MODULE 3

KINEMATICS OF RIGID BODY

A rigid body is an ideal body that does not change its shape when subjected to external forces or couples. The distance between the particles in a rigid body remains same during the motion. Hence, a rigid body can be defined as the group or collection of number of particles at fixed distance from each other.

There are three types of motion of a rigid body

- Translation
- Rotation
- General Plane Motion

Kinematics of rigid body in translation motion

Translation motion occurs when a line (generally axis) in the body remains parallel to its initial orientation throughout the motion.

If the path of motion of any two lines on the rigid body is parallel or is in a straight line, then the translation is termed as rectilinear translation.

If the path of motion of any two lines on the rigid body is along the fixed curve, then the translation is termed as curvilinear translation.

Consider a rigid body that is subjected to translation motion as shown in fig.

Β

The position of points A and B is defined are defined by the position vectors and w.r.t the fixed frame x, y.

Let the translating system is fixed in a body with origin A which is referred to the base point. The position vector of B w.r.t A is denoted by .

By vector addition, we get,

Taking, the first time derivative of the above equation, we get,

According to the definition of rigid body, the term is constant

Hence,

And, gives the absolute velocities and respectively with respect to the fixed frame.

So, we get,

This shows that, for the rigid body in translation motion, the velocity of every particle on the body is same.

Now, taking the time derivative of the velocity equation, it simply yields

Where, and are the absolute accelerations of points A and B on the rigid body w.r.t fixed frame.

This shows that, for the rigid body in translation motion, the velocity and acceleration of every particle on the body is same.

Kinematics of a body rotating about fixed axis

When a rigid body is subjected to the rotation about a fixed axis, as shown in fig below, all the particles of the body, except those which lie on the axis of rotation, move along circular paths whose center lies on the axis of rotation.

Consider, the point P on the rigid body defined by the position vector r. As the rigid body rotates about the fixed axis, then the point P travels a circular path of radius r with center point O as shown in fig below

If the body rotates with a differential angular displacement , then the point P displaces the distance , which is given by

Differentiating the above equation w.r.t time, we get

The term, is referred to as the angular velocity of the rotating body

And gives the velocity of point P

Hence,

The direction of velocity is tangent to the circular path.

Considering the magnitude and direction, velocity is the cross product of angular velocity and the position vector

Where, is directed from axis to the point P

To find acceleration, differentiate velocity vector w.r.t time

Here, the term refers to the angular acceleration of the rigid body

And is the velocity

Hence,

Where, is the tangential acceleration and is normal component of acceleration Hence,

Magnitude of both components is given by,

Tangential component refers to the time rate of change of velocityβs magnitude. Hence, if velocity of point P is increasing, is in same direction of , but if velocity of point P is decreasing, is in direction opposite to that of

The normal component of acceleration refers to time rate of change of the velocityβs direction. Hence, the direction of the normal component is always from point P towards the center O. Hence it is also called as centripetal acceleration.

Β

Rotation of a Body about a Fixed Axis

As a preliminary, letβs look at a body firmly attached to a rod fixed in space, and rotating with angular velocityΒ Ξ©Β radians/sec. About that axis.Β Youβll recall from freshman physics that the angular momentum and rotational energy are

But you also know that both angular velocity and angular momentum are vectors. Obviously, for this example, the angular velocity is a vector pointing along the axis of rotation,Β Ξ©β= (0, 0,Ξ© z).Β One might be tempted to conclude that the angular momentum also points along the axis, butΒ this is not always the case. An instructive example is provided by two massesΒ mΒ at the ends of a rod of lengthΒ 2aΒ held at a fixed angleΒ ΞΈΒ to theΒ zΒ axis, which is the axis of rotation.

Lz =2ma2sin2ΞΈβ Ξ©.

But notice that, assuming the rod is momentarily in theΒ xzΒ plane, as shown, then

Lx=β2ma2cos2ΞΈβ Ξ©.

The total angularΒ momentumΒ is not parallel to the total angularΒ velocity!Β

In fact, as should be evident, the total angular momentum is rotating around the constant angular velocity vector, so the axis must be providing a torque. This is why unbalanced car wheels stress the axle.

General Motion of a Rotating Rigid Body

Weβll follow the Landau notation (which itself tends to be bilingual between coordinatesΒ (x,y,z)Β andΒ (x1,x2,x3).) Β Notice that weβll label the components byΒ (x1, x2, x3), notΒ (r1, r2, r3)Β even though we callΒ the vectorΒ rβ.Β Β Again, weβre following Landau.

We take a fixed, inertial (or lab) coordinate system labelledΒ (X,Y,Z)Β and in this system the rigid bodyβs center of mass, labelledΒ O, is atΒ Rβ.Β We have a Cartesian set of axes fixed in the body, origin at the center of mass, and coordinates in this system,Β vectors fromΒ OβΒ to a point in the body denoted byΒ , are labelledΒ (x,y,z)Β orΒ (x1,x2,x3).

A vector from theΒ externalΒ inertial fixed origin to a point in the body is then

Suppose now that in infinitesimal timeΒ dt, the center of mass of the body movesΒ Β and the body rotates through.Β Then a particle atΒ Β as measured from the center of mass will move through

Relative to the external inertial frame.

Therefore, the velocity of that particle in the fixed frame, writing the center of mass velocity and the angular velocity as

Β

Now, in deriving the above equation, we have not used the fact that the originΒ OΒ fixed in the body is at the center of mass. (That turns out to be useful shortly.)Β What if instead we had taken some other originΒ Oβ²Β fixed in the body? Would we find the angular velocityΒ Β Β aboutΒ Oβ²Β to be the same asΒ The answer turns out to beΒ yes, but we need to prove it! Here's the proof:

If the position ofΒ Oβ²Β relative toΒ OΒ is Β (a vector fixed in the body and so moving with it) then the velocityΒ ofΒ Oβ²Β is

Β

This means that if we describe the motion of any particle in the body in terms of some origin fixed in the body, plus rotation about that origin,Β the angular velocity vector describing the bodyβs motion is the same irrespective of the origin we choose.Β So we can, without ambiguity, talk about the angular velocity of the body. Β

Angular Velocity

Β

The angular velocity of Fi0 with respect to Fi is

Β

Note the ordering of the superscripts in this expression. Also, note that this vector

Quantity has the same components in either frame; that is,

Β

Β

This simple expression results because it is a simple rotation. For the 2" rotation from FiΒ to Fi , the angular velocity vector is

Β

Finally, for the 1" rotation from Fi00 to Fb, the angular velocity vector is

Β

Thus, the angular velocities for simple rotations are also simple angular velocities. Now, angular velocity vectors add in the following way: the angular velocity of Fb with respect to Fi is equal to the sum of the angular velocity of Fb with respect to Fi , the angular velocity of Fi with respect to Fi , and the angular velocity of Fi with respect to Fi. Mathematically,

However, this expression involves vectors, which are mathematically abstract objects. In order to do computations involving angular velocities, we must choose a reference frame, and express all these vectors in that reference frame and add them together. In different reference frames. To add them, we must transform them all to the same frame. In most attitude dynamics applications, we use the body frame, so for this example, we develop the expression

Β

Now we have all the angular velocity vectors of Eq. (3.65) expressed in Fb and can add them together:

Where = [1 2 3]T , and S() obviously depends on which Euler angle sequence is used. For a given Euler angle sequence, it is relatively straightforward to develop the appropriate S(). Often it is clear what angular velocity vector and reference frame we are working with, and we drop the sub and superscripts on !. Thus if the Β΅_'s are known, then they can be integrated to determine the Β΅'s, and then the components of ~! can be determined. This integration is entirely analogous to knowing x_ , y_, and z_, and integrating these to determine the position x, y, and z. In this translational case however, one usually knows x_ , etc., from determining the velocity; i.e., v = [x_ y_ z_]T . The velocity is determined from the kinetics equations of motion as in Eq. (3.3). Similarly, for rotational motion, the kinetics equations of motion are used to determine the angular velocity, which is in turn used to determine the Β΅_'s, not vice versa. In the next section, we develop relationships between the time derivatives of the attitude variables and the angular velocity.

Rotation of a rigid body

An ordinary object does not simply rotate, it wobbles, shakes, and bends, so to simplify matters we shall discuss the motion of a non-existent ideal object which we call aΒ rigid body. This means an object in which the forces between the atoms are so strong, and of such character, that the little forces that are needed to move it do not bend it. Its shape stays essentially the same as it moves about. If we wish to study the motion of such a body, and agree to ignore the motion of its center of mass, there is only one thing left for it to do, and that is toΒ turn. We have to describe that. Suppose there is some line in the body which stays put (perhaps it includes the center of mass and perhaps not), and the body is rotating about this particular line as an axis. How do we define the rotation? That is easy enough, for if we mark a point somewhere on the object, anywhere except on the axis, we can always tell exactly where the object is, if we only know where this point has gone to. The only thing needed to describe the position of that point is anΒ angle. So rotation consists of a study of the variations of the angle with time.

Β

In order to study rotation, we observe the angle through which a body has turned. Of course, we are not referring to any particular angleΒ insideΒ the object itself; it is not that we draw some angleΒ onΒ the object. We are talking about theΒ angular change of the positionΒ of the whole thing, from one time to another.

Β

First, let us study the kinematics of rotations. The angle will change with time, and just as we talked about position and velocity in one dimension, we may talk about angular position and angular velocity in plane rotation. In fact, there is a very interesting relationship between rotation in two dimensions and one-dimensional displacement, in which almost every quantity has its analog. First, we have the angleΒ ΞΈΒ which defines how far the body has goneΒ around; this replaces the distanceΒ s, which defines how far it has goneΒ along. In the same manner, we have a velocity of turning,Β Ο=dΞΈ/dt, which tells us how much the angle changes in a second, just asΒ v=ds/dtΒ describes how fast a thing moves, or how far it moves in a second. If the angle is measured in radians, then the angular velocityΒ ΟΒ will be so and so many radians per second. The greater the angular velocity, the faster the object is turning, the faster the angle changes. We can go on: we can differentiate the angular velocity with respect to time, and we can callΒ Ξ±=Β dΟ/dt=Β d2ΞΈ/dt2Β the angular acceleration. That would be the analog of the ordinary acceleration.

Β

Now of course we shall have to relate the dynamics of rotation to the laws of dynamics of the particles of which the object is made, so we must find out how a particular particle moves when the angular velocity is such and such. To do this, let us take a certain particle which is located at a distanceΒ rΒ from the axis and say it is in a certain locationΒ P(x,y)Β at a given instant, in the usual manner (Fig.Β 18β1). If at a momentΒ ΞtΒ later the angle of the whole object has turned throughΒ ΞΞΈ, then this particle is carried with it. It is at the same radius away fromΒ OΒ as it was before, but is carried toΒ Q. The first thing we would like to know is how much the distanceΒ xΒ changes and how much the distanceΒ yΒ changes. IfΒ OPΒ is calledΒ r, then the lengthΒ PQΒ isΒ rΞΞΈ, because of the way angles are defined. The change inΒ x, then, is simply the projection ofΒ rΞΞΈΒ in theΒ x-direction:

It should not be mysterious that the value of the magnitude of this velocity isΒ Οr; in fact, it should be self-evident, because the distance that it moves isΒ rΞΞΈΒ and the distance it moves per second isΒ r ΞΞΈ/Ξt, orΒ rΟ.

3 dimensional

Translation

rA/B Β remains constant and therefore its time derivative is zero. Β Thus, all points in the body have the same velocity and the same acceleration.

Β

Fixed axis theorem

Any point such as A which is not on the axis moves in a circular arc in a plane normal to the axis and has a velocity and an acceleration

Β

ROTATION ABOUT A FIXED POINT

Β

When a body rotates about a fixed point, the angular velocity vector no longer remains fixed in direction, and this change calls for a more general concept of rotation.

Rotation and Proper Vectors

Consider a solid sphere which is cut from a rigid body confined to rotate about the fixed point P. The x-y-zaxis here are taken as fixed in space and do not rotate with the body.

Β

In part (a) of the figure, two successive 90orotations of the sphere about, first, the x-axis and second, the y axis result in the motion of a point which is initially on the y-axis in position 1, to positions 2 and 3, successively.

On the other hand, if the order of the rotations is reversed, the point undergoes no motion during the y-rotation but moves to point 3during the 90orotation about x-axis. Thus, the two cases do not produce the same final position, and it is evident from this one special example that finite rotations do not generally obey the parallelogram law of vector addition and are not commutative. (a+bβ b+a)

Thus, finite rotations may not be treated as proper vectors.

Infinitesimal rotations do obey the parallelogram law of vector addition. Infinitesimal rotations are dπ1 and dπ1

Β

Β

Suppose thatΒ Ξ³:[0,1]βSO(3)Ξ³:[0,1]βSOβ‘(3)Β is a path in the space of orientation preserving rotations ofΒ R3R3. It is classical that we can find a correspondingΒ Ο:[0,1]βR3Ο:[0,1]βR3, the angular velocity, such that ifΒ r(t)=Ξ³(t)r(0)r(t)=Ξ³(t)r(0), thenΒ rβ²(t)=Ο(t)Γr(t)rβ²(t)=Ο(t)Γr(t)Β (whereΒ ΓΓΒ is the cross product). Phrased differently,Β d/dt (Ξ³(t))Ξ³β1(t) Β is skew-symmetric with respect to an orthonormal basis. The correspondence between these two statements comes from the formula

It is worth pointing out that this formula underlies an isomorphism of Lie algebrasΒ (R3,Γ)β so3(R3,Γ)β so3, given byΒ xβ¦adx

Assuming whatever niceness we need (likely just continuity), givenΒ Ο(t)Ο(t), there is a unique solution to the differential equation

In Dynamics it is often useful to describe motion relative to a coordinate frame that moves and that rotatesβthat is, not by referring to an inertial or Newtonian frame that is fixed to the Earth. This can happen if itβs needed to describe or analyse motion of components of a vehicle, that itself is moving in a fixed frame. Here we develop the so-called five-term acceleration equation in two dimensions, for the purpose of easier visualization. What is developed here can easily be extended to 3-D. This is one of the very nice results in Dynamics, because different types of acceleration come from this derivation, and the derivation itself simply involves repeated use of the product rule when differentiating vectors.

We start by describing the motion to be analysed, which involves every possibility of motion when there are two framesβone stationary and one moving and rotatingβused to describe the motion. The figure shows the scenario

A conventional 2-D kinematic slab is shown. It is moving in a fixed reference frame, denoted with capital letters. On the slab is fixed a small-letter frame with A as its origin. This frame moves with the slab. Besides moving around in XY-space, the slab is also rotating at an angular speed Ξ©, which is also changing at a rate Ξ©Μ. On the moving, rotating slab another point B has a motion of its own described in the little-letter coordinate system. Thus B moves relative to A, and this motion is described relative to the slab.

Β

Position

Β

We are interested in the motion of B in the XY-frame, since the absolute motion is what Newtonβs Second Law applies to. We start with

Β

πβπ΅ = πβπ΄ + πβπ΅/π΄

Β

To see how πβπ΅ changes with time, it is useful to break it up into magnitudes and directions using unit vectors.

Β

Thus Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β πβπ΅ = ππ΄ππΌΜ + ππ΄ππ½Μ + ππ΅/π΄βπ₯ πΜ + ππ΅/π΄βπ¦ π Μ

Β

Since πβπ΅/π΄ is a vector described in the little-letter coordinate system, it is more natural to express it in terms of πΜ and πΜ components.

Velocity

Β

By definition π£βπ΅ = πβΜπ΅

Β

We use the product rule to differentiateΒ Β πβ Β So

Β

π£βπ΅ = πβΜπ΅= πΜπ΄ππΌΜ + ππ΄ππΌΜΜ + πΜπ΄ππ½Μ + ππ΄ππ½ΜΜ + πΜπ΅ /π΄ βπ₯ πΜ + ππ΅ /π΄ βπ₯πΜΜ+ πΜπ΅ /π΄βπ¦ πΜ + ππ΅ /π΄ βπ¦ πΜΜ

Β

All πΜ are changes in the magnitudes of the corresponding π vectors, which are speed-ups or slow-downs. They are velocities, by definition. Thus

Β

π£βπ΅ = π£π΄ππΌΜ + π£π΄ππ½Μ + π£π΅/π΄ βπ₯ πΜ + π£π΅ /π΄βπ¦ πΜ + ππ΅/π΄βπ₯πΜΜ+ ππ΅/π΄βπ¦ πΜΜ

Β

Now we need to find expressions for πΜΜ and πΜΜ Since they are unit vectors, they do not change length; they only change direction. They are fixed to the rotating slab AB Β so their changing depends solely on the slab rotation. The figure below shows this rotation in a magnified view.

Β

Β

After a short βt, the slab has turned a small angle βπ counter-clockwise. The change in the unit vector πΜ is shown as ΞπΜ . The change in πΜ is shown as ΞπΜ. As Ξt is made shorter and shorter, ΞπΜ becomes perpendicular to πΜ , that is, rotated 90Β° counter-clockwise to πΜ , thus in the πΜ-direction. ΞπΜ is directed in the βπΜ-direction. In the limit, the magnitude of ΞπΜ approaches the arc length. Since πΜ is a unit vector, the arc length is 1*βπ So ΞπΜ = Ξπ πΜ. From this, in the limit,

πΜΜ= ΞπΜ / Ξπ‘ = Ξπ / Ξπ‘ πΜ = Ξ© πΜ

AndΒ Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β πΜΜ= ΞπΜ / Ξπ‘ = Ξπ / Ξπ‘ (β π) = βΞ© πΜ

Β

Now we can substitute these into the equation for π£β .

Β

π£βπ΅ = π£π΄ππΌΜ + π£π΄ππ½Μ + π£π΅/π΄βπ₯ πΜ + π£π΅/π΄βπ¦ πΜ + ππ΅/π΄βπ₯ Ξ© πΜ β ππ΅/π΄βπ¦ Ξ© πΜ

Β

Reorganizing

Β

π£βπ΅ = π£π΄ππΌΜ + π£π΄ππ½Μ + (π£π΅/π΄βπ₯ β ππ΅/π΄βπ¦ Ξ©) πΜ + (π£π΅/π΄βπ¦ + ππ΅/π΄βπ₯ Ξ©) π Μ

Β

On the right-hand side, after the first two terms, all of the terms have βB/Aβ in them. This indicates that the location and velocity of B is expressed in the little-letter coordinate system. Since we already have a lower-case x or y in the subscript, the βB/Aβ is somewhat redundant, and we shall dispense with it. Thus

Β

π£βπ΅ = π£π΄π πΌΜ + π£π΄π π½Μ + (π£π₯ β ππ¦ Ξ©) πΜ+ (π£π¦ + ππ₯ Ξ©) πΜ

Β

Note that there are two velocities in the little-letter system. In the term π£π₯ β ππ¦ Ξ©, for example, π£π₯ is the rate of lengthening of the πβπ₯ vector, while ππ₯ Ξ© is a tangential velocity due purely to the rotation of the slab. Note that even if B were fixed to the plate, so that π£π₯ = 0 Β B still has a velocity due to the rotation of the plate and the location of B away from the xy-origin.

We can use this equation to reconstruct a vector version of π£βπ΅ that does not include unit vectors.

π£βπ΅ = π£βπ΄ + π£βπ₯π¦ + Ξ©β Γ πβπ₯π¦

Since Ξ©β Γ πβπ₯π¦ is due to Ξ©β, we can call this velocity π£βΞ© Then

Β

π£βπ΅ = π£βπ΄ + π£βπ₯π¦ + π£βΞ©

Β

This is the three-term velocity equation. Recall that the subscript xy is a shorthand for B/A . If B is a fixed point on a component, then vxy = 0 . The term Ξ©βΓ πβπ₯π¦ is thus π£βπ΅/ . Thus the derived formula is the well-known relative velocity formulae

Β

π£βπ΅ = π£βπ΄ + π£βπ΅/π΄

Β

π£βπ΅/π΄ = πβπ΄π΅ Γ πβπ΅/π΄

Β

Weβve seen so far how the three-term velocity equation results from vector differentiation. A physical picture of what is at play is shown in the figure below.

π£βπ΄ is just the translational movement of the xy-origin in the XY-frame. π£βπ₯π¦ is the translational motion of point B in the xy-frame, that is, on the slab. π£βΞ© is the motion of B due to the rotation of the slab.

Β

Acceleration

Β

By definition πβπ΅ = π£βΜπ΅

We start with the three-term velocity equation in its detailed form above, so that we can see what accelerations result from what differentiations.

π£βπ΅ = π£π΄ππΌΜ + π£π΄ππ½Μ + (π£π΅/π΄βπ₯ β ππ΅/π΄βπ¦ Ξ©) πΜ + (π£π΅/π΄βπ¦ + ππ΅/π΄βπ₯ Ξ©) π Μ

π£βΜπ΅ = π£Μπ΄π πΌΜ + π£π΄π πΌΜΜ+ π£Μπ΄π π½Μ+π£π΄π π½ΜΜ+ (π£Μπ₯ β πΜπ¦ Ξ© β ππ¦ Ξ©Μ) πΜ+ (π£π₯ β ππ¦ Ξ©) πΜΜ+ (π£Μπ¦ + πΜπ₯ Ξ© + ππ₯ Ξ©Μ) πΜ + (π£π¦ + ππ₯ Ξ©) πΜ

Β

Again πΌΜ and π½Μ belong to the fixed frame, so they donβt change and are 0. Also we can substitute in the results for πΜΜ and πΜΜ previously calculated. πΜ is π£, and π£Μ is Β .

Β

π£βΜπ΅ = ππ΄π πΌΜ + ππ΄π π½Μ + (ππ₯ β π£π¦ Ξ© β ππ¦ Ξ©Μ) πΜ+ (π£π₯ β ππ¦ Ξ©) Ξ© πΜ+ (ππ¦ + π£π₯ Ξ© + ππ₯ Ξ©Μ) πΜ + (π£π¦ + ππ₯ Ξ©)(βΞ© πΜ)

Β

πβπ΅ = ππ΄π πΌΜ + ππ΄ππ½Μ + (ππ₯ β 2π£π¦ Ξ© β ππ¦ Ξ©Μβ ππ₯ Ξ©2) πΜ+ (ππ¦ + 2π£π₯ Ξ© + ππ₯ Ξ©Μβ ππ¦ Ξ©2) πΜ

Β

This is the five-term acceleration equation.

Β

The five accelerations, though they arose simply from applying the product rule to vector differentiation, can be ascribed definitive physical meaning. Letβs define them first with a shorthand notation and then discuss their characteristics.

Β

1. πβπ΄ = ππ΄π πΌΜ + ππ΄π π½Μ, the acceleration of point A in the fixed XY-frame

2. πβπ₯π¦ = ππ₯ πΜ+ ππ¦ πΜ, the acceleration of B within the moving, rotating xy-frame

3. πβπΆ = β2π£π¦ Ξ© πΜ+ 2π£π₯ Ξ© πΜ, the Coriolis acceleration of B

4. πβπ‘ = βππ¦ Ξ©Μ πΜ+ ππ₯ Ξ©Μ πΜ, the tangential acceleration of B

5. πβπ = βππ₯ Ξ©2 πΜβ ππ¦ Ξ©2 πΜ, the normal acceleration of B

Β

So πβπ΅ = πβπ΄ + πβπ₯π¦ + πβπΆ + πβπ‘ + πβπ

Β

πβπ΄ And πβπ₯π¦ are both independent accelerations and do not involve the rotation of the slab. Normally in an analysis they will have to be given or observed. The other three accelerations do involve the rotation of the slab. πβπ‘ involves a speed-up of the tangential velocity, π£βΞ©, simply because the spin speed is increasingβ¦or a slow-down in π£βΞ© because the spin speed is decreasing. πβπ is due to the constant change in direction of π£βΞ© as it seemingly rotates about A. Note that this is directed opposite to πβπ₯π¦ Β which is πβπ΅/ . The third term, Coriolis acceleration, is the most puzzling and misunderstood of the five. It results from a combination of movements: movement on the slab while it is rotating. It was not even known until long after Newton developed his three laws and Euler extended them to rigid bodies.

The three accelerations involving Ξ©β can also be cast as cross products.

Β

3. πβπΆ = 2 Ξ©β Γ π£βπ₯π¦ the Coriolis acceleration of B

4. πβπ‘ = Ξ©β Μ Γ πβπ₯π¦, the tangential acceleration of B

5. πβπ = Ξ©β Γ (Ξ©β Γ πβπ₯π¦), the normal acceleration of B

Β

That these are equivalent to the unit-vector versions above can be seen as follows. The magnitude equivalency can be verified by inspection. The directions can be seen in that in 2-D, positive Ξ©β and Ξ©β point out of the slab. Pre-crossing them with a vector produces a vector that is rotated 90Β° to the right of the second vector in the cross-product. Also, since both Ξ©β and Ξ©β Μ point in the π Μ direction (if they are positive), they will convert πΜ-components into πΜ-components and πΜ-components into βπΜ-components. Note that πβπΆ and πβπ‘ have y-components in the βπΜ-direction and x-components in the πΜ-direction. With πβ , there is a double pre-crossing by Ξ©β . Thus the direction is turned 180Β° around from πβπ₯π¦.

In the second pre-cross, the x-components that had been turned into the πΜ-direction get turned another 90Β° into the βπΜ-direction. Note that in the unit-vector version of πβπ x-components stay as πΜ-components, and y-components stay as πΜ- components. But both are now negative

Β

Β

The figure above shows an example of the five accelerations. πβπ΄ and πβπ₯π¦ are independent and could be drawn in any direction. The Coriolis acceleration is perpendicular and to the left of π£βπ₯π¦ since the slab is rotating counter-clockwise. πβπ‘ is perpendicular and to the left of the line from A to B since Ξ©β Μ is counter-clockwise. πβπ leads from B back toward the apparent center of rotation at A .

Though this is explained in 2-D for visualization purposed, it is equally valid for 3-D.

Also noteworthy is that the center of rotation of the slab does not have to be at A . It does look to an observer at A as if the slab is rotating about him. But the rotation point does not even have to be fixed. It could change with time, like an instant center does, in the general case. It could even be, if we take the special case of B not moving on the slab, that B is the center of rotation. But all of the above is still true, and we can center our rotating coordinate system at any point on the slab. I would like to work out this counter-case at some point, just to demonstrate it.

MODULE 3

KINEMATICS OF RIGID BODY

A rigid body is an ideal body that does not change its shape when subjected to external forces or couples. The distance between the particles in a rigid body remains same during the motion. Hence, a rigid body can be defined as the group or collection of number of particles at fixed distance from each other.

There are three types of motion of a rigid body

- Translation
- Rotation
- General Plane Motion

Kinematics of rigid body in translation motion

Translation motion occurs when a line (generally axis) in the body remains parallel to its initial orientation throughout the motion.

If the path of motion of any two lines on the rigid body is parallel or is in a straight line, then the translation is termed as rectilinear translation.

If the path of motion of any two lines on the rigid body is along the fixed curve, then the translation is termed as curvilinear translation.

Consider a rigid body that is subjected to translation motion as shown in fig.

Β

The position of points A and B is defined are defined by the position vectors and w.r.t the fixed frame x, y.

Let the translating system is fixed in a body with origin A which is referred to the base point. The position vector of B w.r.t A is denoted by .

By vector addition, we get,

Taking, the first time derivative of the above equation, we get,

According to the definition of rigid body, the term is constant

Hence,

And, gives the absolute velocities and respectively with respect to the fixed frame.

So, we get,

This shows that, for the rigid body in translation motion, the velocity of every particle on the body is same.

Now, taking the time derivative of the velocity equation, it simply yields

Where, and are the absolute accelerations of points A and B on the rigid body w.r.t fixed frame.

This shows that, for the rigid body in translation motion, the velocity and acceleration of every particle on the body is same.

Kinematics of a body rotating about fixed axis

When a rigid body is subjected to the rotation about a fixed axis, as shown in fig below, all the particles of the body, except those which lie on the axis of rotation, move along circular paths whose center lies on the axis of rotation.

Consider, the point P on the rigid body defined by the position vector r. As the rigid body rotates about the fixed axis, then the point P travels a circular path of radius r with center point O as shown in fig below

If the body rotates with a differential angular displacement , then the point P displaces the distance , which is given by

Differentiating the above equation w.r.t time, we get

The term, is referred to as the angular velocity of the rotating body

And gives the velocity of point P

Hence,

The direction of velocity is tangent to the circular path.

Considering the magnitude and direction, velocity is the cross product of angular velocity and the position vector

Where, is directed from axis to the point P

To find acceleration, differentiate velocity vector w.r.t time

Here, the term refers to the angular acceleration of the rigid body

And is the velocity

Hence,

Where, is the tangential acceleration and is normal component of acceleration Hence,

Magnitude of both components is given by,

Tangential component refers to the time rate of change of velocityβs magnitude. Hence, if velocity of point P is increasing, is in same direction of , but if velocity of point P is decreasing, is in direction opposite to that of

The normal component of acceleration refers to time rate of change of the velocityβs direction. Hence, the direction of the normal component is always from point P towards the center O. Hence it is also called as centripetal acceleration.

Β

Rotation of a Body about a Fixed Axis

As a preliminary, letβs look at a body firmly attached to a rod fixed in space, and rotating with angular velocityΒ Ξ©Β radians/sec. About that axis.Β Youβll recall from freshman physics that the angular momentum and rotational energy are

But you also know that both angular velocity and angular momentum are vectors. Obviously, for this example, the angular velocity is a vector pointing along the axis of rotation,Β Ξ©β= (0, 0,Ξ© z).Β One might be tempted to conclude that the angular momentum also points along the axis, butΒ this is not always the case. An instructive example is provided by two massesΒ mΒ at the ends of a rod of lengthΒ 2aΒ held at a fixed angleΒ ΞΈΒ to theΒ zΒ axis, which is the axis of rotation.

Lz =2ma2sin2ΞΈβ Ξ©.

But notice that, assuming the rod is momentarily in theΒ xzΒ plane, as shown, then

Lx=β2ma2cos2ΞΈβ Ξ©.

The total angularΒ momentumΒ is not parallel to the total angularΒ velocity!Β

In fact, as should be evident, the total angular momentum is rotating around the constant angular velocity vector, so the axis must be providing a torque. This is why unbalanced car wheels stress the axle.

General Motion of a Rotating Rigid Body

Weβll follow the Landau notation (which itself tends to be bilingual between coordinatesΒ (x,y,z)Β andΒ (x1,x2,x3).) Β Notice that weβll label the components byΒ (x1, x2, x3), notΒ (r1, r2, r3)Β even though we callΒ the vectorΒ rβ.Β Β Again, weβre following Landau.

We take a fixed, inertial (or lab) coordinate system labelledΒ (X,Y,Z)Β and in this system the rigid bodyβs center of mass, labelledΒ O, is atΒ Rβ.Β We have a Cartesian set of axes fixed in the body, origin at the center of mass, and coordinates in this system,Β vectors fromΒ OβΒ to a point in the body denoted byΒ , are labelledΒ (x,y,z)Β orΒ (x1,x2,x3).

A vector from theΒ externalΒ inertial fixed origin to a point in the body is then

Suppose now that in infinitesimal timeΒ dt, the center of mass of the body movesΒ Β and the body rotates through.Β Then a particle atΒ Β as measured from the center of mass will move through

Relative to the external inertial frame.

Therefore, the velocity of that particle in the fixed frame, writing the center of mass velocity and the angular velocity as

Β

Now, in deriving the above equation, we have not used the fact that the originΒ OΒ fixed in the body is at the center of mass. (That turns out to be useful shortly.)Β What if instead we had taken some other originΒ Oβ²Β fixed in the body? Would we find the angular velocityΒ Β Β aboutΒ Oβ²Β to be the same asΒ The answer turns out to beΒ yes, but we need to prove it! Here's the proof:

If the position ofΒ Oβ²Β relative toΒ OΒ is Β (a vector fixed in the body and so moving with it) then the velocityΒ ofΒ Oβ²Β is

Β

This means that if we describe the motion of any particle in the body in terms of some origin fixed in the body, plus rotation about that origin,Β the angular velocity vector describing the bodyβs motion is the same irrespective of the origin we choose.Β So we can, without ambiguity, talk about the angular velocity of the body. Β

Angular Velocity

Β

The angular velocity of Fi0 with respect to Fi is

Β

Note the ordering of the superscripts in this expression. Also, note that this vector

Quantity has the same components in either frame; that is,

Β

Β

This simple expression results because it is a simple rotation. For the 2" rotation from FiΒ to Fi , the angular velocity vector is

Β

Finally, for the 1" rotation from Fi00 to Fb, the angular velocity vector is

Β

Thus, the angular velocities for simple rotations are also simple angular velocities. Now, angular velocity vectors add in the following way: the angular velocity of Fb with respect to Fi is equal to the sum of the angular velocity of Fb with respect to Fi , the angular velocity of Fi with respect to Fi , and the angular velocity of Fi with respect to Fi. Mathematically,

However, this expression involves vectors, which are mathematically abstract objects. In order to do computations involving angular velocities, we must choose a reference frame, and express all these vectors in that reference frame and add them together. In different reference frames. To add them, we must transform them all to the same frame. In most attitude dynamics applications, we use the body frame, so for this example, we develop the expression

Β

Now we have all the angular velocity vectors of Eq. (3.65) expressed in Fb and can add them together:

Where = [1 2 3]T , and S() obviously depends on which Euler angle sequence is used. For a given Euler angle sequence, it is relatively straightforward to develop the appropriate S(). Often it is clear what angular velocity vector and reference frame we are working with, and we drop the sub and superscripts on !. Thus if the Β΅_'s are known, then they can be integrated to determine the Β΅'s, and then the components of ~! can be determined. This integration is entirely analogous to knowing x_ , y_, and z_, and integrating these to determine the position x, y, and z. In this translational case however, one usually knows x_ , etc., from determining the velocity; i.e., v = [x_ y_ z_]T . The velocity is determined from the kinetics equations of motion as in Eq. (3.3). Similarly, for rotational motion, the kinetics equations of motion are used to determine the angular velocity, which is in turn used to determine the Β΅_'s, not vice versa. In the next section, we develop relationships between the time derivatives of the attitude variables and the angular velocity.

Rotation of a rigid body

An ordinary object does not simply rotate, it wobbles, shakes, and bends, so to simplify matters we shall discuss the motion of a non-existent ideal object which we call aΒ rigid body. This means an object in which the forces between the atoms are so strong, and of such character, that the little forces that are needed to move it do not bend it. Its shape stays essentially the same as it moves about. If we wish to study the motion of such a body, and agree to ignore the motion of its center of mass, there is only one thing left for it to do, and that is toΒ turn. We have to describe that. Suppose there is some line in the body which stays put (perhaps it includes the center of mass and perhaps not), and the body is rotating about this particular line as an axis. How do we define the rotation? That is easy enough, for if we mark a point somewhere on the object, anywhere except on the axis, we can always tell exactly where the object is, if we only know where this point has gone to. The only thing needed to describe the position of that point is anΒ angle. So rotation consists of a study of the variations of the angle with time.

Β

In order to study rotation, we observe the angle through which a body has turned. Of course, we are not referring to any particular angleΒ insideΒ the object itself; it is not that we draw some angleΒ onΒ the object. We are talking about theΒ angular change of the positionΒ of the whole thing, from one time to another.

Β

First, let us study the kinematics of rotations. The angle will change with time, and just as we talked about position and velocity in one dimension, we may talk about angular position and angular velocity in plane rotation. In fact, there is a very interesting relationship between rotation in two dimensions and one-dimensional displacement, in which almost every quantity has its analog. First, we have the angleΒ ΞΈΒ which defines how far the body has goneΒ around; this replaces the distanceΒ s, which defines how far it has goneΒ along. In the same manner, we have a velocity of turning,Β Ο=dΞΈ/dt, which tells us how much the angle changes in a second, just asΒ v=ds/dtΒ describes how fast a thing moves, or how far it moves in a second. If the angle is measured in radians, then the angular velocityΒ ΟΒ will be so and so many radians per second. The greater the angular velocity, the faster the object is turning, the faster the angle changes. We can go on: we can differentiate the angular velocity with respect to time, and we can callΒ Ξ±=Β dΟ/dt=Β d2ΞΈ/dt2Β the angular acceleration. That would be the analog of the ordinary acceleration.

Β

Now of course we shall have to relate the dynamics of rotation to the laws of dynamics of the particles of which the object is made, so we must find out how a particular particle moves when the angular velocity is such and such. To do this, let us take a certain particle which is located at a distanceΒ rΒ from the axis and say it is in a certain locationΒ P(x,y)Β at a given instant, in the usual manner (Fig.Β 18β1). If at a momentΒ ΞtΒ later the angle of the whole object has turned throughΒ ΞΞΈ, then this particle is carried with it. It is at the same radius away fromΒ OΒ as it was before, but is carried toΒ Q. The first thing we would like to know is how much the distanceΒ xΒ changes and how much the distanceΒ yΒ changes. IfΒ OPΒ is calledΒ r, then the lengthΒ PQΒ isΒ rΞΞΈ, because of the way angles are defined. The change inΒ x, then, is simply the projection ofΒ rΞΞΈΒ in theΒ x-direction:

It should not be mysterious that the value of the magnitude of this velocity isΒ Οr; in fact, it should be self-evident, because the distance that it moves isΒ rΞΞΈΒ and the distance it moves per second isΒ r ΞΞΈ/Ξt, orΒ rΟ.

3 dimensional

Translation

rA/B Β remains constant and therefore its time derivative is zero. Β Thus, all points in the body have the same velocity and the same acceleration.

Β

Fixed axis theorem

Any point such as A which is not on the axis moves in a circular arc in a plane normal to the axis and has a velocity and an acceleration

Β

ROTATION ABOUT A FIXED POINT

Β

When a body rotates about a fixed point, the angular velocity vector no longer remains fixed in direction, and this change calls for a more general concept of rotation.

Rotation and Proper Vectors

Consider a solid sphere which is cut from a rigid body confined to rotate about the fixed point P. The x-y-zaxis here are taken as fixed in space and do not rotate with the body.

Β

In part (a) of the figure, two successive 90orotations of the sphere about, first, the x-axis and second, the y axis result in the motion of a point which is initially on the y-axis in position 1, to positions 2 and 3, successively.

On the other hand, if the order of the rotations is reversed, the point undergoes no motion during the y-rotation but moves to point 3during the 90orotation about x-axis. Thus, the two cases do not produce the same final position, and it is evident from this one special example that finite rotations do not generally obey the parallelogram law of vector addition and are not commutative. (a+bβ b+a)

Thus, finite rotations may not be treated as proper vectors.

Infinitesimal rotations do obey the parallelogram law of vector addition. Infinitesimal rotations are dπ1 and dπ1

Β

Β

Suppose thatΒ Ξ³:[0,1]βSO(3)Ξ³:[0,1]βSOβ‘(3)Β is a path in the space of orientation preserving rotations ofΒ R3R3. It is classical that we can find a correspondingΒ Ο:[0,1]βR3Ο:[0,1]βR3, the angular velocity, such that ifΒ r(t)=Ξ³(t)r(0)r(t)=Ξ³(t)r(0), thenΒ rβ²(t)=Ο(t)Γr(t)rβ²(t)=Ο(t)Γr(t)Β (whereΒ ΓΓΒ is the cross product). Phrased differently,Β d/dt (Ξ³(t))Ξ³β1(t) Β is skew-symmetric with respect to an orthonormal basis. The correspondence between these two statements comes from the formula

It is worth pointing out that this formula underlies an isomorphism of Lie algebrasΒ (R3,Γ)β so3(R3,Γ)β so3, given byΒ xβ¦adx

Assuming whatever niceness we need (likely just continuity), givenΒ Ο(t)Ο(t), there is a unique solution to the differential equation

In Dynamics it is often useful to describe motion relative to a coordinate frame that moves and that rotatesβthat is, not by referring to an inertial or Newtonian frame that is fixed to the Earth. This can happen if itβs needed to describe or analyse motion of components of a vehicle, that itself is moving in a fixed frame. Here we develop the so-called five-term acceleration equation in two dimensions, for the purpose of easier visualization. What is developed here can easily be extended to 3-D. This is one of the very nice results in Dynamics, because different types of acceleration come from this derivation, and the derivation itself simply involves repeated use of the product rule when differentiating vectors.

We start by describing the motion to be analysed, which involves every possibility of motion when there are two framesβone stationary and one moving and rotatingβused to describe the motion. The figure shows the scenario

A conventional 2-D kinematic slab is shown. It is moving in a fixed reference frame, denoted with capital letters. On the slab is fixed a small-letter frame with A as its origin. This frame moves with the slab. Besides moving around in XY-space, the slab is also rotating at an angular speed Ξ©, which is also changing at a rate Ξ©Μ. On the moving, rotating slab another point B has a motion of its own described in the little-letter coordinate system. Thus B moves relative to A, and this motion is described relative to the slab.

Β

Position

Β

We are interested in the motion of B in the XY-frame, since the absolute motion is what Newtonβs Second Law applies to. We start with

Β

πβπ΅ = πβπ΄ + πβπ΅/π΄

Β

To see how πβπ΅ changes with time, it is useful to break it up into magnitudes and directions using unit vectors.

Β

Thus Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β πβπ΅ = ππ΄ππΌΜ + ππ΄ππ½Μ + ππ΅/π΄βπ₯ πΜ + ππ΅/π΄βπ¦ π Μ

Β

Since πβπ΅/π΄ is a vector described in the little-letter coordinate system, it is more natural to express it in terms of πΜ and πΜ components.

Velocity

Β

By definition π£βπ΅ = πβΜπ΅

Β

We use the product rule to differentiateΒ Β πβ Β So

Β

π£βπ΅ = πβΜπ΅= πΜπ΄ππΌΜ + ππ΄ππΌΜΜ + πΜπ΄ππ½Μ + ππ΄ππ½ΜΜ + πΜπ΅ /π΄ βπ₯ πΜ + ππ΅ /π΄ βπ₯πΜΜ+ πΜπ΅ /π΄βπ¦ πΜ + ππ΅ /π΄ βπ¦ πΜΜ

Β

All πΜ are changes in the magnitudes of the corresponding π vectors, which are speed-ups or slow-downs. They are velocities, by definition. Thus

Β

π£βπ΅ = π£π΄ππΌΜ + π£π΄ππ½Μ + π£π΅/π΄ βπ₯ πΜ + π£π΅ /π΄βπ¦ πΜ + ππ΅/π΄βπ₯πΜΜ+ ππ΅/π΄βπ¦ πΜΜ

Β

Now we need to find expressions for πΜΜ and πΜΜ Since they are unit vectors, they do not change length; they only change direction. They are fixed to the rotating slab AB Β so their changing depends solely on the slab rotation. The figure below shows this rotation in a magnified view.

Β

Β

After a short βt, the slab has turned a small angle βπ counter-clockwise. The change in the unit vector πΜ is shown as ΞπΜ . The change in πΜ is shown as ΞπΜ. As Ξt is made shorter and shorter, ΞπΜ becomes perpendicular to πΜ , that is, rotated 90Β° counter-clockwise to πΜ , thus in the πΜ-direction. ΞπΜ is directed in the βπΜ-direction. In the limit, the magnitude of ΞπΜ approaches the arc length. Since πΜ is a unit vector, the arc length is 1*βπ So ΞπΜ = Ξπ πΜ. From this, in the limit,

πΜΜ= ΞπΜ / Ξπ‘ = Ξπ / Ξπ‘ πΜ = Ξ© πΜ

AndΒ Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β πΜΜ= ΞπΜ / Ξπ‘ = Ξπ / Ξπ‘ (β π) = βΞ© πΜ

Β

Now we can substitute these into the equation for π£β .

Β

π£βπ΅ = π£π΄ππΌΜ + π£π΄ππ½Μ + π£π΅/π΄βπ₯ πΜ + π£π΅/π΄βπ¦ πΜ + ππ΅/π΄βπ₯ Ξ© πΜ β ππ΅/π΄βπ¦ Ξ© πΜ

Β

Reorganizing

Β

π£βπ΅ = π£π΄ππΌΜ + π£π΄ππ½Μ + (π£π΅/π΄βπ₯ β ππ΅/π΄βπ¦ Ξ©) πΜ + (π£π΅/π΄βπ¦ + ππ΅/π΄βπ₯ Ξ©) π Μ

Β

On the right-hand side, after the first two terms, all of the terms have βB/Aβ in them. This indicates that the location and velocity of B is expressed in the little-letter coordinate system. Since we already have a lower-case x or y in the subscript, the βB/Aβ is somewhat redundant, and we shall dispense with it. Thus

Β

π£βπ΅ = π£π΄π πΌΜ + π£π΄π π½Μ + (π£π₯ β ππ¦ Ξ©) πΜ+ (π£π¦ + ππ₯ Ξ©) πΜ

Β

Note that there are two velocities in the little-letter system. In the term π£π₯ β ππ¦ Ξ©, for example, π£π₯ is the rate of lengthening of the πβπ₯ vector, while ππ₯ Ξ© is a tangential velocity due purely to the rotation of the slab. Note that even if B were fixed to the plate, so that π£π₯ = 0 Β B still has a velocity due to the rotation of the plate and the location of B away from the xy-origin.

We can use this equation to reconstruct a vector version of π£βπ΅ that does not include unit vectors.

π£βπ΅ = π£βπ΄ + π£βπ₯π¦ + Ξ©β Γ πβπ₯π¦

Since Ξ©β Γ πβπ₯π¦ is due to Ξ©β, we can call this velocity π£βΞ© Then

Β

π£βπ΅ = π£βπ΄ + π£βπ₯π¦ + π£βΞ©

Β

This is the three-term velocity equation. Recall that the subscript xy is a shorthand for B/A . If B is a fixed point on a component, then vxy = 0 . The term Ξ©βΓ πβπ₯π¦ is thus π£βπ΅/ . Thus the derived formula is the well-known relative velocity formulae

Β

π£βπ΅ = π£βπ΄ + π£βπ΅/π΄

Β

π£βπ΅/π΄ = πβπ΄π΅ Γ πβπ΅/π΄

Β

Weβve seen so far how the three-term velocity equation results from vector differentiation. A physical picture of what is at play is shown in the figure below.

π£βπ΄ is just the translational movement of the xy-origin in the XY-frame. π£βπ₯π¦ is the translational motion of point B in the xy-frame, that is, on the slab. π£βΞ© is the motion of B due to the rotation of the slab.

Β

Acceleration

Β

By definition πβπ΅ = π£βΜπ΅

We start with the three-term velocity equation in its detailed form above, so that we can see what accelerations result from what differentiations.

π£βΜπ΅ = π£Μπ΄π πΌΜ + π£π΄π πΌΜΜ+ π£Μπ΄π π½Μ+π£π΄π π½ΜΜ+ (π£Μπ₯ β πΜπ¦ Ξ© β ππ¦ Ξ©Μ) πΜ+ (π£π₯ β ππ¦ Ξ©) πΜΜ+ (π£Μπ¦ + πΜπ₯ Ξ© + ππ₯ Ξ©Μ) πΜ + (π£π¦ + ππ₯ Ξ©) πΜ

Β

Again πΌΜ and π½Μ belong to the fixed frame, so they donβt change and are 0. Also we can substitute in the results for πΜΜ and πΜΜ previously calculated. πΜ is π£, and π£Μ is Β .

Β

π£βΜπ΅ = ππ΄π πΌΜ + ππ΄π π½Μ + (ππ₯ β π£π¦ Ξ© β ππ¦ Ξ©Μ) πΜ+ (π£π₯ β ππ¦ Ξ©) Ξ© πΜ+ (ππ¦ + π£π₯ Ξ© + ππ₯ Ξ©Μ) πΜ + (π£π¦ + ππ₯ Ξ©)(βΞ© πΜ)

Β

πβπ΅ = ππ΄π πΌΜ + ππ΄ππ½Μ + (ππ₯ β 2π£π¦ Ξ© β ππ¦ Ξ©Μβ ππ₯ Ξ©2) πΜ+ (ππ¦ + 2π£π₯ Ξ© + ππ₯ Ξ©Μβ ππ¦ Ξ©2) πΜ

Β

This is the five-term acceleration equation.

Β

The five accelerations, though they arose simply from applying the product rule to vector differentiation, can be ascribed definitive physical meaning. Letβs define them first with a shorthand notation and then discuss their characteristics.

Β

1. πβπ΄ = ππ΄π πΌΜ + ππ΄π π½Μ, the acceleration of point A in the fixed XY-frame

2. πβπ₯π¦ = ππ₯ πΜ+ ππ¦ πΜ, the acceleration of B within the moving, rotating xy-frame

3. πβπΆ = β2π£π¦ Ξ© πΜ+ 2π£π₯ Ξ© πΜ, the Coriolis acceleration of B

4. πβπ‘ = βππ¦ Ξ©Μ πΜ+ ππ₯ Ξ©Μ πΜ, the tangential acceleration of B

5. πβπ = βππ₯ Ξ©2 πΜβ ππ¦ Ξ©2 πΜ, the normal acceleration of B

Β

So πβπ΅ = πβπ΄ + πβπ₯π¦ + πβπΆ + πβπ‘ + πβπ

Β

πβπ΄ And πβπ₯π¦ are both independent accelerations and do not involve the rotation of the slab. Normally in an analysis they will have to be given or observed. The other three accelerations do involve the rotation of the slab. πβπ‘ involves a speed-up of the tangential velocity, π£βΞ©, simply because the spin speed is increasingβ¦or a slow-down in π£βΞ© because the spin speed is decreasing. πβπ is due to the constant change in direction of π£βΞ© as it seemingly rotates about A. Note that this is directed opposite to πβπ₯π¦ Β which is πβπ΅/ . The third term, Coriolis acceleration, is the most puzzling and misunderstood of the five. It results from a combination of movements: movement on the slab while it is rotating. It was not even known until long after Newton developed his three laws and Euler extended them to rigid bodies.

The three accelerations involving Ξ©β can also be cast as cross products.

Β

3. πβπΆ = 2 Ξ©β Γ π£βπ₯π¦ the Coriolis acceleration of B

4. πβπ‘ = Ξ©β Μ Γ πβπ₯π¦, the tangential acceleration of B

5. πβπ = Ξ©β Γ (Ξ©β Γ πβπ₯π¦), the normal acceleration of B

Β

That these are equivalent to the unit-vector versions above can be seen as follows. The magnitude equivalency can be verified by inspection. The directions can be seen in that in 2-D, positive Ξ©β and Ξ©β point out of the slab. Pre-crossing them with a vector produces a vector that is rotated 90Β° to the right of the second vector in the cross-product. Also, since both Ξ©β and Ξ©β Μ point in the π Μ direction (if they are positive), they will convert πΜ-components into πΜ-components and πΜ-components into βπΜ-components. Note that πβπΆ and πβπ‘ have y-components in the βπΜ-direction and x-components in the πΜ-direction. With πβ , there is a double pre-crossing by Ξ©β . Thus the direction is turned 180Β° around from πβπ₯π¦.

In the second pre-cross, the x-components that had been turned into the πΜ-direction get turned another 90Β° into the βπΜ-direction. Note that in the unit-vector version of πβπ x-components stay as πΜ-components, and y-components stay as πΜ- components. But both are now negative

Β

Β

The figure above shows an example of the five accelerations. πβπ΄ and πβπ₯π¦ are independent and could be drawn in any direction. The Coriolis acceleration is perpendicular and to the left of π£βπ₯π¦ since the slab is rotating counter-clockwise. πβπ‘ is perpendicular and to the left of the line from A to B since Ξ©β Μ is counter-clockwise. πβπ leads from B back toward the apparent center of rotation at A .

Though this is explained in 2-D for visualization purposed, it is equally valid for 3-D.

Also noteworthy is that the center of rotation of the slab does not have to be at A . It does look to an observer at A as if the slab is rotating about him. But the rotation point does not even have to be fixed. It could change with time, like an instant center does, in the general case. It could even be, if we take the special case of B not moving on the slab, that B is the center of rotation. But all of the above is still true, and we can center our rotating coordinate system at any point on the slab. I would like to work out this counter-case at some point, just to demonstrate it.