METR 420: Dr. Dave Dempsey Atmospheric Dynamics I Dept. of Earth & Climate Sciences, SFSU

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Acceleration, Forces, and Newton’s Second Law in Cylindrical vs. Cartesian Coordinates

Cylindrical coordinates consist of (1) a coordinate plane, plus (2) an axis perpendicular to the plane through the origin. In the coordinate plane, two coordinates describe position: (1) an angle, θ (azimuth angle, measured positive counterclockwise relative to a reference line in the plane), and (2) a distance, R, from the origin within the plane. R and are just polar coordinates, so we can call the coordinate plane in cylindrical coordinates the polar plane. The third coordinate, which is position in a direction perpendicular to the polar plane, is typically assigned the symbol z. To simplify a comparison between cylindrical and Cartesian coordinates, consider Cartesian and cylindrical coordinate systems such that (1) they have the same origin; (2) the cylindrical coordinates’ polar plane coincides with the x-y plane in Cartesian coordinates; and (3) the z-axes of the two systems are aligned. Finally, suppose that both coordinate systems are set up in an inertial frame of reference, so neither is rotating or otherwise accelerating and hence Newton’s 2nd Law applies. Position and its components The position of an object, robj , can be written in terms of its components in Cartesian coordinates as: (1)(a) robj = xobji + yobj j + zobjk The same vector parcel position, robj , can be written in terms of its components in cylindrical coordinates as: (1)(b) robj = Robj R(θobj )+ zobjk where R(θobj ) is a radial unit vector (that is, outward from the origin) in the

θ

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polar plane. It is written as a function of θobj , the angular coordinate of the

object’s position, because the direction of R depends on θobj . In particular,

R points toward the location in the polar plane where the parcel’s 3-D position vector projects onto the polar plane. That location needs two pieces of information—that is, two coordinates—to describe it: Robj and θobj .

contains the information about θobj . Velocity and its components By definition, the velocity of the object in an absolute frame of reference, ca , is the rate at which the object’s (vector) position changes with respect to (wrt) time. In an inertial (“absolute”) frame of reference, this can be written in terms of its components in Cartesian coordinates by differentiating Eq. (1)(a):

(2) (a)

ca ≡ Dr Dt =

D xi( )Dt

+D yj( )Dt

+D zk( )Dt

= DxDt

i + DyDt

j + DzDt

k + x DiDt

+ y DjDt

+ z DkDt

= DxDt

i + DyDt

j + DzDt

k

≡ ca( )x i + ca( )y j + ca( )z k

Note the following about Eq. (2)(a):

-- For convenience, we’ve dropped the subscript “obj” from the object’s position vector and position coordinates, but they still refer the position of the object.

-- The product rule applies when differentiating the product of each scalar component of position and its corresponding unit vector (e.g., xi ).

-- A unit vector always points in a direction parallel to its corresponding coordinate axis. In Cartesian coordinates, the coordinate axes are straight lines, so the directions in which the three unit vectors point don’t vary from place to place. Moreover, unit vectors always have a magnitude (length) of 1 (with no dimensions). Hence, following the object around, i never changes either magnitude or direction, so Di Dt = 0, and the same is true for the other two unit vectors, j and k .

R

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Similarly, differentiating Eq. (1)(b) gives us the velocity expressed in terms of its components in cylindrical coordinates:

(2)(b)

ca ≡ Dr Dt =

D RR( )Dt

+D zk( )Dt

= DRDt

R + R DRDt

+ DzDt

k + z DkDt

= DRDt

R + R DRDt

+ DzDt

k

≡ ca( )R R + R DRDt

+ ca( )z k

In cylindrical coordinates, as in Cartesian coordinates, the unit vector k always points in the same direction (parallel to the z-axis, which is straight), so following the object, k changes neither magnitude nor direction and so Dk Dt = 0.

However, we can’t say the same about DR Dt , because the direction

of R depends on the object’s angular coordinate of position, θ , which can vary. As a result, to express ca explictly in terms of its components in cylindrical coordinates, we need to be able to write DR Dt in terms of its components in cylindrical coordinates.

First, note that DR Dt is defined as follows:

DRDt

≡ limΔt→0

R(t + Δt)− R(t)Δt

≡ limΔt→0

ΔRΔt

Hence, the direction if DR Dt is determined by the direction of ΔR in

the limit as Δt becomes very small. Because unit vectors always have a length of 1, it follows that in the limit as Δt becomes infinitesimally small, ΔR can only be in a direction perpendicular to R . (ΔR could have a component parallel to R only if R changed length.) More generally, unit

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vectors can change only in directions normal to themselves. In this case, since R always lies within a polar plane, changes in R must also lie in the polar plane, and so DR Dt can’t have a component parallel to k . Hence, it must have only a θ component. That is, DR Dt must be in the tangential direction, because R can change only in a tangential direction.

What is the magnitude of DR Dt ? It should be easy to convince

yourself that the faster the parcel changes its angular position, θ , the faster R should change direction. In fact, it is possible to show formally that:

(3) DRDt

= DθDt

θ

Substituting Eq. (3) into Eq. (2)(b), we can write:

(2)(c) ca = ca( )R R + R Dθ

Dtθ + ca( )z k

= ca( )R R + ca( )θ θ + ca( )z k

Here we’ve taken advantage of the relationship between the tangential component of linear velocity, ca( )θ , and the angular velocity:

(4) ca( )θ = RDθDt

Note that the velocity in Eq. (2)(c) expressed in terms of cylindrical coordinates has three explicit components, just as it does in Cartesian coordinates (though only the components in the k direction are the same in the two systems).

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Acceleration and its components By definition, an object’s acceleration is the rate at which the object’s (vector) velocity changes wrt/time. In an inertial (“absolute”) frame of reference, the acceleration,

aa , can be written in terms of its components in Cartesian coordinates as: (5)(a)

aa ≡ Dca Dt =

D ca( )x i( )Dt

+D ca( )y j( )

Dt+D ca( )z k( )

Dt

=D ca( )xDt

i +D ca( )yDt

j +D ca( )zDt

k + DiDt

ca( )x +DjDt

ca( )y +DkDt

ca( )z

=D ca( )xDt

i +D ca( )yDt

j +D ca( )zDt

k

≡ aa( )x i + aa( )y j + aa( )z k

where we’ve again invoked the fact that Di Dt = 0 in Cartesian coordinates and that the same is true for the other two unit vectors.

Similarly, the acceleration expressed in terms of its components in cylindrical coordinates would be: (5)(b) aa ≡ Dca Dt =

D ca( )R R( )Dt

+D ca( )θ θ( )

Dt+D ca( )z k( )

Dt

=D ca( )RDt

R + ca( )RDRDt

+D ca( )θDt

θ + ca( )θDθDt

+D ca( )zDt

k + ca( )zDkDt

=D ca( )RDt

R + ca( )R RDθDt

θ +D ca( )θDt

θ + ca( )θDθDt

+D ca( )zDt

k

=D ca( )RDt

R +ca( )R ca( )θ

Rθ +

D ca( )θDt

θ + ca( )θDθDt

+D ca( )zDt

k

Here we’ve invoked the fact that Dk Dt = 0 in cylindrical coordinates (in the 2nd step above), and we’ve substituted Eqs. (3) (in the second step) and (4) (third step).

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Note, though, that we can’t say that in general that Dθ Dt = 0, because, like R , the direction of θ can vary from place to place. (This makes sense, since R and θ are perpendicular to each other in the polar plane, so if one varies in direction from place to place in the plane, so must the other.) In fact, like R , the direction of θ depends on θ .

To express

aa entirely in terms of its components in cylindrical coordinates, we need to be able to write Dθ Dt in terms of its components. As with DR Dt , this can be done formally, but we again note first that Dθ Dt cannot have a component parallel to θ (because that would require that θ change length, which unit vectors don’t do). In fact, Dθ Dt must be in the radial direction, because θ changes only in the radial direction.

What is the magnitude of Dθ Dt ? Like DR Dt , you should sense

intuitively that the faster the parcel changes its angular position, θ , the faster θ should change direction. (This makes sense because θ is always perpendicular to R , so the faster R changes, the faster θ must also change.) In fact, it is possible to show that:

(6) DθDt

= − DθDt

R

Substituting Eq. (6) into Eq. (5)(b), and solving Eq. (4) for Daθ Dt and

substituting it into Eq. (5)(b) as well, gives: (5)(c) aa =

D ca( )RDt

R + ca( )R ca( )θ θ +D ca( )θDt

θ − ca( )θDθDt

R +D ca( )zDt

k

=D ca( )RDt

− ca( )θDaθDt

⎡

⎣⎢

⎤

⎦⎥ R +

D ca( )θDt

+ca( )R ca( )θ

R⎡

⎣⎢

⎤

⎦⎥θ +

D ca( )zDt

k

=D ca( )RDt

−ca( )θ

2

R

⎡

⎣⎢⎢

⎤

⎦⎥⎥R +

D ca( )θDt

+ca( )R ca( )θ

R⎡

⎣⎢

⎤

⎦⎥θ +

D ca( )zDt

k

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By analogy with Cartesian coordinates, we can define short-hand symbols for the scalar components of acceleration in cylindrical coordinates: (5)(d) aa = aa( )R R + aa( )θ θ + aa( )z k

where

aa( )R ≡D ca( )RDt

−ca( )θ

2

R

aa( )θ ≡D ca( )θDt

+ca( )R ca( )θ

R

aa( )z ≡D ca( )zDt

Note in particular that the (scalar) radial component of acceleration, aa( )R , is not defined as the rate at which the radial component of velocity,

ca( )R , changes wrt/time. Rather, there is a second contribution, from Dθ Dt, which is also in the radial direction. This term arises from the fact that one of the coordinate axes in cylindrical coordinates, the θ axis, is curved, not straight, and as a result, the direction of θ varies from place to place, and (instantaneous) variations in θ are in the radial direction. Similarly, the (scalar) tangential component of acceleration, aa( )θ , is not defined as the rate at which the tangential component of velocity, ca( )θ ,

changes wrt/time. Rather, there is a second contribution, from DR Dt , which is also in the tangential direction. This term also arises from the fact that the θ axis is curved, and as a result the direction of R varies from place to place, and (instantaneous) changes in R are in the tangential direction.

Because of their origins, the terms

D ca( )θ Dt

Fθ∑θca

and ca( )R ca( )θ R are

called curvature terms. They are part of the scalar components of acceleration in the radial and tangential directions, respectively. They

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compensate for the fact that the orientations of the radial and tangential axes vary from place to place, so the components of c projected onto these axes—namely, ca( )θ and ca( )R —will vary from place to place even when the vector itself ( c ) doesn’t change. Newton’s Second Law Applied in an inertial (“absolute”) frame of references, Newton’s 2nd Law can be written generally as:

(7) aa ≡

DcaDt

=F m∑

Breaking both the acceleration and the forces into their components in cylindrical coordinates, we can write Newton’s 2nd Law as: (8) aa( )R R + aa( )θ θ + aa( )z k = FR m( )∑ R + Fθ m( )∑ θ + Fz m( )∑ k By applying R ⋅ (a dot product) to both sides of Eq. (8), and recalling that (a) R ⋅θ = R ⋅ k = 0 because unit vectors are perpendicular to each other, and (b) R ⋅ R =1, we get the scalar version of Newton’s 2nd Law in the radial direction in cylindrical coordinates.

(9)(a) aa( )R ≡D ca( )RDt

−ca( )θ

2

R= FR m( )∑

Similarly, applying θ ⋅ and k ⋅ to Eq. (8), we get the scalar versions of the Newton’s 2nd Law in the tangential and z-directions, respectively:

(9)(b) aa( )θ ≡D ca( )θDt

+ca( )R ca( )θ

R= Fθ m( )∑

(9)(c) aa( )θ ≡D ca( )zDt

= Fθ m( )∑

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For reference, an analogous set of scalar equations in Cartesian coordinates would look like this:

(10)

aa( )x ≡D ca( )xDt

= Fx m( )∑

aa( )y ≡D ca( )yDt

= Fy m( )∑

aa( )z ≡D ca( )zDt

= Fz m( )∑

There are no curvature terms in Cartesian coordinates, because all three coordinate axes are straight. Conservation of angular momentum A common principle taught in first-semester physics courses is the principle of conservation of angular momentum. This principle makes the most sense in a polar or cylindrical coordinate system. It says that if there are no net force acting on an object in the tangential direction (and hence no torques)—that is, if Fθ∑ = 0—then the angular momentum of the object is conserved. If there is no net tangential force ( Fθ∑ = 0), then it follows from Newton’s 2nd Law that the tangential acceleration will be zero. From Eq. (9)(b):

(11) aa( )θ ≡D ca( )θDt

+ca( )R ca( )θ

R= 0

Multiplying Eq. (11) by R, invoking and substituting the definition ca( )R ≡ DR Dt , and applying the product rule (in reverse), we can show that

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R ×

D ca( )θDt

+ca( )R ca( )θ

R⎡

⎣⎢

⎤

⎦⎥ = R

D ca( )θDt

+ DRDt

ca( )θ

=D R × ca( )θ⎡⎣ ⎤⎦

Dt= 0

This is statement of conservation of angular momentum per unit mass in the absence of torques:

(12) D R × ca( )θ⎡⎣ ⎤⎦

Dt= 0

where R× ca( )θ is the angular momentum per unit mass of an object. According to Eq. (11) (and Eq. (12)), if there is no net tangential force (and hence no torque), then:

(23) D ca( )θDt

= −ca( )R ca( )θ

R

That is, if there are no tangential forces, but if a parcel is already moving with both radial and tangential components to its velocity (for example, spiraling inward or outward, or simply moving in a straight line inward or outward at an angle between purely radial and purely tangential directions), then its tangential component of velocity will change. This is what happens when a spinning ice skater pulls his/her arms toward his/her body. The skater’s arms already have a tangential velocity component (the skater is spinning), and they acquire a (negative) radial component when the skater pulls his/her arms inward. As a result, if the skater was already spinning in a positive θ direction (so that ca( )θ > 0), then Eq. (23) applied to the arms says that ca( )θ of the arms will increase (the arms will move faster in the tangential direction). This manifests itself as faster spinning, too. Similarly, moving the arms back outward ( ca( )R > 0) will reduce the tangential velocity of the arms. It’s important to keep in mind that a change in the tangential

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component of velocity, ca( )θ , does not by itself imply a tangential acceleration. That’s because ca( )θ could change solely due to a change in the orientation of the θ axis. ca( )θ is just the projection of ca onto the θ axis, so if that axis changes orientation, ca( )θ will change, even if ca doesn’t. The curvature term in the tangential direction compensates for this artifact of the coordinate system and is hence a necessary part of the tangential acceleration, along with D ca( )θ Dt . In the absence of a net tangential force, the sum of these terms remains zero (that is, tangential acceleration is zero), even as the tangential component of velocity changes in response to radial motions.

R(t2 )

R(t1)

Figure 1: An object traveling at a constant velocity, , at two different locations, and , at two different times, t1 and t2. Although is constant (so the object travels in a straight line at a steady speed), the projection ( ) of onto the radial axis at each of the two locations is different, and so is ( ), so each component must change over time (at rates and , respectively). However, there is no acceleration. It follows that and

are not the components of acceleration. Instead, curvature terms must be added to each to account for the change in direction of the unit vectors, and , between the two locations. These are part of the acceleration (and cancel the effect that changing the orientation of the axes has on the projection of the velocity onto those axes and hence the scalar components of the velocity).