Planar Motion in normal-tangential coordinates
• General planar motion can be expressed in terms of unit vector 𝑢
which is tangent to the particle path, and the normal vector 𝑢 which is
normal to the particle path.
• The arc length s(t) is defined as the distance travelled by the particle
along its path from t=0 to t.
• Velocity vector 𝒗 𝒕 = 𝑣 𝑡 𝑢 ; 𝑣 𝑡 =
( )
• Curvature of the path is given by 𝜅 =
• Acceleration is given by 𝒂 = 𝒗̇ = 𝑣̇ 𝑢 + 𝑣
• But
So
= 𝑣̇ 𝑢 + 𝑣
is normal to 𝑢
• So we can define the unit normal as:
• 𝑢 (𝑠) is called the principal unit normal and points towards the
concave side of the curve
Planar Motion: Normal and Tangential Coordinates
• We can also write:
where ρ(s) is the radius of curvature
of the curve.
• The expression for acceleration can be written as:
• In 3D coordinates, the third unit vector is the bi-normal given by 𝒖𝒃 = 𝒖𝒕 × 𝒖𝒏
• The radius of curvature of a trajectory y(x) in Cartesian coordinates is given by
• The osculating circle is the instantaneous circular trajectory
created by the radius of curvature of the path.
• The normal component of acceleration is always in the radially
inward direction of the osculating circle
Time Derivative of a Unit Vector
• Consider a unit vector 𝒖 . Since a unit vector cannot change
in magnitude, it can only change in its orientation. Let’s fix the
coordinate system at the base of 𝑢 and let θ(t) be the angle it
makes with the x axis at time t. This θ(t) is changing with time
as the unit vector changes its orientation. This is shown in the
figure.
• 𝒖 = 𝑐𝑜𝑠𝜃𝚤̂ + 𝑠𝑖𝑛𝜃𝚥̂ ⇒ 𝒖̇ = 𝜃̇ −𝑠𝑖𝑛𝜃𝚤̂ + 𝑐𝑜𝑠𝜃𝚥̂
• Note that 𝒖 𝒖̇ = 0. So the vector representing the rate of
change of 𝒖 is perpendicular to the unit vector.
• The angular velocity of the unit vector, as it rotates, is 𝝎 =
𝜃̇ 𝑘 and is perpendicular to both 𝒖 and 𝒖̇
• It can now be easily be shown by cross-product that:
𝒖̇ = 𝝎 × 𝒖 Universal result (2D or 3D case)
• Time derivative of a unit vector is the angular velocity of the
unit vector crossed with the unit vector.
Time Derivative of a Vector
• Any vector can be represented by its magnitude and its
unit vector.
𝑨 = 𝐴𝒖
• So its time derivative is given by (Universal Result)
𝑨̇ = 𝑨̇𝒖𝑨 + 𝑨𝒖̇ 𝑨 = 𝑨̇𝒖𝑨 + 𝝎𝑨 × (𝐴𝒖𝑨 )
• Then in n-t coordinate system, acceleration is:
• But previously we have shown:
𝒗
• Noting that 𝑢 = 𝑢 × 𝑢 we have 𝝎𝒗 = 𝒖𝒃
𝝆
• Thus the angular rate of rotation of the velocity vector is
equal to the speed of the particle divided by the radius
of the osculating circle. Its direction is along the
binormal.
Planar Motion in r-θ (polar) coordinates
• If a particle moves in a curved line, then its position vector
can be written in terms of its distance r from the origin and
the angle θ it makes with the x-axis. Or we can write:
𝒓 = 𝑟𝒖
where 𝒖 is the unit vector along the 𝒓 direction at a given θ
value.
• Differentiating: 𝒗 = 𝑟̇ 𝒖 + 𝑟𝒖̇ = 𝑟̇ 𝒖 + 𝑟𝝎 × 𝒖
• Since θ increases in the counterclockwise direction we have
𝝎 = 𝜃̇ 𝑘 . This implies that 𝝎 × 𝒖 = r𝜃̇ 𝑘 × 𝒖 = r𝜃̇ 𝒖
• Here 𝒖 = 𝑘 × 𝒖 is the unit vector associated with the θ
coordinate and is perpendicular to the line connecting the
particle to the origin as shown.
• Finally then velocity in polar coordinates: 𝒗 = 𝑟̇ 𝒖 + r𝜃̇ 𝒖
• Differentiating and using the relation 𝑨̇ = 𝑨̇𝒖𝑨 + 𝝎𝑨 × (𝐴𝒖𝑨 )
we get 𝒂 = 𝑟̈ − 𝑟𝜃̇ 𝒖 + 𝑟𝜃̈ + 2𝑟̇ 𝜃̇ 𝒖
Special Case: Circular Motion
• If we set the origin as the center of rotation, then
clearly 𝒖 = 𝒖 and 𝒖 = −𝒖 . Further 𝑟̇ = 𝑟̈ =
0 𝑎𝑛𝑑 𝜌 = 𝑟. So we get
• 𝒗 = 𝑣𝒖 = 𝑟𝜃̇ 𝒖 . So speed 𝑣 = 𝑟𝜃̇
• The acceleration vector gives
• Note that only for circular motion does n-t and r-θ
coordinates align, and that too when the origin is
taken as the center of the circle.
Examples
Revisit the projectile problem and derive the equations describing the motion of
a projectile released at 𝐴, using polar coordinates. In addition, derive the
expressions for the initial conditions of the motion in the same polar coordinate
system, given that the projectile is launched from point 𝐴 at speed 𝑣0 in the
direction shown in Fig.
We know that for polar coordinates
Comparing
Coupled non-linear ODE system
We know that for polar coordinates
Initial Conditions:
Final Initial Conditions
So,
Examples
Relative Motion
• Understanding relative motion is important when a
problem involves multiple moving objects, and is also
important in rigid body motion analysis.
• Consider the case where we have the reference fixed
cartesian coordinates XY and frame xy which is fixed
with particle A and translating with it. Note that all
time x and X are parallel to each other, so is y and Y.
• Then 𝒓 = 𝒓 + 𝒓 / where 𝒓 / is the position vector
of B relative to A (i.e. in the x-y system) while rA and rB
are the position vector of the two particles in the X-Y
system.
• The time derivatives give:- 𝒗 = 𝒗 + 𝒗 / and 𝒂 =
𝒂 + 𝒂 / with similar interpretations
Constrained Motion
• Different connected points (on a bar or a rope)
are geometrically constrained to move in a
certain way.
• Consider a rigid bar supported in a wall. As the
bar slides, the end points move along the wall
(vertically) and along the floor (horizontally) and
the distance between the points remain the
same.
• 𝑥 = 𝐿𝑠𝑖𝑛𝜃 𝑎𝑛𝑑 𝑦 = 𝐿𝑐𝑜𝑠𝜃 . So velocity and
acceleration of the two points can be expressed
in terms of the inclination angle
Example
Since 𝑥 = 𝑥
Doing the differentiation
Particle Motion in 3D: Cartesian Coordinates
• In cartesian coordinates we have a
straightforward extension of the 2D case
with the coordinate relations added in.
• 𝒓 𝑡 = 𝑥 𝑡 𝚤̂ + 𝑦 𝑡 𝚥̂ + 𝑧(𝑡)𝑘
• 𝒗 𝑡 = 𝑥̇ 𝑡 𝚤̂ + 𝑦̇ 𝑡 𝚥̂ + 𝑧̇ (𝑡)𝑘
• 𝒂 𝑡 = 𝑥̈ 𝑡 𝚤̂ + 𝑦̈ 𝑡 𝚥̂ + 𝑧̈ (𝑡)𝑘
Particle Motion in 3D: Cylindrical Coordinates
• Cylindrical coordinates are an extension of the polar (Rθ) coordinates with the z axis acting as the third
coordinate system as shown. So coordinate system is Rθ-Z.
• Unit vector along Z: 𝑢 = 𝑢 × 𝑢
• Note that here, R coordinate is no longer the position
vector of the particle, but the component of the actual
position vector r in the R-θ (or X-Y) plane. Similarly θ is
the angle that this projected line makes with the X-axis.
• Since the z axis is fixed, the extension from polar to
cylindrical coordinates is simply adding the
corresponding z coordinate to the 2D r-θ expressions.
• 𝒓 𝑡 = 𝑅(𝑡)𝑢 + 𝑧(𝑡)𝑢 ; 𝒗 𝑡 = 𝑅̇𝑢 + 𝑅𝜃̇ 𝑢 + 𝑧̇ 𝑢 ;
𝒂 𝑡 = 𝑅̈ − 𝑅𝜃̇ 𝑢 + 𝑅𝜃̈ + 2𝑅̇ 𝜃̇ 𝑢 + 𝑧̈ 𝑢
Particle Motion in 3D: Spherical Coordinates
• Also an extension of the polar (r-θ) coordinates where
we keep the r coordinate as the position vector, θ
coordinate as the angle the projection of r makes in the
x-y plane, and a new angular Φ coordinate that gives
the orientation of the position vector r with respect to
the z coordinate. (r-θ-Φ) coordinates. The unit vectors
are 𝑢 , 𝑢 , 𝑢
• Note that the unit vector in the z direction is related to
the angular coordinates as: 𝑢 = 𝑐𝑜𝑠𝜙𝑢 − 𝑠𝑖𝑛𝜙𝑢
• Position vector 𝒓 = 𝑟𝑢
• Velocity vector 𝑣 = 𝑟̇ 𝑢 + 𝑟𝑢̇ ; 𝐵𝑢𝑡 𝑢̇ = 𝝎𝒓 × 𝑢
• Angular velocity of the unit position vector 𝑢 of r are
rotations with respect to θ (hence in z direction) and
rotation with respect to Φ (which is along the θ
direction). Hence
𝝎𝒓 = 𝜙̇𝑢 + 𝜃̇𝑢 = 𝜙̇𝑢 − 𝜃̇ 𝑠𝑖𝑛𝜙𝑢 + 𝜃̇ 𝑐𝑜𝑠𝜙𝑢
Particle Motion in 3D: Spherical Coordinates
• So we finally get 𝑢̇ = 𝝎𝒓 × 𝑢 = 𝜙̇𝑢 + 𝜃̇ 𝑠𝑖𝑛𝜙𝑢
• Thus the expression for velocity in the spherical
coordinate system is 𝒗 = 𝑣 𝑢 + 𝑣 𝑢 + 𝑣 𝑢 ; 𝑤ℎ𝑒𝑟𝑒
𝑣 = 𝑟̇ ; 𝑣 = 𝑟𝜙̇; 𝑣 = 𝑟𝜃̇𝑠𝑖𝑛𝜃
• The acceleration is obtained from 𝒂 = 𝒗̇ and using
appropriate expressions for differentiation of the unit
vectors as shown below
𝑢̇ = 𝝎𝒓 × 𝑢 ; 𝑢̇ = 𝝎𝒓 × 𝑢 ; 𝑢̇ = 𝝎𝒓 × 𝑢
• This finally gives: 𝐚 = 𝑎 𝑢 + 𝑎 𝑢 + 𝑎 𝑢 where
𝑎 = 𝑟̈ − 𝑟𝜙̇ − 𝑟𝜃̇ 𝑠𝑖𝑛 𝜙
𝑎 = 𝑟𝜙̈ + 2𝑟̇ 𝜙̇ − 𝑟𝜃̇ 𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜙
𝑎 = 𝑟𝜃̈𝑠𝑖𝑛𝜙 + 2𝑟̇ 𝜃̇𝑠𝑖𝑛𝜙 + 2𝑟𝜙̇𝜃̇ 𝑐𝑜𝑠𝜙
Example
In cylindrical coordinates
So we get
It is given that
Example
Take off speed
Speed after 60 seconds
P’
From the right triangle OAP’
P’
A