MEE 206: Engineering Mechanics II – Femi Koya
KINEMATICS OF PARTICLES Part 2
CURVILINEAR MOTION
In circular motion, the particle moves in a path
of constant radius so that its displacement is
given by 𝑟∆𝜃 and it motion is described by:
𝑟∆𝜃
𝑑𝜃
∆𝑡
Velocity, 𝑉 = 𝑙𝑖𝑚∆𝑡→0
= 𝑟 𝑑𝑡
This is elegantly written as, 𝑟𝜃̇ , alternatively as
𝑟𝜔. And in vector form, it is 𝑟 ∆ 𝜔
The acceleration is derived as,
𝑑𝑣
𝑟𝑑𝜃̇
𝑎=
=
= 𝑟𝜃̈
𝑑𝑡
𝑑𝑡
It is also written as, 𝑟𝛼.
And in vector form as, 𝑟 ∆ 𝜔 ∆ 𝜔 , or say -𝜔2 𝑟
along the radius. This acceleration is centerseeking and it is called the centripetal
acceleration (𝜔2 𝑟).
Curvilinear motion is the description of motion
along a curve: it thus combines of linear and
circular motions in a single description, where
the radius of rotation is not fixed.
Coordinates Systems in Curvilinear Motion
Unlike Cartesian Coordinates, where the
directions of x and y axes are usually placed in
fixed orientations in a particular problem, the
coordinates in curvilinear motions, though unit
vectors, are normally not fixed; hence; they
are differentiable.
In plane curvilinear, one axis is taken along the
rotating radius and the other perpendicular to
it. So that the distance ∆𝑆 is given by:
∆𝑆 = ∆𝑅𝑟̅ + 𝑅∆𝜃𝑝̅
Where, 𝑟̅ and 𝑝̅ are mutually perpendicular
unit vectors.
Therefore,
𝑉 = 𝑙𝑖𝑚∆𝑡→0 [
So that, 𝑽 =
𝑑𝑅
̅
𝒓̅ + 𝜔𝑅𝒑
𝑑𝑡
∆𝑆
]
∆𝑡
Check: if the curve has a constant radius, it is
then a circle; the radial component becomes
zero, and we are left with the perpendicular
(tangential component, as we say in
secondary school).
If we differentiate further, we get the
acceleration. The differentiation of unit vector
was a subject in a prerequisite course.
𝒂=(
𝑑2 𝑅
𝑑𝑅
̅
− 𝑅𝜔2 ) 𝒓̅ + (2𝜔
+ 𝑅𝛼) 𝒑
2
𝑑𝑡
𝑑𝑡
We have so far worked in (𝑟̅ , 𝑝̅ ) coordinates.
Other equivalent coordinates are:
(ii)
(𝑒̂𝑟 , 𝑒̂𝜃 )
: polar coordinates
(iii)
(𝑒̂𝑛 , 𝑒̂ 𝑇 )
: normal-tangential
Note that, we could also write the equations
as follows:
̅
𝑽 = 𝑅̇ 𝒓̅ + 𝑅𝜃̇ 𝒑
̅
𝒂 = (𝑅̈ − 𝑅𝜃̇ 2 )𝒓̅ + (2𝜔𝑅̇ + 𝑅𝜃̈ )𝒑
Practically, the first term is the positive radial
acceleration, the second term, though radial,
but pointing to the center, is the centripetal
acceleration. The third term is the Coriolis,
while the fourth term is the tangential
acceleration.
The n-t Coordinates System is particularly
useful when resolving motion of a vehicle. The
road being fixed, and in relation to the x-y
Coordinates has radius curvature 𝜌 defined
by,
𝑑2 𝑦
1
𝑑𝑥 2
=
3⁄
𝜌
𝑑𝑦 2 2
⌈1 + ( ) ⌉
𝑑𝑥
Steps in Solving Problems
➢ List all given data in consistent units (e.g.,
S.I. Units), and identify what is to be found
Page 1 of 2
➢ Identify the appropriate coordinate system
(Cartesian (e.g., in motion under gravity, or
Path variables)
➢ In Cartesian Coordinates, split motion into
components, e.g., x and y.
• Along each axis: is the particle in
constant acceleration?
▪ Yes: use relevant equations
▪ No: write the appropriate
𝑑𝑉
relations, e.g., 𝑎 = 𝑉 𝑑𝑆
State and apply boundary
conditions
▪ If the motion was explicitly
written in vector, you may,
from start, write out the
position vector.
➢ In solving motion on path-variable
coordinates, mastering the equations of
motion in this lecture is extremely
important.
➢ In some problems, both Cartesian and
Path-Variable Coordinates are mixed, but
it is critical to note that the particle is
essentially same and both relations must
be equal in any direction!
➢ Sometimes, it is useful to recognize 𝜔2 𝑅 as
▪
𝑉2
, or
𝑅
𝑉2
. This, of course, you are familiar
𝜌
with, where 𝜌 is called the radius of
curvature.
Examples and Assignments
1. Questions 9 – 12 in the last set of Home
Work were on Plane Curvilinear Motion
under Gravity. Please review.
2. The x- and y-components of a particle
moving with plane curvilinear motion are
𝑡3
given by 𝑥 = 2𝑡 2 + 3𝑡 and 𝑦 = 3 − 8, where
x and y are in meters and t in seconds.
Determine the magnitudes of the velocity
and acceleration, as well as the angles
which the vectors make with the x-axis
when t = 3 s.
𝑡3
Hints: Write S = (2𝑡 2 + 3𝑡) i + ( 3 − 8) j
3. At the bottom A of a vertical inside loop
the magnitude of the total acceleration of
an airplane is 3g. If the airspeed is 800 kmh1 and increasing at the rate of 20 kmh-1per
second, calculate the radius of curvature
of the path.
Hint: Given 𝑉 = 800 kmh-1, 𝑉̇ = 20 kmh-1 per
second.
The
magnitude
of
the
⌊𝑎
⌋
acceleration 𝑟 + 𝑎𝑡 = 3g. We rewrite the
equation for the acceleration, noting that
𝑑𝑅
= 0 as follows:
𝑑𝑡
̅
𝒂 = (𝑅̈ − 𝑅𝜃̇ 2 )𝒓̅ + (2𝜔𝑅̇ + 𝑅𝜃̈ )𝒑
𝑉
Since, 𝑅̇ = 0, 𝑅̈ = 0. Also, 𝜃̇ = 𝜔 =
𝑅
𝑉̇
And, 𝛼 = 𝜃̈ = 𝜔̇ = 𝑅
Hence,
𝑉2
̅
) 𝒓̅ + (𝑉̇ )𝒑
𝑅
Make sure all parameters are in consistent units.
𝒂 = (−
4. A jet plane flying at a constant speed v at an
altitude h = 10 km is being tracked by radar
located at O directly below the line of flight, If the
angle θ is decreasing at the rate of 0,020 rad/s
when 𝜃 = 60o, determine the value of 𝑟̈ at the
instant and the magnitude of the velocity of the
plane.
Ans. 𝑟̈ = 4.62 m/s2 v= 960 km/h.
Hint: Set up the (r, 𝜃) for the radar at a point on
the plane. Resolve the velocity v of the plane
along these coordinates.
5. Car A is moving with constant speed v on
the straight and level highway. The police
officer in the stationary car P attempts to
measure the speed v with radar. If the
radar measures “line-of-sight” velocity,
with what velocity v’ will the officer
observe? Evaluate your general expression
for the values v = 115 km/h, L = 150 m, and
D = 6 m, and draw any appropriate
conclusions
Ans:
𝑉′
=
𝑉𝐿
√𝐿2 + 𝐷2
𝑉 ′=
114.9 km/h
Page 2 of 2