Post on 25-Feb-2022
transcript
CHAPTER 3
MOTION IN TWO AND THREE DIMENSIONS
• General properties of vectors • displacement vector • position and velocity vectors • acceleration vector • equations of motion in 2- and 3-dimensions
• Projectile motion • path of a projectile • time of flight • range
Examples of motion in 2- and 3-dimensions:
• any non-linear motion (circular motion, orbits of
planets, etc.)
• flight of projectiles (artillery shells, golf balls,
footballs, etc.)
Motion in 2- and 3-dimensions requires the use of
vectors. Basically, there are two main types of physical
quantities we deal with:
Scalars have size (magnitude) only and they can be
added, subtracted, multiplied and divided according to
the rules of simple math, i.e., Examples
include time, ordinary numbers, mass, distance (not
displacement) and speed (not velocity).
+, −, ×, ÷ .
Examples of vector quantities:
• Displacement. (The result of going is not
the same as going ).
• Force. (The result of pushing an object up is not the
same as pushing it down.)
• Velocity.
• Acceleration.
You cannot use the rules of simple math with vectors. As
an example take the displacement vector …
E → W
W → E
Military Trail
Glades Road
Spanish River Blvd
Palmetto Park Road
US 1 A1A
A B
C D
E
F
BBQ at Spanish River Park! Follow the map.
Military Trail
Glades Road
Spanish River Blvd
Palmetto Park Road
US 1 A1A
A B
C D
E
F
You have several choices:
A B
C
6.0 km
4.2 km
Displacement (resultant)
7.3 km
Note that:
but . . . . vector equation.
The displacement vector is 7.3 km,
AB+ BC = 10.2 km ≠ AC
35!
35! N of E.
AB! "!!
+ BC! "!!
= AC! "!!
Military Trail
Glades Road
Spanish River Blvd
Palmetto Park Road
US 1 A1A
A B
C D
E
F
55!
A
D C
A E
F C
In all three cases the displacement vectors are the same
and so are the distances.
55!( E of N) AD! "!!
+ DC! "!!
= AC! "!!
AE! "!!
+ EF! "!
+ FC! "!
= AC! "!!
All parallel vectors with the same lengths are equivalent.
These two vectors are not equivalent.
ADDITION
SUBTRACTION
!A +!B =!C
!A + (−
!B) =
!D
i.e., !A −!B =!D
Multiplication by a scalar.
A vector multiplied by a scalar n is another vector
parallel to but with magnitude (size) n times as large:
i.e.,
and
!C = n
!A
!A
!C
!A
!C = n
!A .
When working with vectors we will either
• draw them to scale, or
• use trigonometry and components:
!A
Ax
Ay
θ
y
x
In component form: and
The components are: and
where
!A = (Ax ,Ay)
!A = Ax
2 + Ay2 .
Ax =!A cosθ
Ay =
!A sinθ,
tanθ =
Ay
Ax.
Ax Bx
By
Ay
!A
!B
!C
y
xθ
Adding two vectors in component form:
In component form: and
The resultant is
where
Also and
and
!A = (Ax ,Ay)
!B = (Bx ,By).
!C = (Cx ,Cy),
!C = (Ax + Bx )2 + (Ay + By)2 .
Cx = (Ax + Bx ) Cy = (Ay + By),
tanθ =
Cy
Cx =
Ay + By
Ax + Bx
⎛
⎝⎜
⎞
⎠⎟ .
!A +!B =!C
Question 3.1: A yacht sails from a port (A) for 7.07 km in
a north-easterly direction to point B. Then, it sails in a
south-easterly direction for a further 4.24 km to point C.
(a) What is the displacement of the ship relative to the
port (A) when it reaches point C?
(b) If the yacht travels at 10.0 km/h, what is its average
velocity for the journey from A to C?
!A
!B
!C
Ax Bx
By
Ay
θ
(a) We have to determine The components of
and are:
But
and N of E.
!C =!A +!B.
Ax =!A cos45" = (7.07 km)(0.707) = 5.0 km.
Ay =
!A sin 45" = (7.07 km)(0.707) = 5.0 km.
Bx =!B cos(−45") = (4.24 km)(0.707) = 3.0 km.
By =
!B sin(−45") = (4.24 km)( − 0.707) = −3.0 km.
!B
!A
!C = (Cx ,Cy) = (Ax + Bx ,Ay + By)
= (8.0 km,2.0 km).
∴!C = (8.0 km)2 + (2.0 km)2 = 8.24 km
θ = tan−1 Cy
Cx
⎛
⎝⎜
⎞
⎠⎟ = tan−1 2.0 km
8.0 km⎛⎝⎜
⎞⎠⎟= 14.0!
(b) Total distance traveled by yacht was
and the speed was So, the time taken was
The average velocity is defined as
Velocity is a vector (in the same direction as the
displacement) so to be precise the average velocity of the
yacht is in a direction N of E.
7.07 km + 4.24 km = 11.31 km,
10.0 km/h.
11.31 km10.0 km/h
= 1.13 h.
displacement
time= 8.24 km
1.13 h= 7.29 km/h.
7.29 km/h 14.0!
In vector addition
if and
then and
So:
!A = (Ax ,Ay)
!B = (Bx ,By)
!A = Axi + Ay j
!B = Bxi + By j.
!A ±!B = (Ax ± Bx )i + (Ay ± By) j+ (Az ± Bz )k.
Often, when we use vectors in component form, it is
more convenient to use “unit vectors”. Unit vectors
indicate direction only (and have magnitude ).
i k
j
i = j = k = 1.
unit vectors
= 1
Question 3.2: If and
what are:
(a) and
(b)
!L = 6i + 3j− k,
!M = 4i − 5j+ 8k
!N = i − 4 j− 4k,
!L + 2
!M −
!N,
!L + 2
!M −
!N ?
(a)
(b)
!L + 2
!M −
!N = (6i + 3j− k) + 2(4i − 5j+ 8k)
−(i − 4 j− 4k)
= 6i + 3j− k + 8i −10 j+16k − i + 4 j+ 4k
= 13i − 3j+19k.
!L + 2
!M −
!N = (13)2 + (−3)2 + (19)2 = 539
= 23.2.
Question 3.3: If and
what is the angle between !A = 3.8i +1.5 j
!B = 1.4i + 3.5 j,
!A and
!B?
From earlier, tanθA =
AyAx
⎛ ⎝
⎞ ⎠ , where θA is the angle
between ! A and the i direction (x-axis).
∴θA = tan−1 Ay
Ax⎛ ⎝
⎞ ⎠ = tan−1 1.5
3.8( ) = 21.5".
Similarly, from the i direction (x-axis),
θB = tan−1 By
Bx⎛ ⎝
⎞ ⎠ = tan−1 3.5
1.4( ) = 68.2".
So, the angle between ! A and
! B is
θB − θA = 68.2" − 21.5" = 46.7".
I will show you another way to solve this problem later.
! A
! B
21.5"
68.2"
i
j
0 1 2 3 4
1
2
3
4
Position and velocity vectors
The position vector of a point
with coordinates is (x, y,z)
!r = xi + yj+ zk.
Consider an object with position vector
that moves from to Then, if
and
The average velocity vector is
(parallel to ).
The instantaneous velocity vector at any point is
i.e.,
!r = xi + yj+ zk
P1(t1) P2(t2 ).
Δ!r = !r2 −
!r1 Δt = t2 − t1
!vav = Δ!r
Δt Δ!r
!r
!v = LimitΔt→0
Δ!rΔt
⎛⎝⎜
⎞⎠⎟ !r
= d!rdt
= dxdt
i + dydt
j+ dzdt
k,
!v = vx i + vy j+ vzk.
The velocity vector at the point is the tangent to the
position-time graph at and is parallel to the
instantaneous direction of motion.
The instantaneous speed is the magnitude of the
instantaneous velocity vector.
!r
!r
Acceleration vector
Consider an object with the velocity vector
that moves from to Then, if
and
the average acceleration vector is
(parallel to ).
The instantaneous acceleration vector is
i.e.,
v = vx i + vy j+ vzk
P1(v1, t1) P2(v2, t2 ).
Δ!v = !v2 −
!v1 Δt = t2 − t1
!aav = Δ!v
Δt Δ!v
!a = Limit
Δt→0
Δ!vΔt
⎛⎝⎜
⎞⎠⎟= d!v
dt=
dvxdt
i +dvy
dtj+
dvzdt
k,
!a = ax i + ay j+ azk.
Since ! a is parallel to Δ
! v , the direction of the acceleration is towards the “inside” of a curve. Note: on a curve or turn, there is always an acceleration even though the magnitude of the velocity (i.e., speed) may be constant.
! v
! a
! v
! a
Special case of circular motion
Shown here are the position and velocity vector diagrams of an object in uniform circular motion, i.e., motion at constant speed (v) and constant radius (r). The average acceleration from P1 to P2 is:
aav =
Δ! v Δt
=2v sin θ
2( )Δt
,
where Δt is the time to travel from P1 to P2. Now,
Δt =
arc length P1P2v
=rθv
.
∴aav =
2v2 sin θ2( )
rθ.
θ •
•
P1 t1( )
P2 t2( ) ! v 2
! v 1
! v 1 = ! v 2 = v
! r 1
! r 2
! r 1 = ! r 2 = r
−! v 1
! v 2
Δ! v
θ
Δ! v = 2v sin θ
2( ) Δ! v = ! v 2 − ! v 1
The instantaneous acceleration at the point P1 occurs when
Δt → 0, i.e., as θ → 0. Then the radial acceleration is
a r = Limitθ→0
2v2 sin θ2( )
rθ
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ =
2v2
rLimitθ→0
sin θ2( )
θ
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ .
Put φ = θ2, then
a r = Limitφ→0
sin φ2φ
⎡
⎣ ⎢
⎤
⎦ ⎥
= Limitφ→0
12sin φφ
⎡
⎣ ⎢
⎤
⎦ ⎥ =
12
.
∴ar =
2v2
r12
=v2
r.
Note that the direction of the acceleration is parallel to Δ! v ,
i.e., it is radial. It is called the centripetal (i.e., center-directed) acceleration.
We’ll return to this topic in chapter 5.
The instantaneous acceleration at the point P1 occurs when
Δt → 0, i.e., as θ → 0. Then the radial acceleration is
a r = Limitθ→0
2v2 sin θ2( )
rθ
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ =
2v2
rLimitθ→0
sin θ2( )
θ
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥ .
Put φ = θ2, then
a r = Limitφ→0
sin φ2φ
⎡
⎣ ⎢
⎤
⎦ ⎥
= Limitφ→0
12sin φφ
⎡
⎣ ⎢
⎤
⎦ ⎥ =
12
.
∴ar =
2v2
r12
=v2
r.
Note that the direction of the acceleration is parallel to Δ! v ,
i.e., it is radial. It is called the centripetal (i.e., center-directed) acceleration.
We’ll return to this topic in chapter 5.
Put , then
But
φ = θ
2 ar =
2v2
rLimitφ→0
sinφ2φ
⎡
⎣⎢
⎤
⎦⎥.
Limitφ→0sinφ2φ
⎡
⎣⎢
⎤
⎦⎥ = Limitφ→0
12
sinφφ
⎡
⎣⎢
⎤
⎦⎥
= 12
Limitφ→0sinφφ
⎡
⎣⎢
⎤
⎦⎥ =
12
.
Question 3.4: An object experiences a constant
acceleration of At time , it is at
rest at Find
(a) the velocity and position vectors at any later time
t, and
(b) the equation describing the path of the object in
the xy-plane.
!a = (6i + 4 j) m/s2. t = 0
!r" = (10,0) m.
When :
(a) From chapter 2:
Also:
(b) But
Hence, so
i.e.,
t = 0 !r" = (10,0) m and !v" = (0,0) m/s,
!a = (ax ,ay) = (6i + 4 j) m/s2.
v = v! + at ⇒ "v = "v! +"at.
∴!v = (6i + 4 j)t = (6ti + 4tj) m/s.
(!r − !r") =
!v"t +12!at2 ⇒ !r = !r" +
!v"t +12!at2.
∴!r = 10i + 1
2(6i + 4 j)t2 = (10 + 3t2)i + 2t2 j m.
!r = (x, y) = xi + yj.
∴x = (10 + 3t2) and y = 2t2.
t2 = y
2 x = 10 + 3
y2
⎛⎝⎜
⎞⎠⎟
,
y = 2
3x − 20
3.
Question 3.5: A ship departs from point A heading to a
port 52 km due north at point B. The speed of the boat
relative to the water is 10 km/h. If there is a steady
current flowing at 2 km/h in a north-westerly direction,
(a) what is the proper heading to make the trip from
A to B?
(b) How long does the journey take?
(c) HOME PROBLEM: If the captain of the ship had
failed to take into account of the wind, how far due west
of B would the ship had been?
(a) Draw the velocity vector diagram. The resultant path
is the result of the ship setting
out along but being re-directed
by the current along .
So,
Since the resultant heading is due
north, the resultant velocity
component in the E-W direction is zero, i.e.,
east of north.
(b) The resultant speed of the ship is
Since the journey time is
θ
45!
A
B
C N
D
!vAB = !vAC + !vCB.
∴(10 km/h)sinθ = (2 km/h)cos45!,
i.e., θ = sin−1 (2 km/h)cos45!
10 km/h
⎛
⎝⎜
⎞
⎠⎟ = 8.1!
(10 km/h)sinθ − (2 km/h)sin45! = 0
= (10 km/h)cos8.1! + (2 km/h)sin45! = 11.3 km/h.
AB = 52 km,
52 km
11.3 km/h= 4.60 h ⇒ 4 h 36 min.
AB! "!!
AC! "!!
CB! "!!
!vAB = !vAD + !vDB = !vAC cosθ + !vCB sin45"
Projectile motion
• thrown objects (baseball, arrows, shells, etc.)
• bouncing balls, etc.
• Investigated first by Galileo.
• Historically important for the military.
vy
vy
vx
vx !v
!v
!v
!v"
v!x
v!y
(x!, y!) x
y
θ!
Important concept: we can treat the x and y components
separately. The initial velocity components are:
If we ignore air resistance and other drag forces, the only
force acting on the projectile is the gravitational force, so
the acceleration components of the projectile are:
v!x = "v! cosθ! and v!y = "v! sinθ!.
ax = 0 and ay = −g.
Analysis of projectile motion Since!!!!!!!!!!!!!!!!!!!!
•
Since
•
The components of the displacements are:
• and
•
These latter two equations represent a parabola.
The four equations listed above are the equations of
motion for a projectile (in the absence of air resistance).
Although the equations for the horizontal and vertical
components are treated separately, they are connected by
time (t).
ax = 0,
vx (t) ⇒ constant = v!x .
ay = −g,
vy(t) = v!y − gt.
x(t) = x! + v!t
y(t) = y! + v!yt − 1
2gt2.
If a rock is dropped from the same height as a rifle at the
same time it is fired, no matter how powerful the rifle, in
the absence of air resistance and assuming the ground is
horizontal and level, the bullet and the rock will hit the
ground at the same time, since all objects fall the same
distance in the same time, i.e., a distance of
in t seconds. 12
gt2
At every point on the path: tanθ =
vyvx
.
At maximum height: vy = v!y − gt = 0.
∴v!y = v! sinθ! = gt .
So, the time to reach maximum height is
t =
v! sin θ!g
.
If the take-off and landing are in the same horizontal plane, the total time of flight is:
T = 2
v! sin θ!g
.
vxˆ i
vyˆ j
Range
θ
" v
If the take-off and landing are in the same horizontal plane the horizontal range is:
R = v!xT
where T is the total flight time.
∴R = (v! cosθ!) × 2
v! sin θ!g
=
v!22sin θ! cosθ!g
.
i.e., R =
v!2
gsin 2θ!.
Maximum range ⇒ when θ! = 45!.
But sin 2θ = 2sin θcosθ and sin θ = cos(90! −θ)
∴sin 2θ = 2cos(90! − θ)cosθ
i.e., R(90! − θ) = R(θ).
So ... R(30! ) = R(60!), etc.
75!
15!
60!
30! 45!
Question 3.6: The sketch above shows the trajectories of
two golf balls, A and B. If the balls reach the same
height, which one had the greater initial velocity? Ignore
air resistance.
The balls reach the same height, so the vertical components of the initial velocities are equal,
i.e., vA sin θA = vB sin θB,
where vA and vB are the initial velocities of A and B,
respectively. Clearly, the take-off angle for A θA( ) is greater than that for B θB( ), then
sin θA > sin θB, so vB > vA,
i.e., the initial velocity of B is greater than that of A.
•
•
A
B
Question 3.7: A cannonball is fired from the top of a
40 m high tower with an initial speed of 42.2 m/s at an
angle of above the horizontal.
(a) How long is the cannonball in the air before it
hits the ground?
(b) When it hits the ground, how far is it from the
base of the tower?
(c) What is its speed when it hits the ground?
30!
∴4.91t2 − 21.1t − 40 = 0,
t =
21.1± (−21.1)2 − 4 × 4.91× (−40)2 × 4.91
. i.e.,
The appropriate solution is
(b) Range
∴ t = +5.72 s and t = −1.42 s.
t = +5.72 s.
R = v!xt = (36.5 m/s)(5.72 s) = 209 m.
(c)
The final speed is
But
and
Hence,
Vx
Vy
!V
!v = vx
2 + v2y .
vx = v!x = 36.5 m/s,
v2
y = v!y2 + 2(−g)(y − y!)
= (21.1 m/s)2 + 2(−9.81 m/s2)(−40 m)
= (1230 m/s)2.
∴vy = 35.1 m/s.
v = (36.5 m/s)2 + (35.1 m/s)2 = 50.6 m/s.
DISCUSSION QUESTION:
Question 3.8: During the Roman invasion of Britain, a
group of Roman soldiers (at A) locate a British camp on
the top of a hill (at B). Using a trebuchet they want to lob
a rock from A onto the British camp, a horizontal distance
of 500 m away. If the maximum angle of elevation of the
trebuchet is and the height of the hill is 16 m,
(a) what initial speed of the rock is required in order
to hit the camp?
(b) What is the time of flight of the rock?
(c) With what speed does the rock strike the camp?
(d) What is the maximum height (h) achieved by the
rock?
60!,
v!x = v! cos60! = 0.50v!,
v!y = v! sin60! = 0.866v!,
(x!, y!) = (0,0); (x,y) = (500 m, 16 m).
(a) The horizontal range is i.e.,
Also,
i.e.,
(b) From above,
v!xt, t = 500 m
0.50v!= 1000 m
v!.
(y − y!) = v!yt − 12
gt2.
∴16 m = 0.866v!(1000 m)
v!− 1
2(9.81 m/s2) (1000 m)
v!
⎛
⎝⎜⎞
⎠⎟
2
= (866 m) − (4.91 m/s2)(106 m2)
v!2
.
∴v!2 = (4.91 m/s2)(106 m2)
850 m= 5776.5 (m/s)2,
v! = 76.0 m/s.
t = 1000 m
v!= 1000 m
76.0 m/s= 13.16 s.
(c) The rock strikes the camp with speed
But
(d) The maximum height is h, where
v = vx
2 + vy2 .
vx = v!x = 0.50v! = 38.0 m/s,and
vy2 = v!y
2 − 2g(y − y!)
= (0.866 × 76.0 m)2 − 2(9.81 m/s2)(16 m − 0)
= 4017.8 (m/s)2, i.e., vy = −63.4 m/s.
∴v = (38.0 m/s)2 + (−63.4 m/s)2
= 73.9 m/s.
vy = 0.
∴0 = v!y2 − 2g(h − y!), i.e., v!y
2 = 2gh.
∴h =v!y
2
2g = (0.866 × 76.0 m/s)2
2(9.81 m/s2)= 220.8 m.