12.6: Velocity and Acceleration in Polar Coordinates - Mathematics

12.6: Velocity and Acceleration in Polar Coordinates - Mathematics

The position of any point in a cylindrical coordinate system is written as

[ <f r>= r hat <f r>+ z hat <f z>]
where (hat <f r>= (cos heta, sin heta, 0)). Note that (hat heta) is not needed in the specification of (<f r>) because ( heta), and (hat <f r>= (cos heta, sin heta, 0)) change as necessary to describe the position. However, it will appear in the velocity and acceleration equations because

In summary, identities used here include

Returning to the position equation and differentiating with respect to time gives velocity.

[ <f v>quad = quad (r hat <f r>+ z hat<f z>) quad = quad (dot r , hat <f r>+ r omega hat<oldsymbol< heta>> + dot z , hat<f z>) ]
This could also be written as

[ <f v>= (v_r , hat <f r>+ v_ heta hat<oldsymbol< heta>> + v_z , hat<f z>) ]
where (v_r = dot r, v_ heta = r , omega,) and (v_z = dot z).

Differentiating again to get acceleration.

The (- r , omega^2 , hat<f r>) term is the centripetal acceleration. Since ( omega = v_ heta / r ), the term can also be written as (- (v^2_ heta / r) , hat <f r>).

The (2 dot r omega , hat<oldsymbol< heta>>) term is the Coriolis acceleration. It can also be written as (2 , v_r , omega , hat<oldsymbol< heta>>) or even as ( (2 , v_r , v_ heta / r ) hat<oldsymbol< heta>>), which stresses the product of (v_r) and (v_ heta) in the term.

Centripetal Accelerations in the Tire

The centripetal acceleration of a tire traveling at 70 mph is remarkably high. 70 mph is 31.3 m/s, and this is (v_ heta). For a tire with a 0.3 m radius, the centripetal acceleration is

Cylindrical Acceleration Example

This example uses the function, (r=[17-cos(4 heta)]/16), with ( heta = t), and calculates acceleration components.

[ egin ddot r & = & cos(4 heta) end ]
So the acceleration vector is

2nd Cylindrical Acceleration Example

A bar is rotating at a rate, (omega). A collar starts at (R_o) and is being flung off with zero friction. So the radial acceleration is zero. Therefore

[ a_r = ddot r - r , omega^2 = 0 ]
This is a 2nd order differential equation, whose solution is

[ r = A e^ + B , e^ <-omega , t>]
Assume the initial conditions are (r(0) = R_o) and (dot r(0) = 0). This leads to

[ A = B = ]
So the solution is

[ r = R_o cosh(omega , t) ]
Remember, this gives zero net radial acceleration for the case where (omega = constant).

Recall that the circumferential acceleration is

[ a_ heta = r , alpha + 2 dot r omega ]

(alpha) is zero because (omega) is a constant. (dot r) is

[ dot r = R_o , omega sinh(omega , t) ]
and this all combines to give

[ a_ heta = 2 R_o , omega^2 sinh(omega , t) ]
which is a large circumferential acceleration due entirely to the Coriolis effect, even though (omega) is constant.

12.6: Velocity and Acceleration in Polar Coordinates - Mathematics

For plane motion, many problems are better solved using polar coordinates, r and &theta. This requires the development of position, velocity and acceleration equations based on, r and &theta.

Before velocity and acceleration can be determined in polar coordinates, position needs to be defined. Unlike rectilinear coordinates (x,y,z), polar coordinates move with the point and can change over time. Even though the r coordinate is moving, the position vector r, is measured in the r direction, giving

r = r er

Similar to other coordinate systems, velocity can be determined by taking a time derivative of position,

Since the coordinate system is moving, the time derivative of the unit vector, er, is not zero. Using a derivation similar to that found in the theory of n-t coordinate systems, expressions for the derivatives of the unit radial and unit transverse vectors can be determined as,

The derivative of the e&theta unit vector includes a negitve sign due to it changing inward as it moves (inward is a negative r-direction). The chain rule can then be used to express the time derivative of the unit radial vector as

Substituting the above relationships into the velocity equation gives

A time derivative of the velocity will give an expression for the acceleration,

The chain rule can be used to express the time derivative of the unit transverse vector as

By substituting this into the previous equation, and rearranging, gives the acceleration in terms of radial and transverse components,

It is important to note that both r and &theta directions have multiple terms. Acceleration in the r and &theta direction is not just d 2 r/dt 2 and d 2 &theta/dt 2 , respectively, since the coordinates are moving.

Mr. Kennedy: Another Maths Enthusiast

Referring to the diagram above the polar unit vector ( hat < r >) has the following relation to the rectangular unit vectors ( hat < i >) and ( hat < j >)

$ hat = left(cos heta ight) hat + left( sin heta ight ) hat $

The polar unit vector ( hat < heta >) is always perpendicular to ( hat < r >). From the diagram above ( hat < heta >) has the following relation to the rectangular unit vectors ( hat < i >) and ( hat < j >)

$ hat < heta>= left( -sin heta ight) hat + left( cos heta ight) hat $

In polar coordinates the position of an object ( R ) distance from the origin as represented in the diagram above is modelled
$ mathbf = R hat $

The velocity and acceleration in polar coordinates is derived by differentiating the position vector.

Differentiating ( hat < r >) yields

$ frac>

= left(-frac
sin heta ight) hat + left( frac
cos heta ight ) hat = frac
left[ left( -sin heta ight) hat + left( cos heta ight) hat ight] $
$ frac>
= frac
hat < heta>$

which implies the velocity is

Differentiating the velocity yields acceleration.

$ frac>

= left(-frac
cos heta ight) hat + left( – frac
sin heta ight ) hat = -frac
left[ left( cos heta ight) hat + left( sin heta ight) hat ight] $
$ frac>
= -frac
hat $

Plane Curvilinear Motion - Polar Coordinates

The polar coordinate system is defined by the coordinates r and θ . Just like the n-t coordinate axes, the r and θ axes are attached to and move with the particle.

  • Position: ( r ) The vector that starts at the origin of the x-y coordinatesystem and points to the particle.


The r-θ axes are body-fixed. That means that they are attached to and move with the particle. We know that velocity is the time rate of change of position as shown below. As you can see, the velocity equation contains the time derivative of a unit directions vector. The time derivative of a unit direction vector in a body-fixed coordinate system is generally not zero.

v = d r /dt = d(r e r )/dt = (dr/dt) e r + r(d e r /dt)

Using the values for the time derivative of the unit direction vectors, we get the following polar coordinate velocity equation.


The acceleration is the time rate of change of velocity as shown below. Again, the derivatives of the unit direction vectors are not generallyzero.

Using the above expressions for the time derivatives of the unit direction vectors, we get the following polar coordinate velocity equation.

2 Answers 2

Consider the picture below.

In Cartesian coordinates $hat r=cos hetahat i+sin hetahat j,$ and $hat heta=-sin hetahat i+cos hetahat j.$ Therefore $frac=hat heta$

We'll use one upper dot for the 1st derivative with respect to $:t:$, for example

Now, let a system of coordinates $:left(x,y ight):$ in the plane as in above Figure and $:mathbf,mathbf:$ the unit basic vectors along axis $:Ox,Oy:$ respectively. The position vector $mathbfleft(t ight)$ of the particle may be expressed as follows : egin mathbfleft(t ight)= left[r cos heta left(t ight) ight]mathbf+left[r sin heta left(t ight) ight]mathbf ag <03>end

Note that all quantities as position vector $:mathbf:$, velocity vector $:mathbf:$, angle $: heta:$ and as we see bellow the unit vectors $:mathbf_,mathbf_< heta>:$ are functions of time and so it's convenient to omit $:t:$.

So (03) yields egin mathbf= rleft[left(cos heta ight)mathbf+ left(sin heta ight)mathbf ight]= rmathbf_ ag <04>end where by definition egin mathbf_ equiv left(cos heta ight)mathbf+ left(sin heta ight)mathbf ag <05>end is a unit vector along $:mathbf :$, as in Figure. The velocity vector is egin mathbf=dfrac<>>

= dot>=dotmathbf_+rdot>_=dotmathbf_+rdot< heta>left[left(-sin heta ight)mathbf+ left(cos heta ight)mathbf ight]=dotmathbf_+ rdot< heta>mathbf_ < heta> ag <06>end where by definition egin mathbf_ < heta>equiv left(-sin heta ight)mathbf+ left(cos heta ight)mathbf ag <07>end is a unit vector normal to $:mathbf :$, as in Figure.

12.6: Velocity and Acceleration in Polar Coordinates - Mathematics

Polar (Radial/Transverse) Coordinates

This coordinate system is convenient to use when the distance and direction of a particle are measured relative to a fixed point or when a particle is fixed on or moves along a rotating arm.

Coordinates are chosen such that:

er is a unit vector pointing radially outward from the origin toward the particle, and e is a unit vector pointing perpendicular to the radial line in the direction of increasing .

er = cos i + sin j
e = -sin i + cos j

Therefore, position of the particle in polar coordinates is given by

velocity of the particle is given by

and acceleration is given by

Again, velocity will always be tangent to the path, and acceleration will generally have components both normal and tangent to the path.

Lecture Materials

Below you will find blank copies of the lecture outlines that you will find in my lecture videos. Some students tell me they like to print or download these to help them take notes on. After we finish pass a topic, I will also post a filled in handwritten copy of my notes on this page.

NOTE: Please ignore any homework due dates one page one of these outlines, instead always go off the course calendar and the due dates on Webassign.

12.1 Lecture Outline - 12.1 Notes: Intro to 3D, axes, coordinate planes, distance, and spheres.
12.2 Lecture Outline - 12.2 Notes: Intro to vectors: addition, magnitude, scalar multiplication, unit vector Then Intro to dot products.
12.3 Lecture Outline - 12.3 Notes: Dot Products: Definition, Big Theorems/Facts, Orthogonality, Angle between vectors, Projections Then Intro to Cross-Product.
12.4 Lecture Outline - 12.4 Notes: Cross Products: Defintion, Computing/Checking, Big Facts, Right-Hand Rules, Area of Parallelogram Then intro to Lines.
12.5 Lecture Outline - 12.5 Notes 1: Lines and Planes in 3D.
12.5 Notes 2: Lines and Planes in 3D - How to approach problems.
12.6 Summary: Intro to Surfaces in 3D, traces, then 7 important names: Cylinders, Paraboloids (Two Types: Elliptical or Hyperbolic), Hyperboloids (Two Types: One Sheet or Two Sheets), Cones, Spheres/Ellipsoids. You must know how to identify all these shapes and generally know what they look like.
Ch. 12 - Quick Fact Sheet
Exam 1 Overview, Review, and Study Tips
EXAM 1 - on Chapter 12.

13.1 Lecture Outline - 13.1 Notes: Intro to vector curves: how to visualize (surface of motion), thinking in terms of points or position vectors.
13.2 Lecture Outline - 13.2 Notes: Calculus on vector curves: Tangent/Derivative Vector, Unit Tangent, Tangent Line, Intergral/Antidervative.
13.3 Lecture Outline - 13.3 Notes: Measurement on 3D curves: Unit Tangent, Principal Unit Normal, Arc Length, Curvature.
13.4 Lecture Outline - 13.4 Notes: Acceleration and Velocity in 3D: Antidertivative to go from acceleration to velocity to position, tangent and normal components of acceleration.
Ch. 12 & 13 Fact Sheet - Vector Tools and Vector Calculus on 3D curves
EXAM 2 - on Chapter 13.

14.1/3 Lecture Outline - 14.1/3 Notes: Intro to 3D Surfaces and Partial Derivatives Domain, Traces, Level Curves, Contour Map, partial derivatives and interpretting.
14.3/4 Lecture Outline - 14.3/4 Notes: More on partial derivatives as well as Tangent Planes and linear approximation
14.4/7 Lecture Outline - 14.7 Notes 1: Discussion of Critical Points and Local Max/Min
14.7 Lecture 2 Outline - 14.7 Notes 2: Discussion of Global Max/Min (boundaries of a region)
Ch. 14 Full Review
EXAM 3 - on Chapter 14.

15.1 Lecture Outline - 15.1 Notes: Intro to double integrals.
15.2 Lecture Outline - 15.2 Notes: Double integrals over general regions, reversing order, setting up, evaluating.
10.3 Lecture Outline - 10.3 Notes: Polar Coordinates (a tool we need in order to work with circular regions).
15.3 Lecture Outline - 15.3 Notes: Double integrals over polar regions, how to integral above circular regions!
15.4 Lecture Outline - 15.4 Notes: Center of Mass
Exam 4 Facts - Ch. 15 Review
EXAM 4 - on Chapter 15.

TN 1 Lecture Outline - TN 1 Notes: tangent lines and intro to error bounds
TN 2-3 Lecture Outline - TN 2-3 Notes: higher order Taylor polynomials and Taylor's inequality
TN 4 Lecture Outline - TN 4 Notes: Taylor Series
TN 5 Lecture Outline - TN 5 Notes: Manipulating Taylor Series
Taylor Notes Fact Sheet
EXAM 5 - on Tayloy Polynomials/Series

12.6: Velocity and Acceleration in Polar Coordinates - Mathematics

That's fine, but we usually want the velocity expressed in the directions of r and θ, rather than the x and y coordinates. To switch to the new orthonormal basis, take the dot product of the above velocity vector with the unit radial vector and the unit tangent vector. These vectors are cos(θ),sin(θ) and -sin(θ),cos(θ) respectively. After taking dot products, the velocity, measured along r and θ, is r′ and r×θ′. The radial speed is the change in r, and the tangential speed is the change in θ times r. When the particle is far from the origin, a small change in θ makes a big difference.

Differentiate again, using the chain rule, to get acceleration. Dot this with the radial and tangential unit vectors, as we did with velocity. I'll spare you the algebra. The acceleration away from the origin is r′′-r×θ′ 2 . The first term is the radial acceleration, and the second is the centripital acceleration. If the partical is tracing a perfect circle, the second term gives the acceleration needed to keep the particle in its orbit.

The acceleration in the tangential direction is rθ′′+2r′θ′. The first term is angular acceleration magnified by the radial distance, and the second term is the coriolis force. It takes extra force to keep a skater spinning at the same rate, as she extends her arms. Her hands, moving outward, represent r′×θ′.

12.6: Velocity and Acceleration in Polar Coordinates - Mathematics

In the Curvilinear Motion: Rectilinear Coordinates section, it was shown that velocity is always tangent to the path of motion, and acceleration is generally not.

If the component of acceleration along the path of motion is known, motion in terms of normal and tangential components can be analyzed.

Consider a point P moving along a curved path.

The position vector r specifies the position of P with respect to the reference point O, and s measures the position of P along the path relative to the reference point O'. The unit tangent vector, et, is tangent to the path at point P.


Velocity of a Particle in a Plane

The speed of the point along the path is found by taking the derivative of the position,

Since the velocity is tangent to the path, it can be expressed in terms of et,

v = ds/dt et = v et

Relationship Between Normal
and Tangent Direction

Relationship between et and en

Radius of Curvature Definition

Similar to rectilinear coordinates, acceleration is obtained by differentiating the velocity (two parts) as

a = dv/dt = dv/dt et + v det/dt

Since the acceleration is not, in general, tangent to the path, it is useful to express it in terms of components that are normal and tangent to the path. To do so, the time derivative of the unit tangent vector, et, will be found.

Let et(t) be the unit tangent vector at time t, and et(t + &Deltat) be the unit tangent vector at time t + &Deltat. If et(t) and et(t + &Deltat) are drawn from the same origin, they form two radii of length 1 on the unit circle. The magnitude of &Deltaet is given by the equation

Watch the video: Calculus 3 Lecture: Velocity and Acceleration of Vector Functions (December 2021).