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Chapter 6
Rigid Body Dynamics
6.1 Introduction
In practice, it is often not possible to idealize a system as a particle. In this section, we construct a more sophisticated description of the world, in which objects rotate , in addition to translating. This general branch of physics is called ‘Rigid Body Dynamics.’
Rigid body dynamics has many applications. In vehicle dynamics, we are often more worried about controlling the orientation of our vehicle than its path – an aircraft must keep its shiny side up, and we don’t want a spacecraft tumbling uncontrollably. Rigid body mechanics is used extensively to design power generation and transmission systems, from jet engines, to the internal combustion engine, to gearboxes. A typical problem is to convert rotational motion to linear motion, and vice-versa. Rigid body motion is also of great interest to people who design prosthetic devices, implants, or coach athletes: here, the goal is to understand human motion, to protect athletes from injury or improve their performance, or to design devices that replicate the complicated motion of a human joint correctly. For example, Professor Crisco’s orthopaedics lab at Brown studies human motion and the forces they generate at human joints, to help understand how injuries occur and how they can be prevented.
The motion of a rigid body is often very counter-intuitive. That’s why there are so many toys that exploit the properties of rigid bodies: the motion of a spinning top; a boomerang; the ‘rattleback’ and a Frisbee can all be explained using the equations derived in this section.
Here is a quick outline of how we analyze motion of rigid bodies.
- A rigid body is idealized as an infinite number of small particles, connected by two-force members.
- We already know the equations of motion for a system of particles (Section 4 of the notes):
The force-momentum equation 1
N ext i i i i i
d d m dt dt (^) = ∑ =^ = ∑
p F v
The moment – angular momentum equation 1
N ext i i i i i i i
d d m dt dt (^) =
∑ ×^ =^ =^ ∑ ×
h r F r v
The work-kinetic energy equation 1
N ext i i i i i i i
dT d m dt dt (^) =
∑ F^ ⋅^ v^ =^ =^ ∑ v^ ⋅ v
- These equations tell us how a rigid body moves. But to use them, we would need to keep track track of an infinite number of particles! To simplify the problem, we set up some mathematical methods that allow us to express the position and velocity of every point in a rigid body in terms of the position r G , velocity v G and acceleration a G of its center of mass, and its rotation tensor R (quantifying its orientation) and its angular velocity ω , and angular acceleration α. This allows us to write the linear momentum, angular momentum, and kinetic energy of a rigid body in the form
p = M v (^) G h = r G (^) × M v G (^) + I (^) G ω
2^ G^ G^ 2 G
T = M v ⋅ v + ω I ⋅ ω
where M is the total mass of the body and I G is its mass moment of inertia.
- We can then derive the rigid body equations of motion:
iext G i iext G G G^ [^ G ] i i
∑ F^ =^ M^ a^ ∑ r^ ×^ F^ =^ M r^ ×^ a^ +^ I^ α^ +^ ω^ × I^ ω
6.2 Describing Motion of a Rigid Body
We describe motion of a particle using its position, velocity and acceleration. We can describe the position of a rigid body in the same way - we could specify the position, velocity and acceleration of any convenient point in the body (we usually use the center of mass). But we also need a way to describe the orientation of a rigid body, and its rotational motion.
In this section, we define the various mathematical quantities that we use to describe rotation, angular velocity, and angular acceleration.
6.2.1 Describing rotations: The Rotation Tensor (or matrix)
Rotations are quantified by a mathematical object called a rotation tensor. It is defined as follows:
- Choose some convenient initial orientation of the rigid body (eg for the rectangular prism in the figure, we chose to make the faces perpendicular to the { , , i j k } directions.
- When the body is rotated, every line in the body (eg the sides) moves to a new orientation, without changing its length. We can describe this orientation change as a mapping. Let A and B be two arbitrary points in the body. Let p (^) A , p (^) B be the initial positions of these points, and let r A (^) , r B be their final positions. We introduce the ‘rotation tensor^1 ’ R which has the property that r B (^) − r A (^) = R p ( (^) B − p A )
When we solve problems, we always express vectors as components in some basis. When we do this, R becomes a matrix. For example, if p (^) B − p (^) A = x 0 (^) i + y 0 (^) j + z 0 k r B (^) − r A = x i + y j + z k we would write
0 0 0
xx xy xz yx yy yz yz zy zz
x R^ R^ R x y R R R y z (^) R R R z
^ ^
^
= ^
^
Here, R 11 (^) , R 12 ,... are a set of nine numbers (or sometimes formulas). Following the usual rules of matrix-
vector multiplication, this is just a short-hand notation for
0 0 0 0 0 0 0 0 0
xx xy xz yx yy yz zx zy zz
x R x R y R z y R x R y R z z R x R y R z
The subscripts on R are meant to you help remember what each element in the matrix does – for example, R xx maps the x 0 onto x, Rxy maps the y 0 onto x , and so on.
(^1) By definition, a ‘second order tensor’ maps a vector onto another vector. In actual calculations R is always just a
matrix, but ‘tensor’ sounds better.
k
j i
A
B
A
B
k
j i
p B -p A
r B -r A
In robotics, game engines, and vehicle dynamics the axis-angle representation of a rotation is often stored as a quaternion. We won’t use that here, but mention it in passing in case you come across it in practice. A
quaternion is four numbers [ q 0 , q x , q y , qx ]that are related to n and θ through the formulas:
0 cos(^ / 2) x x sin(^ / 2)^ y y sin(^ / 2)^ z z sin^ / 2
q q n q n q n
θ θ θ θ
Mapping the coordinate axes
In some problems we might know what happens to vectors that are parallel to the { i,j,k } directions in the initial rigid body (eg we might know what happens to the sides of our rectangular prism). For example, we might know that { , , i j k }map to (unit) vectors a b c , ,.
In that case we can write down each of a b c , , as components in { , , i j k } a = a x i + a (^) y j + az k b = bx i + by j + bz k c = cx i + c (^) y j + cz k
and use the formula
x x x y y y z z z
a b c a b c a b c
R
A sequence of rotations
Suppose we rotate an object twice (perhaps about two different axes). How do we describe the result of two rotations? That’s not hard. Suppose we do the first rotation with one mapping (1) (^) ( ) r B (^) − r A (^) = R p (^) B − p A
Now we rotate our body again – this maps (^) r B (^) − r A onto some new vector (^) u B (^) − u (^) A :
( ) (2)( ) u (^) B − u (^) A = R r B (^) − r A
We can therefore write
( ) (2)^ (1)( ) u (^) B − u (^) A = R R p (^) B − p A
We see that Sequential rotations are matrix products
R = R^ (2)^ R (1)
Health warning: Matrix products (and hence sequences of rotations) do not commute
R^ (1)^ R (2)^ ≠ R (2)^ R (1) For example, the figure below shows the change in orientation caused by (a) a 90 degree positive rotation about i followed by a 90 degree positive rotation about k (the figure on the left); and (b) a 90 degree positive rotation about k followed by a 90 degree positive rotation about i (the figure on the right).
k j i k j i
a
b
c
j b
k
k
k
j i
i
i
j j
(1) k rotation (2) i rotation
(1) i rotation (2) (^) k rotation
Orthogonality of R
The rotation tensor (matrix) has a very important property: If you multiply R by its transpose, the result is always the identity matrix. Another way to say this is that The transpose of R is equal to its inverse
Let’s try this with the 2D rotation matrix 2 2
2 2
cos sin cos sin cos sin 0 1 0 sin cos sin cos (^0) sin cos 0 1
cos sin cos sin 1 0 sin cos sin cos 0 1
T
T
−^ ^ +
−^ ^ +
RR
R R
A matrix or tensor with this property is said to be orthogonal.
Why is this? It turns out that a length-preserving mapping must be an orthogonal tensor. To see this, let’s calculate the length of the rotated vector r B (^) − r A (^) = R p ( (^) B − p (^) A ). We need to remember two vector/matrix
operations:
- We can calculate the length of a vector by dotting it with itself and taking the square root
- For a vector u and a matrix R, we know (or can show!) that ( Ru ) ( ⋅ Ru ) = u ⋅( R T Ru )
This means
( r B^ −^ r A^ ) (⋅^ r B^ −^ r A^ ) =^ { R p (^^ B −^ p^ A )} {^ ⋅^ R p (^^ B −^ p^ A )}^ =^ (^ p^ B −^ p^ A )^ ⋅^ { R T^ R p (^^ B − p A )}
But we want the length of r B (^) − r A to equal the length of p B (^) − p (^) A , which means we need R to satisfy
(2) (1)
= = ^ ^ − ^ = ^ −
R R R
- Find the axis-angle representation for the combined rotation in problem (2).
We can calculate the axis and angle of this rotation using the formulas
1 2cos 2cos 2
cos cos cos 1 cos 1 cos 1 cos 1 ( 1) 0 ( 1) 0 ( 1) 1 1 ( 1) 1 ( 1) 1 ( 1) (^2)
xx yy zz
xx yy zz
R R R
R R R
n i j k
i j k j k
To decide which of these two choices to use we notice that R (^) yz = − 1 , which tells us that n ny z < 0. The answer is
therefore 1 , ( ) 2
θ = π n = j − k
It is incredibly difficult to visualize the effect of a rotation about an arbitrary axis (at least for me). In fact this formula looks wrong – how can a 180 degree rotation end up tipping the box on its side? But the answer is right, as the animation (which will only show up in the html version of the notes) shows.
6.2.2 Describing rotational motion: The angular velocity vector and spin tensor
We described the location of a particle in space using its position vector, and its motion using velocity. We need to come up with something similar to velocity for rotations.
Definition of an angular velocity vector Visualize a spinning object, like the cube shown in the figure. The box rotates about an axis – in the example, the axis is the line connecting two cube diagonals. In addition, the object turns through some number of revolutions every minute. We would specify the angular velocity of the shaft as a vector
ω , with the following properties:
- The direction of the vector is parallel to the axis of the shaft (the axis of rotation). This direction would be specified by a unit vector n parallel to the shaft.
- There are, of course, two possible directions for n. By convention, we always choose a direction such that, when viewed in a direction parallel to n ( so the vector points away from you) the shaft appears
n Axis of
rotation
to rotate clockwise. Or conversely, if n points towards you, the shaft appears to rotate counterclockwise. (This is the `right hand screw convention’)
Viewed along n Viewed in direction opposite to n
3. The magnitude of the vector is the angular speed d θ / dt of the object, in radians per second. If
you know the revs per minute n turned by the shaft, the number of radians per sec follows as
d θ / dt = 120 π n. The magnitude of the angular velocity is often denoted by ω = d θ/ dt
The angular velocity vector is then
d
dt
ω = n =ω n.
Since angular velocity is a vector, it has components ω = ω x i + ω y j +ω z k in a fixed Cartesian basis.
As always, in two dimensions, everything is very simple. In this case objects can only rotate about the k axis, and we can write the angular velocity vector as d dt
ω = k
where θ is the counterclockwise angle of rotation of any line embedded in the body.
Writing down angular velocities:
For 2D problems, we always know the direction of the angular velocity and can just use ω = ω z k to write it
down (of course if we know the value or a formula for ω z we can use it).
For 3D problems, we can often use vector addition to write down ω. We can illustrate this with a simple
example:
Example: The propeller on the aircraft shown in the figure spins (about its axis) at 2000 rpm. The aircraft travels at speed 200 km/hr in a turn with radius 1 km. What is the angular velocity vector of (i) the body of the aircraft, and (ii) the propeller? Express your answer in the normal-tangential- vertical basis.
(i) The circumference of the circle is
s = 2 π R = 2 πkm.^ The airplane completes a full circle in^ t = s / V = (2 π / 200) × 3600 = 36 π sec.^ A full
turn is 2 π radians, so the aircraft body turns at a rate 2 π / (36π ) = (1 / 18) k rad/sabout the k axis.
There is an important formula relating W and ω. Let r B (^) − r A be a vector joining any two points in a rigid
body. Then W r ( (^) B (^) − r A (^) ) = ω × ( r B (^) − r A )
You can see this by just multiplying out the definition of W and comparing the result to the cross product: if r B (^) − r A (^) = x i + y j + z k , then
z y y z B A z x z x y x x y
x z^ x y x z z y x
−^ −
W r r
Hopefully you can see that this is the same as the cross product!
6.2.3 The angular acceleration vector
Angular acceleration is the time derivative of angular velocity d dt
ω α
For 3D, we can use
x y z x y z
d d d dt dt dt
ω^ ω ω
For 3D, we can’t express the angular accelerations or velocities as derivatives of rotation angles, because these can’t be defined for a general motion.
For a 2D problem, the direction of angular velocity and acceleration are known, so we have
α = α z k ω =ω z k
The components are related by 2 2
z z z z
d d d dt (^) dt d
For 2D problems, we can use all the usual rules of calculus to go from angular acceleration to angular velocity to angle, and vice-versa (just like distance-speed-acceleration formulas for straight line motion).
6.2.3 Relative velocity and acceleration of two points in a rigid body
We now know how to describe rotational motion. Our next order of business is to discuss a couple of very important formulas that we use to analyze the motion of a system of rigid bodies, and also to derive formulas for the angular momentum and kinetic energy of a rigid body..
Consider a rigid body: Let ω be the (instantaneous) angular velocity of the body, and W the corresponding spin tensor
A
B
k
j
i
r B -r A
v A ,a A
v B ,a B
Let A and B be two arbitrary points in a rigid body, and let r A (^) , r B and v (^) A , v B , a (^) A , a (^) B be their (instantaneous) position, velocity and acceleration vectors.
Then the relative position and velocity of A and B are related by ( ) ( )
B A B A B A B A
− = × −
v v ω r r v v W r r
The relative acceleration of A and B are related their relative positions and velocity by a B (^) − a (^) A = α × ( r B (^) − r A (^) ) + ω × ( v (^) B − v (^) A ) = α × ( r B (^) − r A (^) ) + ω × (^) [ ω × ( r B (^) − r A )]
For 2D problems only: we can simplify these, because we know ω is always parallel to the k direction. Therefore
v B − v A = ω z k × ( r B − r A )
a B − a A = α z k × r B − r A − ω z r B − r A
Proof: These fornulas are easy to prove. Remember the mapping:
B A^ (^ B A )^ B A (^ B A )^ (^ B A )
d d dt dt
R
r r R p p v v r r p p
Also, r B (^) − r A (^) = R p ( (^) B (^) − p (^) A ) ⇒ R T^ ( r B (^) − r A (^) ) = R T R p ( (^) B − p (^) A ) = ( p B (^) − p A )
Hence
B A T^ (^ B A )^ (^ B A )
d dt
R
v v R r r W r r
Remember that W r ( (^) B (^) − r A (^) ) = ω × ( r B (^) − r A ), so the acceleration formula then follows as
B A^ (^ B A )^ (^ B A )^ (^ B A )^ (^ B A )^ (^ B A )
d d d dt dt dt
− = − = × − + × − = × − + × −
ω a a v v r r ω r r α r r ω v v
6.3 Analyzing motion in connected rigid bodies
The formulas in 6.2.3 are used to analyze motion in machines. A typical problem is illustrated in the figure. An actuator moves point B on the car jack shown in the figure horizontally with constant velocity V. What are the velocity and acceleration of the platform (CF)?
You could probably solve this rather simple example with elementary trig, but we need a more systematic method for general problems, especially to analyze 3D motion. Here’s the general procedure
- Define variables to denote the unknown angular
j
A i
B
r B -r A
θ ω αz z
v A ,a A
v B ,aB
j
A B i
D C
F
V
L
L
θ
- We don’t know the angular acceleration of BC, so we introduce α (^) zBC as an unknown and use the rigid body formula for member BC to write down an equation for the acceleration of C ( ) 2 ( ) (2 2 ) 32 2 2
C B zBC C B zAB C B zBC C BC zBC
− = × − − − = × − −
a a k r r r r k i j 0 a j i j
- Point C can only move horizontally, so it can’t have any vertical acceleration. This means
that the j component of acceleration is zero:
zBC zBC C
⇒ a = i
Example 2: For a more complicated example, we can solve the car jack problem posed at the start of this section. An actuator moves point B on the car jack shown in the figure horizontally with constant velocity V. What are the velocity and acceleration of the platform (CF)?
The system contains 3 rigid bodies (AC, BD, CF^3 ). We don’t know the angular velocities or accelerations of any of
them, so we denote them by unknowns ω zAC ,
ω (^) zBD ω (^) zCF , α (^) zAC , α (^) zBD α zCF
Calculating the velocity:
- We start at point(s) with known velocity: A is stationary, and the velocity of B is given: v (^) A = 0 v (^) B = V i
- Point E lies on both member AC and on member BD. We use the rigid body formulas to write down an equation for the velocity of E on each member (notice we use the 2D equations): ( ) ( )
E A zAC E A E B zBD E B
ω ω
− = × −
− = × −
v v k r r v v k r r
- The two members AC and BD are pinned together at E and so must have the same velocity. We can eliminate v (^) E and write out the position vectors in i,j components ( cos30 sin 30 ) ( cos30 sin 30 ) ( sin 30 ) cos30 sin 30 cos
zAC zBD zAC zAC zBD zBD
L L V L L
L V L L L
× + − = × − +
k i j i k i j i j i j
The i,j components give two equations for ω zAC , ω zBD
sin sin cos cos 2 sin cos cos 0 / (2 sin ) / (2 sin )
zAC zBD zAC zBD zAC zAC zBD
L V L
L L
L V
V L V L
- We can now use the rigid body formulas for members AC and BD to find the velocities of C and D
(^3) You may be wondering why only a single point was defined at C and E, but there are two points at D and F. That’s
because at C and E the members are pinned together, but there is a roller at D. At E, members AC, BD always have the same velocity and acceleration – we can just use a single variable to denote the velocity of this point. The same is true at C. Members CF and BD touch at F and D, but point D on AB does not have the same horizontal velocity as point F CF, so we need to be able to distinguish between them.
j
A B i
D C
F
V
L
L
θ
( ) (2 cos 2 sin ) cot 2 sin
( ) (2 cos 2 sin ) cot 2 sin
C A zAC C A C
D B zBD D B D
V
L L V V
L
V
V L L V
L
− = × − ⇒ = × + = −
− = × − ⇒ = + × + = −
v v k r r v k i j i j
v v k r r v i k i j j
- We can use the rigid body formula for CF to relate the velocities of C and F ( ) cot 2 cos
F C zCF F C F V^ V^ zCF L
ω θ ω θ
− = × −
v v k r r v i j j
- Point D on CD and point F on CF must have the same vertical velocity (the roller at D allows their horizontal velocities to differ). This can be expressed as cot 2 cos cot 0
F D zCF zCF
V θ ω L θ V θ
v j v j
- All points on CF therefore have the same velocity (equal to the velocity of C)
v CF = V i − V cot θ j
Calculating the acceleration.
- We can now calculate the accelerations. We start at a known point: Points A and B have zero acceleration.
- We can use the rigid body formula to calculate the acceleration of E on each of AC and BD: 2
2
E A zAC E A zAC E A
E B zAD E B zAD E B
α ω
α ω
− = × − − −
− = × − − −
a a k r r r r
a a k r r r r
- The two members are connected at E and so must have the same acceleration there. This shows that 2
2
2
2
( cos sin ) ( cos sin )
( cos sin ) ( cos sin )
( cos sin ) ( cos sin )
( cos sin ) ( cos sin )
zAC zAC
zAD zAD
zAC zAC
zAD zAD
L L L L
L L L L
L L L L
L L L L
α θ θ ω θ θ
α θ θ ω θ θ
α θ θ ω θ θ
α θ θ ω θ θ
× + − +
= × − + − − +
k i j i j
k i j i j
j i i j
j i i j
- The i,j components give two equations for the unknown angular accelerations: 2 2
2 2
2 2 2 2 2 2 3
2 2 3
sin cos sin cos
cos sin cos sin
2 sin cos (cos sin ) cos / (4 sin )
cos / (4 sin )
zAC zAC zAD zAD
zAC zAC zAD zAD
zAC zAD zAC zAC
zAD zAC
L L L L
L L L L
L L L V L
V L
α θ ω θ α θ ω θ
α θ ω θ α θ ω θ
α θ θ ω θ θ ω α θ θ
α α θ θ
- We can use the rigid body acceleration formulas to calculate the velocities of D and C:
- Contact between two objects without relative slip (sliding) at the contact (friction forces must act to prevent the slip, in general): The velocities of the touching objects must be equal at the contact point. The tangential components of acceleration must also be equal (the normal components of acceleration differ) v (^) B = v (^) A a (^) B ⋅ t = a (^) A ⋅ t
n A (^) B
t
6.3.2 The Rolling Wheel
Wheels are everywhere. They can be analyzed using the general rigid body equations, but it’s helpful to be able to avoid all the tedious cross products. In this section we summarize special formulas for velocity and acceleration of points on a wheel.
Motion of a wheel rolling without slip on a stationary surface
It is surprisingly difficult to visualize the motion of a wheel. The figure above might help: it shows the trajectory of one point on the circumference of the wheel. The point traces quite a complicated path. The important thing to notice is:
If a wheel rolls without slip on a stationary surface, the point touching the surface is stationary
Each point is only in contact with the ground for an instant, and while it touches the ground it has a large vertical acceleration, but it is instantaneously stationary. We know this from the list of constraints in Sect 6.3.1, of course, but it’s still not an easy thing to visualize.
More generally, the ground need not necessarily be stationary (or the wheel could touch another surface). In this case we know that the contacting points on two bodies in rolling contact have equal velocity at the contact****.
Angular velocity-linear velocity formula: With this insight, we can use the rigid body formulas to calculate the instantaneous velocity vector for any point on the wheel. Assume that
- The wheel rolls with angular velocity ω = ω z k counterclockwise rotation is positive.
- The center of the wheel moves with velocity v O (^) = vxO i
The rolling wheel formula gives
v xO = − ω zR
j
i
O
C
ωz
R v xO
To see this, you can simply use the rigid body formula to go from the contact point (which is stationary) to O v O (^) − v (^) C = ω × ( r O (^) − r C (^) ) ⇒ v O (^) = ω z k × ( R j ) = −ω zR i
More generally, we can calculate the velocity of any point on the wheel we might be interested in. In fact, we can just write down the velocity of any point in the wheel by noticing that instantaneously all points are in circular motion about the contact point (just imagine the disk is rotating about C). See if you can show all the following:
- v (^) A = − ω (^) z R ( i + j )
- v (^) D = − ω z 2 R i
- v (^) B = − ω (^) z R ( i − j )
Notice that the direction of the velocity at each point is always perpendicular to the line connecting to the point to C.
Angular acceleration-linear acceleration formula: Assume that
- The wheel rolls with angular acceleration α = α z k counterclockwise rotation is positive.
- The center of the wheel moves with acceleration a O (^) = axO i
The rolling wheel formula gives a xO = − α zR
You can derive this formula in two different ways:
(1) Differentiate the velocity formula vxO = − ω zR with respect to time
(2) Use the rigid body formula: 2
2
( (^) O C ) ( (^) O C ) (^) z ( (^) O C )
O C z R^ zR
ω
α ω
− = × − − −
a a α r r r r
a a i j We know that the i component of acceleration at point C has to be the same as the i component of acceleration of the ground (i.e. zero). (The j components don’t have to be equal). We also know that O has no j acceleration, because it remains at the same height above the ground. Therefore 2
2
xO yC z z
xO z yC z
a a R R
a R a R
i j i j
We can calculate the acceleration of any other point on the disk using the rigid body formula.
Example: The block AB has horizontal acceleration a and horizontal speed v. Calculate the angular velocity and angular acceleration of the rollers. Then, calculate the linear velocity and acceleration of O
To solve problems like this we use two ideas: (1) the formulas relating velocity and accelerations of points on the disk; and (2) the tangential velocity and acceleration of contacting points are equal.
j
i
A B
O
C
ωz D
θ
R
j
i
O
C
ωz
R
αz
a xO
j
i
O
C
ωz
αz R
O
R
A B v,a
Example 1: The left gear in the figure rotates with counterclockwise
angular velocity ω z 1. The large gear has radius R 1 and N 1 teeth, the
small one has radius R 2 and N 2 teeth. Calculate the angular velocity
of the smaller gear.
Note:
- The velocities of the two touching gears are equal at C
- The gear rotation/velocity formula gives 2 1 1 2 2 2 1
z z z
R
R R
R
ω j = −ω j ⇒ ω = −
Notice that we assume both gears rotate counterclockwise. The formula tells us that the second gear has a negative angular velocity – this means that it is actually rotating clockwise. The animation at the top of this section confirms that this indeed is the case.
Example 2: An ‘epicyclic’ gearbox is a special arrangement of gears that has many applications. The sketch shows a simple example. The gearbox can be driven in three different places: one drive shaft is connected to the central sun gear (A); the other is attached to the ‘planet carrier’, which is joined to the center of the ‘pinion gears’ B,C and D. The outer gear (E – called the ‘ring gear’) can also be driven separately.
Epicyclic gearboxes are used in all automatic vehicle transmissions. They are also very useful in ‘split power’ drives, where two motors need to be connected together to drive a single axle. Hybrid vehicles, which have both an electric motor and an internal combustion engine driving the same axle, are one example. You can find a very nice description of the Toyota Prius split power transmission here: the website includes a Flash animation that lets you change the speeds of the motors in the system and visualize the motion of the gears.
The figure shows a schematic diagram illustrating the general geometry and motion of the system. We have four rigid bodies:
- The central sun gear, radius (^) RS , (^) N (^) S teeth, rotating at angular velocity ω zS
- The planet carrier, angular velocity (^) ω zPC
- The ring gear, radius RR , with N (^) R teeth, angular
velocity ω zR
- The planet gear, radius r = ( RR − RS ) / 2, N (^) P = ( N (^) R − NS ) / 2teeth, rotating at angular velocity
ω zP
In any application, we are given the angular velocity of two of the drive shafts (any two of
ω zS , ω zPC , ω zR ), and must calculate the third. The planet gear is not connected to any drive shaft, so we
usually don’t care very much about its angular speed, but we will need to find ω zP to solve for the unknown
one of ω zS , ω zPC , ω zR.
j
i
O C
ωz
R 1
R 2
ωz
O
Ring gear
Sun gear r
RS
RR
zS
ω zR
Planet Carrier
Planet gear
ω zP
ω zPC
This seems a terribly difficult problem, but it can be solved in a very simple way with a trick.
We start by solving a simpler version of the problem. Suppose that the planet carrier is stationary ( ω zPC =0) and the sun gear
rotates with angular speed ω zS (see the animation). What is the
angular velocity of the ring gear?
The sun gear and the planet gear are just a standard gear pair so we know that
S zS S zP zP zS
R
R r r
The two touching points on the planet gear and the ring gear must have the same velocity, so (using the rotating gear formula)
zP zR R R zP R
r r R R
ω j = ω j ⇒ ω =ω
We can eliminate ω zP to get the answer:
S zR zS R
R
R
Now let’s try the harder problem. The animation shows a general situation, where ω zS , ω zPC are both
nonzero. How can we find ω zR now?
This is difficult to analyze because the center of the planet gear is not fixed, so it’s hard for us to visualize the motion, and the standard gear formulas don’t work. But we can simplify the problem by analyzing motion in a reference frame that rotates with the planet carrier. For example, imagine attaching a videocamera to the planet carrier – this camera would show the planet carrier to be stationary, with the surrounding world rotating in the opposite direction. The angular velocity of the planet carrier would be subtracted from all the other angular velocities. In this reference frame, we can use the result we just calculated: ( ) ( )
zR zPC S zS zPC R
R
R
ω ω ω ω
This result is general, and can be re-arranged to tell you the
angular velocities for any given combination of ω zS , ω zPC and
ω zR.
