Lecture 11 – Curvilinear
What you will learn for today?
1. Two statements & scope of curvilinear
2. (x, y) Analysis
3. (n, t) Analysis
4. (r, ) Analysis
5. combination (x, y) & (n, t)
By: Dr. Muhammad Hanif Ramlee
Credit to: Prof. Dato’ Ir. Dr. Mohammed Rafiq Abdul Kadir
Curvilinear Motion

2 Statements
1) Velocity MUST always be tangent to the curvature AND in the direction of motion.
2) Acceleration MUST always act inwards of the curvature.
Curvilinear Motion

Scope
◼
Coordinates / axes considerations
Curvilinear Motion

(x, y) Analysis
◼
◼
More popular special case
Projectiles (free flight)





Trajectory of projectile is assumed to be parabolic
Equations are derived based on constant acceleration analysis
s, v and a are resolved into x and y components
Axes
Analysis involves 2 points
Curvilinear Motion

Y-axis (+) analysis
Main equations:
v y = u y + a yt
v = u + 2a y (s y − s0 y )
2
y
2
y
1 2
s y − s0 y = u y t + a y t
2
Where uy = initial velocity component in y-direction (m/s)
vy = final velocity component in y-direction (m/s)
ay = 9.81 m/s2
s0y = initial displacement component in y-direction (m)
sy = final displacement component in y-direction (m)
t = time taken from initial to final (s)
Curvilinear Motion

X-axis (+→) analysis
Main equations:
vx = u x
sx − s0 x = uxt
Where ux = initial velocity component in x-direction (m/s)
s0x = initial displacement component in x-direction (m)
sx = final displacement component in x-direction (m)
t = time taken from initial to final (s)
x and y analyses are independent of each other (linked by t)
Curvilinear Motion

Characteristics of projectiles and implications of vx = ux
Curvilinear Motion

(+) observations
◼
◼
◼
◼
◼
As the projectiles moves up, the y component of velocity will
decrease in magnitude, i.e. vAY > vBY > vCY until it stops
momentarily before changing directions at D where vDY = 0
As the projectile starts its motion downwards, the y component
of velocity will increase in magnitude, i.e. vGY > vFY > vEY .
The y component of velocity is equal in magnitude (not
direction) when the projectile is at the same level of height, i.e.
vAY = vGY, vBY = vFY, vCY = vEY.
The y component of velocity is equal to zero, vDY = 0 at its
maximum height.
The only acceleration (unless stated otherwise) acting on
projectiles is a = g = 9.81m/s2 acting vertically downwards.
Curvilinear Motion

(+→) observations
◼
Since the acceleration (unless stated otherwise) acting on
projectiles is a = g = 9.81m/s2 acting vertically downwards, i.e.
ax = 0. This also means that the horizontal component (x) of the
velocity vx, is always assumed to be constant throughout the
flight.
Example 1
A projectile is launched from A with an initial velocity vA = 10 m/s as shown.
Determine:
a) The horizontal distance, d.
b) the maximum height, hmax.
Example 1
A projectile is launched from A with an initial velocity vA = 10 m/s as shown. Determine:
a) The horizontal distance, d.
b) the maximum height, hmax.
Solution
Starts by drawing the axes.
Identify 2 points of analysis, AC
Example 1
A projectile is launched from A with an initial velocity vA = 10 m/s as shown. Determine:
a) The horizontal distance, d.
b) the maximum height, hmax.
Solution
Y-axis analysis (+)
uy = vAy = 10 sin 60
vy = vCy
ay = -9.81m/s2
sy = sCy = 5-1.5 = 3.5m
s0y = sAy = 0
t = tAC
using
sy – s0y = uyt + ½ ayt2
3.5 - 0 = 10 sin 60tAC + ½ (-9.81)tAC2
3.5 = 8.66tAC – 4.905tAC2z
rearranging
4.905tAC2 – 8.66tAC + 3.5 = 0
tAC = 0.63 or 1.14s
 tAC = 1.14s
Example 1
A projectile is launched from A with an initial velocity vA = 10 m/s as shown. Determine:
a) The horizontal distance, d.
b) the maximum height, hmax.
Solution
X-axis analysis (+→)
ux = vAx = 10 cos 60
sx = sCx = d
s0x = sAx = 0
t = tAC = 1.14s
using sx – s0x = uxt
d - 0 = 10 cos 60 (1.14)
d = 5.7m
Example 1
A projectile is launched from A with an initial velocity vA = 10 m/s as shown. Determine:
a) The horizontal distance, d.
b) the maximum height, hmax.
Solution
Determine the maximum height, hmax.
Draw the axes.
Identify 2 points of analysis, AB
Example 1
A projectile is launched from A with an initial velocity vA = 10 m/s as shown. Determine:
a) The horizontal distance, d.
b) the maximum height, hmax.
Solution
Y-axis analysis (+)
uy = vAy = 10 sin 60
vy = vBy (max height)
ay = -9.81m/s2
sy = sBy = hmax -1.5
s0y = sAy = 0
t = tAB
using
vy2 = uy2 + 2ay(sy – s0y)
0 = (10 sin 60)2 + 2 (-9.81)(hmax – 1.5)
hmax = 5.32m
Example 1 – implications of t value

Implications of t values
◼
◼
Two +ve values mean that both times are on the same level, as
shown below. Choose the required t, depending on the location,
either B or C.
A +ve and a –ve value means that the required location is below
datum level of initial velocity.
Example 2
A projectile is released from A with velocity as shown. Determine the horizontal
distance, d.
Example 2
A projectile is released from A with velocity as shown. Determine the horizontal distance, d.
Solution
Starts by drawing the axes.
Identify 2 points of analysis, AB
Example 2
A projectile is released from A with velocity as shown. Determine the horizontal distance, d.
Solution
Y-axis analysis (+)
uy = vAy = -6 sin 15
vy = vBy
ay = -9.81m/s2
sy = sBy = -3m
s0y = sAy = 0
t = tAB
using
sy – s0y = uyt + ½ ayt2
-3 - 0 = -6 sin 15tAB + ½ (-9.81)tAB2
-3 = 1.553tAB – 4.905tAB2
rearranging
4.905tAB2 + 1.553tAB - 3 = 0
tAB = -0.954 or 0.639s
 tAC = 0.639s
Example 2
A projectile is released from A with velocity as shown. Determine the horizontal distance, d.
Solution
X-axis analysis (+→)
ux = vAx = 6 cos 15
sx = sBx = d
s0x = sAx = 0
t = tAB = 0.639s
using sx – s0x = uxt
d - 0 = 6 cos 15 (0.639)
d = 3.7m
Example 3
The trajectory of a launched projectile is as shown. Determine the initial velocity,
u.
Example 3
The trajectory of a launched projectile is as shown. Determine the initial velocity, u.
Solution
Starts by drawing the axes.
Identify 2 points of analysis, AB
Identify 2 points of analysis, AC
Example 3
The trajectory of a launched projectile is as shown. Determine the initial velocity, u.
Solution
Y-axis analysis (+) for AB
uy = uy
vy = 0.9 (max height)
ay = -9.81m/s2
sy = 15
s0y = 0
t = tAB
using
vy2 = uy2 + 2ay(sy – s0y)
0 = uy2 + 2 (-9.81)(15)
uy = 17.16m/s
Example 3
The trajectory of a launched projectile is as shown. Determine the initial velocity, u.
Solution
Y-axis analysis (+) for AC
uy = 17.16m/s
vy = vCy
ay = -9.81m/s2
sy = -45m
s0y = 0
t = tAC
using
sy – s0y = uyt + ½ ayt2
-45 - 0 = 17.16tAC + ½ (-9.81)tAC2
-45 = 17.16tAC – 4.905tAC2
rearranging
4.905tAC2 – 17.16tAC - 45 = 0
tAC = -1.75 or 5.25s
 tAC = 5.25s
Example 3
The trajectory of a launched projectile is as shown. Determine the initial velocity, u.
Solution
X-axis analysis (+) for AC
using
sx – s0x = uxt
60 = ux(5.25)
ux = 11.43m/s 
u = 2 17.162 + 11.432 = 20.62m / s
17.16
 = tan −1
= 56.3
11.43
Example 4
A surveillance ship detects a spy plane flying at an altitude of 6km with a constant
velocity of 900km/h. The ship launch a missile with an initial velocity of 540m/s
the instant the plane passes the vertical line AB. Determine the angle T and the
time required for the missile to hit the plane. Both the ship and plane lies on the
same plane.
Example 4
A surveillance ship detects a spy plane flying at an altitude of 6km with a constant velocity of 900km/h. The
ship launch a missile with an initial velocity of 540m/s the instant the plane passes the vertical line AB.
Determine the angle T and the time required for the missile to hit the plane. Both the ship and plane lies on the
same plane.
Solution
From observations:
➢ The time taken for the missile from A to C must be equal to the time taken for the
plane from B to C.
➢ The horizontal distance of the missile from A to C must be equal to the
horizontal distance for the plane from B to C.
Example 4
A surveillance ship detects a spy plane flying at an altitude of 6km with a constant velocity of 900km/h. The
ship launch a missile with an initial velocity of 540m/s the instant the plane passes the vertical line AB.
Determine the angle  and the time required for the missile to hit the plane. Both the ship and plane lies on
the same plane.
Solution
Horizontal components (+→)
The missile
sx = uxt
sx = 540 cos t
equating gives 540 cos  = 250
 = cos-1 (250/540) = 62.4
The plane
sx = uplanet
sx = (900/3.6)t
Example 4
A surveillance ship detects a spy plane flying at an altitude of 6km with a constant velocity of 900km/h. The
ship launch a missile with an initial velocity of 540m/s the instant the plane passes the vertical line AB.
Determine the angle  and the time required for the missile to hit the plane. Both the ship and plane lies on
the same plane.
Solution
Vertical components (+) for the missile
uy = 540 sin 62.4 = 478.5m/s
vy = vCy
ay = -9.81m/s2
sy = 6000m
s0y = 0
t = tAC
using
sy – s0y = uyt + ½ ayt2
6000 - 0 = 478.5tAC + ½ (-9.81)tAC2
6000 = 478.5tAC – 4.905tAC2
rearranging
4.905tAC2 – 478.5tAC + 6000 = 0
tAC = 14.78 or 82.78s
 tAC = 14.78s
(n, t) Analysis

Axes
◼


t-axis
: in the direction of velocity, v
: i.e. tangent to the curvature, in the direction of motion

n-axis
: perpendicular / 90 to the t-axis
: always directed towards the centre of rotation
Velocity Analysis
◼
◼

Unlike (x,y) axes, the (n,t) axes are located at the particle
Velocity is resolved into two components, vt and vn.
Since v is always in the direction of t-axis and vice versa, vt = v and vn = 0
Acceleration Analysis
◼
◼
Acceleration is resolved into two components, an and at.
a = an + at
a vector summation
(n, t) Analysis
 an
◼
Magnitude
:
an = v2 / 

V
: velocity of profile


: radius of curvature
Direction
: always in the (+ve) n-axis direction
◼ an exist due to change in direction of v.
◼
◼
an always exist (except in rectilinear translation and as long as v ≠
0).
(n, t) Analysis
 at
◼
Magnitude
: at = v
: where v is the rate of change of speed
: analysis is similar to rectilinear motion
: acceleration: (+ve) t-axis direction
: deceleration: (-ve) t-axis direction
: constant v: at = 0
◼ at exist due to change in magnitude of v.
◼ at is the rate of change of speed.
◼
Direction
Example 5
Particle A is taking a bend with velocity and acceleration shown. Determine:
a. The radius of curvature, .
b. Rate of change of speed.
Example 5
Particle A is taking a bend with velocity and acceleration shown. Determine:
a. The radius of curvature, .
b. Rate of change of speed.
Solution
Axes
t-axis: same direction as v
n-axis: directing towards centre
of rotation AND/OR on the
side as a.
Example 5
Particle A is taking a bend with velocity and acceleration shown. Determine:
a. The radius of curvature, .
b. Rate of change of speed.
Solution
Acceleration analysis
an
at
Resolution
3 sin 60
3 cos 60
Equation
v2 / 
rate of change of speed
a.
b.
 = 152 / 3 sin 60 = 86.6m
Rate of change of speed, at = 3 cos 60 = 1.5m/s2, angle 40.
Example 6
A car travels along a straight road before taking a 20m radius curve at B as shown.
Determine the acceleration at B for the following situations:
a. vA = 20m/s, aA = 2m/s2 constant
b. vA = 20m/s, constant velocity
Example 6
A car travels along a straight road before taking a 20m radius curve at B as shown. Determine the acceleration
at B for the following situations:
a. vA = 20m/s,
aA = 2m/s2 constant
b. vA = 20m/s,
constant velocity
Solution
Part a.
Rectilinear Analysis
AB
u = vA = 20m/s
v = vB
a = 2m/s2
s = 50m
using
vB2 = vA2 + 2as
= 202 + 2(2)(50)
vB = 24.5m/s
Example 6
A car travels along a straight road before taking a 20m radius curve at B as shown. Determine the acceleration
at B for the following situations:
a. vA = 20m/s,
aA = 2m/s2 constant
b. vA = 20m/s,
constant velocity
Solution
Part a.
(n,t) analysis
1. Axes
Example 6
A car travels along a straight road before taking a 20m radius curve at B as shown. Determine the acceleration
at B for the following situations:
a. vA = 20m/s,
aA = 2m/s2 constant
b. vA = 20m/s,
constant velocity
Solution
Part a.
(n,t) analysis
2. Acceleration analysis
at = 2m/s2 →
an = vB2 /  = 24.52 / 20 = 30m/s2 
aB = 2 2 + 302 = 30.1ms −2
 = tan −1
30
= 86.2
2
Example 6
A car travels along a straight road before taking a 20m radius curve at B as shown. Determine the acceleration
at B for the following situations:
a. vA = 20m/s,
aA = 2m/s2 constant
b. vA = 20m/s,
constant velocity
Solution
Part b.
Rectilinear Analysis
AB
u = vA = 20m/s
v = vB = 20m/s (i.e. constant)
Example 6
A car travels along a straight road before taking a 20m radius curve at B as shown. Determine the acceleration
at B for the following situations:
a. vA = 20m/s,
aA = 2m/s2 constant
b. vA = 20m/s,
constant velocity
Solution
Part b.
(n,t) analysis
1. Axes
Example 6
A car travels along a straight road before taking a 20m radius curve at B as shown. Determine the acceleration
at B for the following situations:
a. vA = 20m/s,
aA = 2m/s2 constant
b. vA = 20m/s,
constant velocity
Solution
Part b.
(n,t) analysis
2. Acceleration analysis
at = 0m/s2
an = vB2 /  = 202 / 20 = 20m/s2 
aB = an = 20m/s2 
(r, ) Analysis

Axes
◼
◼
◼
◼
◼
Similar to (x, y), the (r, ) axes refer to a fixed point (0, 0).
Similar to (n, t), the (r, ) axes analyse 1 point at a time.
Unlike (x, y) axes, the (r, ) axes are located at the particle.
r-axis draw a line from (0, 0) to the point
(+ve) r-axis is the outwards extension of this line.
-axis
perpendicular / 90 to the r-axis
(+ve) -axis is directed towards increasing .
(r, ) Analysis

Velocity Analysis
◼ Velocity is resolved into two components, vr and v.
v = vr + v
a vector summation
•
• = rate of change of r (m/s)
where
vr = r
r
•
v = r 

•

= rate of change of  (rad/s)
Acceleration Analysis
◼ Acceleration is resolved into two components, ar and a.
a vector summation
•
••
2
where
r = rate of change of (m/s ) r
a = ar + a
••
2) •
••
•2
=
rate
of
change
of
(rad/s


ar = r − r 
••
• •
a = r  + 2 r 
Example 7
The movement of collar B is defined by the relationship r = 3t2 – t3 and  = 2t2
where r is in m,  in rad and t in seconds. Determine the magnitude of velocity
and acceleration when t = 2 seconds.
Example 7
The movement of collar B is defined by the relationship r = 3t2 – t3 and  = 2t2 where r is in m,  in rad and t
in seconds. Determine the magnitude of velocity and acceleration when t = 2 seconds.
Solution
r = 3t 2 − t 3
•
dr
r=
= 6t − 3t 2
dt
••
r=
•
•
t=2
dt
•
d
=
=4
dt
r = 6(2) − 3(2) 2 = 0m / s
••
r = 6 − 6(2) = −6m / s
dr
= 6 − 6t
dt
 = 2t 2
•
d
=
= 4t
••
r = 3(2) 2 − 23 = 4m / s
 = 2(2) 2 = 8rad
t=2
•
 = 4(2) = 8rad / s
••
 = 4 = 4rad / s 2
Example 7
The movement of collar B is defined by the relationship r = 3t2 – t3 and  = 2t2 where r is in m,  in rad and t
in seconds. Determine the magnitude of velocity and acceleration when t = 2 seconds.
Solution
VELOCITY
•
vr = r = 0 m / s
•
v = r  = 4(8) = 32m / s
 v = 32m / s
ACCELERATION
•2
••
ar = r − r  = −6 − 4(8) 2 = −262m / s 2
••
• •
a = r  + 2 r  = 4(4) + 2(0)(8) = 16m / s 2
 a = ar2 + a2 =
(− 262)2 + 162 = 262.5m / s 2
Example 8
Particle A is moving along a straight path as shown with a velocity v = 30m/s and decelerates at 5m/s2.
• • •• ••
Determine
observed from O.
r,  , r, 
Example 8
Particle A is moving along a straight path as shown with a velocity v = 30m/s and decelerates at 5m/s2.
• • •• ••
Determine
observed from O.
r,  , r, 
Solution
1. AXES
Example 8
Particle A is moving along a straight path as shown with a velocity v = 30m/s and decelerates at 5m/s2.
• • •• ••
Determine
observed from O.
r,  , r, 
Solution
2. VELOCITY ANALYSIS
Resolution
vr
v
•
Equation
30 cos 30
− 30 sin 30
r = 30 cos 30 = 26m / s
•
30 sin 30
 =−
= −0.15rad / s
100
•
r
•
r
Example 8
Particle A is moving along a straight path as shown with a velocity v = 30m/s and decelerates at 5m/s2.
• • •• ••
Determine
observed from O.
r,  , r, 
Solution
3. ACCELERATION ANALYSIS
Resolution
ar
a
••
− 5 cos 30
5 sin 30
Equation
••
•2
••
• •
r − r
r + 2r
r = −5 cos 30 + 100(−0.15) 2 = −2.08m / s 2
••
5 sin 30 − 2(26)(−0.15)
=
= 0.103rad / s 2
100
Example 9
A radar tracks the movement of a rocket. For the instant shown,  = 25 and r =
90km. If the rocket is moving with a velocity of 1600m/s and the acceleration is
only gavitational, determine
observed from O.
•
•
••
••
r,  , r, 
Example 9
A radar tracks the movement of a rocket. For the instant shown,  = 25 and r = 90km. If the rocket is
moving with a velocity of 1600m/s and the acceleration is only gavitational, determine
observed
• • •• ••
from O.
r,  , r, 
Solution
1. AXES
Example 9
A radar tracks the movement of a rocket. For the instant shown,  = 25 and r = 90km. If the rocket is
moving with a velocity of 1600m/s and the acceleration is only gavitational, determine
observed
• • •• ••
from O.
r,  , r, 
Solution
2. VELOCITY ANALYSIS
Resolution
vr
v
Equation
•
1600 sin 55
1600 cos 55
r
•
r
•
r = vr = 1600 sin 55 = 1310.64m / s
•
=
1600 cos 55
= 0.0102rad / s
90000
Example 9
A radar tracks the movement of a rocket. For the instant shown,  = 25 and r = 90km. If the rocket is
moving with a velocity of 1600m/s and the acceleration is only gavitational, determine
observed
• • •• ••
from O.
r,  , r, 
Solution
3. ACCELERATION ANALYSIS
Resolution
ar
a
••
− 9.81sin 65
9.81cos 65
Equation
•2
••
r − r
••
• •
r + 2 r
r = −9.81sin 65 + 90000(0.0102) 2 = −0.473m / s 2
••
9.81cos 65 − 2(1310.64)(0.0102)
=
= −0.00034rad / s 2
90000
Example 10 {combination (x, y) & (n, t)}
A projectile is launched with an initial velocity, v = 10m/s, 30 to the horizontal as
shown. Determine:
a. The radius of curvature r at A,
b. The radius of curvature r at B, when it reaches maximum height,
c. The radius of curvature r at C, 1 second after launching.
Example 10 {combination (x, y) & (n, t)}
A projectile is launched with an initial velocity, v = 10m/s, 30 to the horizontal as shown.
Determine:
a. The radius of curvature r at A,
b. The radius of curvature r at B, when it reaches maximum height,
c. The radius of curvature r at C, 1 second after launching.
Solution
Part a.
1. AXES
Example 10 {combination (x, y) & (n, t)}
A projectile is launched with an initial velocity, v = 10m/s, 30 to the horizontal as shown. Determine:
a. The radius of curvature r at A,
b. The radius of curvature r at B, when it reaches maximum height,
c. The radius of curvature r at C, 1 second after launching.
Solution
Part a.
2. ACCELERATION
ANALYSIS
Resolution
Equation
an
g cos 60
v2 / 
at
− g sin 60
rate of speed change
102
A =
= 20.39m
9.81cos 60
Example 10 {combination (x, y) & (n, t)}
A projectile is launched with an initial velocity, v = 10m/s, 30 to the horizontal as shown. Determine:
a. The radius of curvature r at A,
b. The radius of curvature r at B, when it reaches maximum height,
c. The radius of curvature r at C, 1 second after launching.
Solution
Part b.
At maximum height, vB = uX = 10 cos 60 = 5m/s →
1. AXES
Example 10 {combination (x, y) & (n, t)}
A projectile is launched with an initial velocity, v = 10m/s, 30 to the horizontal as shown. Determine:
a. The radius of curvature r at A,
b. The radius of curvature r at B, when it reaches maximum height,
c. The radius of curvature r at C, 1 second after launching.
Solution
Part b.
2. ACCELERATION
ANALYSIS
Resolution
Equation
an
g
v2 / 
at
0
rate of speed change
52
B =
= 2.548m
9.81
Example 10 {combination (x, y) & (n, t)}
A projectile is launched with an initial velocity, v = 10m/s, 30 to the horizontal as shown. Determine:
a. The radius of curvature r at A,
b. The radius of curvature r at B, when it reaches maximum height,
c. The radius of curvature r at C, 1 second after launching.
Solution
Part c. Find the velocity vc 1 second after launching to determine the (n, t) axes.
(x, y) analysis
(+ →) vcx = ux = 10 cos 60 = 5m / s
(+ ) v = u + at
cy
y
vcy = 10 sin 60 − 9.81(1) = −1.15m / s
vcy = 52 + 1.152 = 5.13m / s
 = tan −1
1.15
= 12.95
5
Example 10 {combination (x, y) & (n, t)}
A projectile is launched with an initial velocity, v = 10m/s, 30 to the horizontal as shown. Determine:
a. The radius of curvature r at A,
b. The radius of curvature r at B, when it reaches maximum height,
c. The radius of curvature r at C, 1 second after launching.
Solution
Part c.
(n, t) analysis
1. AXES
Example 10 {combination (x, y) & (n, t)}
A projectile is launched with an initial velocity, v = 10m/s, 30 to the horizontal as shown. Determine:
a. The radius of curvature r at A,
b. The radius of curvature r at B, when it reaches maximum height,
c. The radius of curvature r at C, 1 second after launching.
Solution
Part c.
(n, t) analysis
2. ACCELERATION
ANALYSIS
Resolution
an
g cos12.95
at
g sin 12.95
Equation
v2 / 
rate of speed change
5.132
C =
= 2.75m
9.81cos12.95