2-D Motion - UCF Physics

Chapter 3 – 2D and 3D Motion
I. Definitions
II.
Projectile motion
III. Uniform circular motion
IV. Relative motion

I. Definitions
Position vector: extends from the origin of a coordinate system to the particle.
r = xiˆ
+ yj ˆ +
zkˆ
(4.1)
Displacement vector: represents a particle’s position change during a certain
time interval.
∆r
=
r ( )ˆ ( ) ˆ ( ) ˆ
2 − r1
= x2
− x1
i + y2
− y1
j + z2
− z1
k
(4.2)
Average velocity:
v avg
=
∆r
∆t
=
∆x
iˆ
+
∆t
∆y
∆t
ˆj
+
∆z
∆t
kˆ
(4.3)

Instantaneous velocity:
v = v
iˆ
+ v
ˆj
+ v
kˆ
dr
dt
dx
iˆ
+
dt
dy
dt
x y z = = +
ˆj
dz
dt
kˆ
(4.4)
-The direction of the instantaneous velocity of a
particle is always tangent to the particle’s path at
the particle’s position
Average acceleration:
a avg
=
v2
− v
∆t
1
=
∆v
∆t
(4.5)
Instantaneous acceleration:
a =
a
iˆ
+ a
ˆj
+ a
kˆ
dv
dt
dv
dt
dv
iˆ
+
dt
x y
x y z = =
+
ˆj
dv
dt
z
kˆ
(4.6)

II. Projectile motion
Motion of a particle launched with initial velocity, v 0 and free fall acceleration
g.
The horizontal and vertical motions are independent from each other.
- Horizontal motion: a x =0 v x =v 0x = constant
x − x0 = v0xt
= ( v0
cosθ0)
t
(4.7)
Range (R): horizontal distance traveled by a
projectile before returning to launch height.
- Vertical motion: a y = -g = constant
2
1
y − y0 = v0
yt
− gt = ( v0
sinθ0)
t − gt
2
2
1 2
(4.8)
v y
= v 0 sinθ 0 − gt (4.9)
v y = ( v0
sinθ0)
− 2g(
y − y0)
(4.10)
2
2

- Trajectory: projectile’s path.
x
0 = y0
=
0
( 4 .7 ) + ( 4 .8 ) →
y = (tan
θ
0
) x −
t =
v
2 ( v
0
0
gx
cos
x
cos
2
θ
θ
0
)
0
2
→
y
= v
0
sin
( 4 .11 )
θ
0
v
0
x
cos
θ
0
−
1
2
g
⎛
⎜
⎝
v
0
x
cos
θ
0
⎞
⎟
⎠
2
→
- Horizontal range: R = x-x 0 ; y-y 0 =0.
R = ( v
0 = ( v
0
0
cosθ
) t
sinθ
) t −
0
0
→ t =
1
2
gt
2
v
0
R
cosθ
= ( v
0
0
sinθ0)
v
0
R
cosθ
0
−
1
2
⎛
g
⎜
⎝ v
0
R
cosθ
0
⎞
⎟
⎠
2
=
R tanθ
0
−
1
2
g
v
2
0
R
2
cos
2
θ
0
→
R
=
2sinθ0
cosθ0
g
v
2
0
=
2
0
v
g
sin 2θ
0
(4.12)
(Maximum for a launch angle of 45º )
Overall assumption: the air through which the projectile moves has no effect
on its motion friction neglected.

122: A third baseman wishes to throw to first base, 127 feet distant. His best throwing speed is 85 mi/h. (a) If
he throws the ball horizontally 3 ft above the ground, how far from first base will it hit the ground? (b) From
the same initial height, at what upward angle must the third baseman throw the ball if the first baseman is to
catch it 3 ft above the ground? (c) What will be the time of flight in that case?
y
h=3ft
v 0
B3
B1
x
0
x max x B1
=38.7m
⎛ mi ⎞ ⎛ 1h
⎞ ⎛1609m
⎞
⎜85
⎟⋅⎜
⎟⋅⎜
⎟ = 38m
/ s
⎝ h ⎠ ⎝ 3600s
⎠ ⎝ 1mi
⎠
⎛ 0.305m
⎞
⎝ 1foot
⎠
( 3 feet) ⋅⎜
⎟ = 0.91m
h=3ft
Horizontal movement
Vertical movement
xmax
− x0
= v0xt
1 2
y − y0
= v0
yt
− gt
xmax
− 0 = 38t
= (38m
/ s)(0.43s)
= 16.4m
from B3
2
0 − 0.91m
= −4.9t
y
B3
v 0
θ
2
→ t = 0.43s
38.7m
B1
x
The ball will hit ground at 22.3 m from B1
1 2
y − y0
= 0 = v0
yt
− gt → v0
y = 4.9t
= v0
sinθ
→ t
2
38.7m
38.7
v0x
= = v0
cosθ
→ t = = 1s
t
38cosθ
38.7 38sinθ
= →189.63
= 1444sinθ
cosθ
→
38cosθ
4.9
0.13 = 0.5sin 2θ
→θ
=
7.6
=
38sinθ
4.9

N7: In Galileo’s Two New Sciences, the author states that “for elevations (angles of projection) which exceed
or fall short of 45º by equal amounts, the ranges are equal…” Prove this statement.
y
v 0
θ=45º
x=R=R’?
sin( a + b)
= sin a cosb
+ cos asin
b
sin( a − b)
= sin a cosb
− cos asin
b
x
2
0
v
R =
g
2
v0
R'
=
g
θ = 45
θ = 45
1
θ = 45
2
+ δθ
−δθ
2
0
v
R =
g
2
v0
R'
=
g
sin
sin
[ sin 90 cos(2δθ
) + cos90 sin(2δθ
)]
2
0
=
v
Range: R = sin 2θ
0 → dmax
at h
g
2
v
[ 2( 45 + δθ )] = sin( 90 2δθ
)
0
+
g
2
v
[ 2( 45 −δθ
)] = sin( 90 2δθ
)
0
−
2
v
[ sin 90 cos(2δθ
) − cos90 sin(2δθ
)] =
0 cos(2δθ
)
g
2
0
v
= cos(2δθ
)
g
g
0

III. Uniform circular motion
Motion around a circle at constant speed.
Magnitude of velocity and acceleration constant.
Direction varies continuously.
-Velocity: tangent to circle in the direction of motion.
- Acceleration: centripetal
a =
2
v
r
(4.13)
- Period of revolution:
T
=
2π
r
v
(4.14)
v x
v y
ˆ ˆ ( sinθ
)ˆ ( cosθ
) ˆ
⎛ − v ⋅ y
v = vxi
+ vy
j = −v
i + v j =
⎜
⎝ r
dv ⎛ − v dy p ⎞
ˆ
⎛ v dx p ⎞
ˆ ⎛ − v
a = =
⎜
⎟i
+
⎜
⎟ j = ⎜ vy
dt ⎝ r dt ⎠ ⎝ r dt ⎠ ⎝ r
2
2 2 v 2 2 v
a = ax
+ ay
= cos θ + sin θ =
r
r
ay
a directed along radius → tanφ
= =
a
x
2
p
⎞ˆ
⎛ v
⎟i
+ ⎜
⎠ ⎝ r
⎞
ˆ
⎛ v ⋅ x
⎟i
+
⎜
⎠ ⎝ r
v
x
sinθ
= tanθ
cosθ
p
⎞
ˆ
⎟ j
⎠
⎞ ⎛
ˆ
− v
⎟ j = ⎜
⎠
⎝ r
2
⎞ ⎛
cosθ
ˆ
− v
⎟ + ⎜
i
⎠ ⎝ r
2
⎞
sinθ
⎟ ˆ
j
⎠

54. A cat rides a merry-go-round while turning with uniform circular motion. At time t 1
= 2s, the cat’s velocity
is: v 1
= (3m/s)i+(4m/s)j, measured on an horizontal xy coordinate system. At time t 2
=5s its velocity is:
v 2
= (-3m/s)i+(-4m/s)j. What are (a) the magnitude of the cat’s centripetal acceleration and (b) the cat’s
average acceleration during the time interval t 2
-t 1
?
v 1
v 2
x
In 3s the velocity is reversed the cat reaches the opposite
side of the circle
y
v =
3
2
+ 4
2
= 5m
/ s
2πr
πr
T = → 3s
= → r = 4.77m
v 5m
/ s
2 2 2
v 25m
/ s
2
a c = = = 5.23m
/ s
r 4.77m
a
a
avg
avg
v2
− v1
=
∆t
= 3.33m
/ s
( −6m
/ s)ˆ
i − (8m
/ s)
ˆj
=
3s
2
= ( −2m
/ s
2
)ˆ i − (2.67m
/ s
2
) ˆj

IV. Relative motion
Particle’s velocity depends on reference frame
v = v + v
PA
PB
BA
(4.15)
1D
Frame moves at constant velocity
d
dt
d d 0
( vPA)
= ( vPB
) + ( vBA)
→ aPA
= aPB
dt dt
(4.16)
Observers on different frames of reference measure the same acceleration
for a moving particle if their relative velocity is constant.

75. A sled moves in the negative x direction at speed v s
while a ball of ice is shot from the sled with a velocity
v 0
= v 0x
i+ v 0y
j relative to the sled. When the ball lands, its horizontal displacement ∆x bg
relative to the ground
(from its launch position to its landing position) is measured. The figure gives ∆x bg
as a function of v s
.
Assume it lands at approximately its launch height. What are the values of (a) v 0x
and (b) v 0y
?
The ball’s displacement ∆x bs
relative to the sled can also be measured. Assume that the sled’s velocity is not
changed when the ball is shot. What is ∆x bs
when v s
is (c) 5m/s and (d) 15m/s?
v
v
v
s
0g
= −v
iˆ
0, rel
s
= v
= ( v
0x
0x
iˆ
+ v
ˆj
− v )ˆ i + v
s
0 y
0 y
ˆj
Launch velocity relative
to ground
∆x
x
y
2v0
−
g
bg
land
land
y
Displacements relative to ground
− x
− y
= 0 → v
launch
launch
= ∆x
80m
= − → v
20m
/ s
s
bg
= 0 = v
0 y
= ( v
0 y
2v0
= 10m
/ s → 0 =
g
t
0x
flight
= 19.6m
/ s
x
− v ) t
s
flight
− 0.5gt
2
flight
2⋅(10m
/ s)
− → v
g
0x
= 10m
/ s
0 = v
∆x
bg
0 y
t
= ( v
∆x bg
= a+ b v s
flight
0x
− 0.5gt
2v
− vs
)
g
2
flight
0 y
→ t =
a
2v0xv
=
g
2v
0 y
0 y
g
b
2v0
y
− v
g
v
t
2 0 y
2
flight = = 4s
v 10 19.6
2
0 = + = 22m
/ s
g
s
Displacements relative to the sled
∆xbs
= v0 xt
flight = (10m
/ s)
⋅4s
= 40m
Relative to the sled, the displacement does not depend
on the sled’s speed Answer (c)= Answer (d)

(iii) A dog wishes to cross a river to a point directly opposite as shown. It can swim at
2m/s in still water and the river is flowing at 1m/s. At what angle with respect to the
line joining the starting and finishing points should it start swimming?
1m/s
y
v f
θ
start
finish
v
r
v 0
x
v
v
f
v r
= (−1m
/ s)
iˆ
= 2 sin θ iˆ
+ 2 cos
0 θ
= v
r
+ v
0
2 sin θ −1
= 0 → θ = 30
ˆj
= ( −iˆ
+ 2sin θ iˆ
+ 2 cos θ ˆ) j m / s
= ( v
f
ˆ) j m / s
2 cos θ = v
f
→ v
f
=
3m
/ s

(ii) A particle moves with constant speed around the circle below. When it is at
point A its coordinates are x=0, y=3m and its velocity is (5m/s)i. What are its
velocity and acceleration at point B? Express your answer in terms of unit
vectors.
a B
v
=
r
2
2 2
ˆ 2
i
25m
/ s
=
3m
iˆ
= (8.3m
/ s
) iˆ
v B
= (5m
/ s)
ˆj
B
a B
y
A
= (8.3m
/ s
2
) iˆ
v A
= (5m
/ s)
iˆ
x

120. A hang glider is 7.5 m above ground level with a velocity of 8m/s at an angle of 30º below the
horizontal and a constant acceleration of 1m/s 2 , up. (a) Assume t=0 at the instant just described and write
an equation for the elevation y of the hang glider as a function of t, with y=0 at ground level. (b) Use the
equation to determine the value of t when y=0. (c) Explain why there are two solutions to part B. Which one
represents the time it takes the hang glider to reach ground level? (d) how far does the hang glider travel
horizontally during the interval between t=0 and the time it reaches the ground? For the same initial position
and velocity, what constant acceleration will cause the hang glider to reach ground level with zero velocity?
Express your answer in terms of unit vectors.
1
y − y y +
2
2
0 = v0
t + at → y = 7.5 − 4t
0. 5
2
0 1 2
y = → 0 = 7.5 − 4t
+ 0.5t
→ t = 5s,
t = 3s
t
2
y
h=7.5m
30º
v 0 = 8m/s
a = 1ˆj
v = 8cos30
o
iˆ
−8sin 30
ˆj
y(m)
7.5
0
x
0
1 2 3 4 5
t(s)
If the ground was not solid, the glider would swoop down,
passing through the surface, then back up again, with the two
times of passing being t=3s, t=5s.
d = v xt
= (6.93m
/ s)
⋅(3s)
20. 78m
max 0
=
Vertical movement :
y = 0 → v = v iˆ
+ v ˆj
= 0
f
fx
fy
Horizontal
movement :
v
v
2
y
y
= v
= v
2
0 y
0 y
+ 2a
⋅(
y − y
−15a
→ a
+ a t = 0 → 0 = −4
+ 1.1t
→ t = 3.75s
y
y
0
) = 4
2
y
y
= 1.1m
/ s
2
0 = vx
= v0
a = ( −1.85m
/ s
x
2
+ a
x
⋅t
= 6.93 + 3.75a
)ˆ i + (1.1m
/ s
2
) ˆj
x
→ a
x
= −1.85m
/ s
2

40. A ball is to be shot from level ground with certain speed. The figure below shows the range R it will
have versus the launch angle θ 0
at which it can be launched. The choice of θ 0
determines the flight time; let
t max
represent the maximum flight time. What is the least speed the ball will have during its flight if θ 0
is
chosen such as that the flight time is 0.5t max
?
R(m)
200
y − y
→ t =
t flight
0
=
= 0 = v
2v
0
0 y
sinθ0
g
t − 0.5gt
2
= v
→ sinθ
= 1 → t
0
0
sinθ
t − 4.9t
max
v 0 sinθ0
2v0
sinθ0
v0
−
= 0.5tmax
→ = →θ0
= sin 0.5
g
g g
=
2 1
0
2v
=
g
0
2
30
100
θ 0
The lowest speed occurs at the top of the trajectory (half of total time of
flight), when the velocity has simply an x-component.
v
min = vx
at half trajectory = v0x
= v0
cos30
for 0. 5tmax
Graph
R
max
= 240m
for θ = 45
0
v0
R =
g
v
min
2
v0
sin 2θ
0 → 240m
=
g
sin 90
= vx = (48.5m
/ s)cos30
= 42m
/
2
2
v0
=
g
s
→ v
0
= 48.5m
/ s