
e1 =
(cos αi − sin αj) = ẋi − ẋ tan αj
cos α
cos α
h−y
which gives x tan α = h − y and thus ẋ tan α = −ẏ, so that (1) becomes
x
˙ 1 = ẋi − ẋ tan αj = ẋi + ẏj,
de
which confirms that the velocity is the same whether we work with i, j or e1 , e2 .
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(1)
2
Newton’s Laws
We assume, unless stated otherwise, that our frame of reference is at rest.
MATH1301, these laws are
To recap on
Newton’s Laws of motion
First Law Every particle persists in its state of rest or uniform
motion in a straight line unless it is compelled to change that
state by impressed forces;
Second Law The rate of change of motion is proportional to the
motive force impressed; and it is made in the direction of the
straight line through which the force is impressed;
Third Law To every action there is an equal and opposite reaction: the mutual interactions of two bodies on each other is equal
but oppositely directed.
The second law, which is just a quantitative statement of the first law, is usually written in
equation form as
force = mass × acceleration,
or if F is the force acting on the particle of mass m,
F = ma.
Example 3: Particle moving vertically under gravity
Figure 4: Ball moving vertically under gravity
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Referring to figure 4, a ball is thrown upwards with initial speed U . Let y(t) be the height of
the ball at time t. Then Newton’s 2nd law gives
F = ma ⇒ mg(−j) = m × ÿj.
Hence we have ÿ = −g. Integrating once ẏ = −gt + C, C constant. But at t = 0, ẏ = U , so
that U = −g0 + C, i.e. C = U which gives ẏ = U − gt. Let’s check this: As t grows from zero,
ẏ reduces from U to zero at t = U/g, i.e. at the highest point the ball reaches, after which ẏ
becomes negative, that is y starts to decrease. This all agrees with what we expect. To find the
height integrate again y = U t − 21 gt2 + C 0 , where C 0 is a constant. Now use that y = 0 at t = 0
to obtain C 0 = 0 and hence y = U t − 12 gt2 .
Example 4: Ball sliding down an inclined plane
Figure 5: Ball sliding down an inclined plane
Refer to figure 5. We have
total force on ball = R + mg(−j).
The acceleration a = ẍi + ÿj. Hence Newton’s 2nd law gives
R + mg(−j) = m(ẍi + ÿj).
Now write R = R1 i + R2 j so that R1 = R · i = R sin α and R2 = R · j = R cos α
R(sin αi + cos αj) + mg(−j) = m(ẍi + ÿj).
Now equation coefficients (of the linearly independent) vectors i, j we obtain the two scalar
equations
mẍ = R sin α
mÿ = R cos α − mg
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Now eliminate the R:
m sin αÿ = R sin α cos α − mg sin α = m cos αẍ − mg sin α.
This gives ÿ = ẍ cot α − g. Integrate once:
ẏ = ẋ cot α − gt + C.
The constant C is found by using that ẋ = 0 and ẏ = 0 at t = 0, i.e. C = 0, which gives
ẏ = ẋ cot α − gt. Now integrate a second time: y = x cot α − 21 gt2 + C 0 . Now use that at t = 0,
x = 0 and y = h to obtain C 0 = h and hence
1
y = h + x cot α − gt2 .
2
Finally by the geometry, tan α =
h−y
x
and so eliminating x we have (after some algebra)
1
y = h − g sin2 αt2 ,
2
and
1
x = g sin α cos αt2 .
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Example 5: Two slabs on a smooth hill connected by a light inextensible string
Figure 6: Two heavy slabs on a smooth hill connected by a light inextensible string.
π
See figure 6 where α + β =
so that the top angle is a rightangle (e.g. α = β = π/4). Two
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heavy slabs are connected by a light inextensible string. Let d be the distance that the slab
of mass m has moved up the hill. We do not need to worry about the normal reaction at the
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hill top as the only forces that influence the block subsystems are the tension (which must be
the same in magnitude in each part of the string as there is no friction), the weights and the
reaction forces of the slabs on the hill. We have T1 = T e2 and T2 = T (−e1 ). Also R1 = R1 (−e1 )
and R2 = R2 e2 . This gives
¨ 1 = −T e1 + R2 e2 + 2mg sin βe1 − 2mg cos βe2
2mde
¨ 2 = T e2 − mg sin αe2 + mg cos αe1 − R1 e1 .
mde
Equating coefficients of e1 , e2 and using that sin α = cos β and cos α = sin β gives
2mg cos α − T
T − mg sin α
R2
R1
=
=
=
=
2md¨
md¨
2mg sin α
mg cos α.
Adding the first two equations: 3md¨ = 2mg cos α + mg sin α so that since the slabs start at rest,
d(t) =
gt2
(2 cos α + sin α).
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Example 6: Particle projected in vertical moving under gravity
Figure 7: Particle projected in vertical plane under gravity
See figure 7. Take i, j as unit vectors in the x, y directions. Then the position of the particle
(say mass m) at time t is r(t) = x(t)i + y(t)j. Let v = dtd r be its velocity. Then Newton’s law
gives
F = ma = mv̇,
where F = mg(−j) is the total force acting on the particle. Thus
v̇ = −gj,
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(2)
This is a vector differential equation, and may be treated just like a scalar differential equation,
except we must add on an arbitrary constant vector. Thus integrating (2) we have
Z
Z
v̇dt = −gj dt
so v = C − gtj. For C use t = 0, v = U if the particle has initial velocity U : get C = U . Hence
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 0 .
2
If initially r = r0 then C 0 = r0 and so
1
r = r0 + U t − gt2 j.
2
If particle is projected from the origin (r0 = 0) with speed U at an angle α to the horizontal then
U = U cos αi + U sin αj and hence r = U cos αti + U sin αtj − (gt2 /2)j. Equating components
of i, j gives
x = U cos αt
1
y = U sin αt − gt2 .
2
So 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 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 =
gx2
x
x
x
2
(α 6= π/2) and hence y = U sin αt−gt2 /2 = U sin α U cos
− g2 ( U cos
)2 = x tan α− 2U
2 sec α.
U cos α
α
α
So that
gx2
π
(1 + tan2 α) (α 6= ).
y = x tan α −
(3)
2
2U
2
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y
α
x
Figure 8: Parabola of safety (the dotted line in figure)
Parabola/paraboloid of safety
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 airplane wishes to fly over the gun, what is the shape of the area of
space that it must avoid? If the position of the airplane is X, Y (X 6= 0), 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 from the previous paragraph we want (X, Y ) such that the equation (3) at X, Y :
Y = X tan α −
gX 2
(1 + tan2 α)
2U 2
has no real solutions in α, or equivalently tan α. Write as a quadratic in tan α: we get
gX 2
gX 2
2
tan α − X tan α + Y +
= 0.
2U 2
2U 2
This has no real solutions in tan α if
2
X <4
2
gX 2
2U 2
gX 2
Y +
.
2U 2
2
Hence plane is safe if Y > U2g − gX
when X 6= 0. By continuity it is also safe when X = 0.
2U 2
Thus the airplane is safe if it lies above the parabola
Y =
U 2 gX 2
−
.
2g
2U 2
By symmetry p
we can think of an air plane flying over a gun placed at the origin of the (x, y)plane. If r = x2 + y 2 and z is the height of the airplane, the airplane is safe if it flies above
the paraboloid
U 2 g(x2 + y 2 )
z=
−
.
2g
2U 2
Particle projected in vertical moving under gravity and air resistance
Suppose we are told that the air resistance on the particle is kv per unit mass. Now the total
force on the particle is F = mg(−j) + mk(−v). The minus before the v is because the frictional
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force acts in the opposite direction to the velocity. Thus from Newton’s 2nd law we obtain
v̇ = −gj − kv.
We may
rewrite this as v̇ + kv = −gj. If v were a scalar v we would use an integrating factor
R
exp( kdt) = ekt . On reflection one sees that the same integrating factor works for the vector
version:
d
(vekt ) = ekt v̇ + kvekt = ekt (v̇ + kv) = −gjekt .
dt
Integrating we obtain
g
vekt = − ekt j + C.
k
Now use v(0) = U to obtain
g
C=U+ j
k
and tidy up to obtain
g
g
v(t) =
j + U e−kt − j.
k
k
g
Note that as t → ∞ we have v(t) → − k j. Thus the speed |v| of the particle tends to the terminal
velocity g/k. Particle eventually moves vertically at terminal velocity (ẋ(t) → 0, ẏ(t) → −g/k).
[check: y is measured upwards, so it is decreasing, and hence we expect the minus sign in −g/k
since then dy/dt < 0].
To find r(t) we integrate
g
dr
g
v=
= v(t) =
j + U e−kt − j.
dt
k
k
Thus
Z g
−1 g
g
g
r=
j + U e−kt − j dt =
j + U e−kt − tj + C 0 .
k
k
k k
k
0
1 g
Hence since r(0) = 0 we get C = k k j + U and hence
g
U
g
j+
(1 − e−kt ) − tj.
r(t) =
2
k
k
k
using that u = U cos αi + U sin αj we obtain
U cos α
(1 − e−kt )
k
g
U sin α
gt
y(t) =
+
(1 − e−kt ) − .
2
k
k
k
x(t) =
To find the particle path (y as a function of x, assuming α 6= π2 ), we need to eliminate t. It is
α
easiest to find t as a function of x: since x(t) = U cos
(1 − e−kt ) we find
k
1 − e−kt =
and
1
t = log
k
kx
U cos α
U cos α
U cos α − kx
10
.
Figure 9: Particle projected in vertical plane under gravity and under
air resistance proportional to speed
Now substiutute for t in the equation for y:
U sin α
U sin α
g
g
kx
g1
U cos α
g
−kt
+
+
(1 − e ) − t =
−
log
.
y=
k2
k
k
k2
k
U cos α
kk
U cos α − kx
Tidying up gives finally the cartesian equation for the particle path:
g
g
kx
π
y=
+ tan α x + 2 log 1 −
, (α 6= ).
kU cos α
k
U cos α
2
[Note that since x(t) =
U cos α
(1
k
− e−kt ), 1 −
kx
U cos α
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= e−kt > 0 so the log makes sense.]