6 Kinematics of planar rigid bodies - ETH Z

6 Kinematics of planar rigid bodies

A

B2R R

v

C

Figure 6.1:

6.1 In-class problem A piston A has constantvelocity v in the horizontal direction to the right, asindicated in the figure. A wheel, which is connectedto the piston by the connecting rod AB, rolls with-out slipping on a flat plane. Determine the velocityand the acceleration of the center of the wheel forthe configuration shown in the sketch.

Solution

Figure 6.2:

Let the x axis be horizontal to theright and the y axis be vertical and up-ward; the axes are denoted by the unitvectors ex and ey, respectively. The an-gular velocity vector sof each body is per-pendicular to the plane and directed as ez,with unknown magnitudes.To find the acceleration and velocity ofcenter C at a given instant, we first derivean expression that describes its position inthe x-direction with respect to the inertialframe. Subsequently, we take the first and second derivatives of the position with respectto time to calculate the velocity and acceleration, respectively. Note that the center ofthe wheel does not have any motion in the y-direction.We use three variables, x, α, and θ to describe the position of the wheel in ex direction.Later, we account for the fact that these variables are related to each other by two con-straints. This implies that the system has 1 degree of freedom, which can be understoodby inspection as well. The instantaneous position of center C in the ex direction is

xC = xA + xAB + xBC = x+ 2R cosα−R cos θ (6.1)

where xAB and xBC denote the position of points B and C relative to points A and B,respectively.The wheel rolls without slipping, therefore it holds that

vC = vCex; vC = xC = −Rθ. (6.2)

Moreover, the position of the center C in the y-direction can be described as

yC = 2R sinα−R sin θ = R. (6.3)

Eqs. (6.2) and (6.3) are the two constraints applied to the system.Next, we differentiate (6.1) to obtain an expression for the velocity of the center C

xC = x− 2Rα sinα +Rθ sin θ. (6.4)

6-1

6 Kinematics of planar rigid bodies 6-2

Constraint equations must hold for all the time instances. Hence, we can differentiatethem to obtain the constraints at velocity or acceleration levels.Differentiating Eq. (6.3) yields

yC = 2Rα cosα−Rθ cos θ = 0. (6.5)

By substituting (6.2) into (6.5) we can obtain an expression for α

α =cos θθ

2 cosα= − cos θ

2R cosαxC . (6.6)

Substituting Eqs. (6.2) and (6.6) into (6.4) gives the velocity of the center C:

xC = x+ xCcos θ sinα

cosα− xC sin θ ⇒ (1 + sin θ − cos θ tanα)xC = x (6.7)

The values of θ and α for the position shown in Fig. 6.1 can be calculated easily from thegeometry of the problem

θ0 = π; α0 =π

6. (6.8)

Substituting θ0 and α0 into (6.7) yields the velocity of the center C at the given position

vC |θ0,α0 =

√3√

3 + 1xex =

√3√

3 + 1vex, (6.9)

To obtain the acceleration, we differentiate (6.7) with respect to time

(θ cos θ + θ sin θ tanα− α

cos2 αcos θ)xC + (1 + sin θ − cos θ tanα)xC = 0 (6.10)

To obtain the acceleration at the given configuration, we need θ0 and α0, which can becomputed from (6.2) and (6.6)

θ0 = −√

3√3 + 1

v

R; α0 =

1√3 + 1

v

R; (6.11)

Substituting θ0, α0, θ0, and α0 into (6.10) yields

xC = aC |θ0,α0,θ0,α0= − 3

√3 + 4

(√

3 + 1)3v2

Rex (6.12)


6.2 In-class problem Consider the velocity transfer formulavB = vA + ω × rAB, which relates the velocities of two arbitrarypoints A and B and the angular velocity vector of a rigid body.Consider a planar motion and determine the relative accelerationaB − aA by differentiating the equation above with respect totime. Simplify the final formula.

Solution

I. Choose the reference systemSet the reference system as shown on the figure above.

II. Differentiation of the velocity transfer formula with respect to time

The velocity transfer formula is given by:

vB = vA + ω × rAB . (6.13)

Also, recall that

rAB = ω × rAB . (6.14)

The acceleration of point B with respect to point A can be expressed as the time derivativeof (6.13)

aB = aA + ω × rAB + ω × rAB . (6.15)

Denoting the angular acceleration as ε = ω and using (6.14) and (6.15), we obtain

aB = aA + ε× rAB + ω × (ω × rAB) . (6.16)

By using the triple product expansion 2 for the third term, we get

aB = aA + ε× rAB + ω(ω · rAB)− rAB(ω · ω) . (6.17)

Since we are dealing with planar motion, ω is always perpendicular to rAB, therefore

(ω · rAB) = 0 . (6.18)

Substituting ω · ω = |ω|2 and (6.18) into (6.17), yields

aB = aA + ε× rAB − |ω|2rAB . (6.19)

From (6.19), the relative acceleration is

aB − aA = ε× rAB − |ω|2rAB . (6.20)

2a× (b× c) = b(a · c)− c(a · b)


It is important to observe that the ε× rAB term is the tangential acceleration componentof B with respect to A, since its direction is always perpendicular to rAB. And −|ω|2rABis the corresponding normal component, since its direction goes always from B towardsA. Therefore we can also write that

aB = aA + (aB/A)t + (aB/A)n , (6.21)

where (aB/A)t = ε× rAB and (aB/A)n = −|ω|2rABThe concept is illustrated in the figure below.


6.3 In-class problem The planetary gear systemshown in the figure is used in an automatic transmis-sion for an automobile. By locking or releasing cer-tain gears, it can provide different rotational speeds.Consider the case where the ring gear R is held fixed,ωR=0, and the sun gear S is rotating at ωS = 5 rad/s.Determine the angular velocity of each of the planetgears P and shaft A.

Solution

0. CommentThe acutal gearing mechanism is complicated and involves complex contact mechanics.Here, we model all the gears as disks that rotate without slipping around their mutualcontact points.

I. Choose the reference systemSet the right-handed reference system as shown on the next figure.

II. Imposing no-slipping contraintsLet us consider the planet gear first. The velocity transfer formula for point B withrespect to point D is

vB = vD + ωP × rDB. (6.22)

The velocity of point D can also be expressed in terms of the angular velocity of the sungear:

vD = ωS × rFD. (6.23)

We know also that vB = 0.since the planet gear rotates without slipping on the outer, fixed ring. Therefore, (6.22)


becomes

0 = ωS × rFD + ωP × rDB. (6.24)

All the position vectors are aligned with the y axis. The above equations can thereforebe conveniently written in vector components as

0 = ωSez × rFDey + ωP ez × rDBey ,

0 = −ωSrFDex − ωP rDBex. (6.25)

we can now solve for (6.25) for ωP and obtain

ωP = −ωS · rFDrDB

= −5 rad/s (in the ez direction). (6.26)

In order to calculate the velocity of the shaft, we can adopt the same strategy. Thevelocity transfer formula for point C with respect to point B is

vC = 0 + ωP × rBC . (6.27)

the same velocity can be expressed for the shaft

vC = ωA × rFC , (6.28)

where rFC is the position vector in the x − y plane which points from the center of theupper planet gear P to the center of the shaft A. By expressing the vectors in components,performing the vector product and equating (6.27) and (6.28), we can solve for the magnetof the angular velocity of the shaft ωA:

ωA =ωP rBCrAP

= 1.67 rad/s. (6.29)


6.4 In-class problem A ring of radius R is hingedat point O. A disk of radius r rolls without slipping in-side the ring, as shown. Determine the angular veloci-ties of the ring and the disk in terms of the generalizedcoordinates θ, ϕ indicated in the figure.

Solution

I. Choose the reference systemSet the reference system as shown on the figure.

II. Draw the geometryLet us consider first the ring only. We draw it in a displaced configuration:

By comparing the orientation of OA and OA′, the angular velocity of the ring is:

ωring = θk . (6.30)

Now we sketch the reference and the displaced configurations for the whole system. Inorder to determine the position of the disk, we use the angles θ, ϕ and ψ, and thenimpose the rolling without slipping constraint between the disk and the ring to eliminateone angle. The orientation of the disk is easily determined by putting a mark BA on it.

Comparing the orientation of BA and B′A′ between the reference and the displacedconfiguration, the angular velocity of the disk is:

ωdisk = [θ + ϕ− ψ]k . (6.31)


In other words, the angular velocity of the ring is given by three contributions,namely 1. the angular velocity of the ring, 2. the angular velocity due to the rotationabout the center of the ring, and 3. the angular velocity due to rolling without slippingin the ring. Since the disk rolls without slipping, the velocity of the contact point vD′ hasto be the same on the ring and on the disk. It follows that the length of the trajectory ofD′ on the disk and on the ring from the initial configuration are equal.

rψ = Rϕ⇒ ψ =R

rϕ (6.32)

Substituting (6.32) into (6.31) yields,

ωdisk =

[θ +

(1− R

r

)ϕ

]k . (6.33)

Note that, if we impose θ = 0 (fixed ring) we have the case of a disk rolling on a circularguide. The angular velocity of the disk, in this case, is given by

ωdisk =

[(1− R

r

)ϕ

]k. (6.34)


6.5 Homework The crank CB oscil-lates about C, causing the crank OA tooscillate about O. At the instance whenthe linkage attains the position shown withCB horizontal and OA vertical, the mem-ber CB has an angular velocity ωCB = 2rad/s counterclockwise. For this configu-ration, determine the angular velocity andacceleration of links OA and BA.

Solution

0. NotationWe use the following notations for vectors:

rOA =

(rOA)x(rOA)y(rOA)z

= (rOA)xex + (rOA)yey + (rOA)zez

If only one vector’s component different from 0, we use the following notation:

rOA =

0(rOA)y

0

= (rOA)yey = rOAey

I. Choose the reference systemSet the reference system as shown the figure of the assignment.

II. Velocity transfer formulaThe velocity transfer formula is rewritten as

ωOA × rOA = ωCB × rCB + ωBA × rBA . (6.35)

Remember we are asked to consider only a particular snapshot of the system.Using unit vectors to express the variables, we getrOA = rOAey,rCB = rCBex,rBA = (rBA)xex + (rBA)yey,ωOA = ωOAez,ωCB = ωCBez = 2ez,ωBA = ωBAez.

From the geometry we know that

rOA =

(0

100

)rCB =

(−75

0

), rBA =

(−175

50

).

Substituting the unit vector expressions into (6.35) gives

ωOAez × rOAey = ωCBez × rCBex + ωBAez × ((rBA)xex + (rBA)yey) ,


−ωOArOAex = ωCBrCBey + ωBA(rBA)xey − ωBA(rBA)yex . (6.36)

Matching coefficients of the respective ex and ey terms gives

ωBA(rBA)y − ωOArOA = 0 and ωCBrCB + ωBA(rBA)x = 0 . (6.37)

Substituting the known values and solving (6.37), we getωOA = −3/7 rad/s and ωBA = −6/7 rad/s.

The minus signs indicate that the vectors ωBA and ωOA are in the negative z-direction.Hence, the angular velocities are clockwise.

III. Angular accelerationsThe acceleration of point A with respect to point O and point B with respect to point Ccan be written as

aA = aO + εOA × rOA − ω2OArOA , (6.38)

aB = aC + εCB × rCB − ω2CBrCB . (6.39)

Since O and C are fix points, aO = 0 and aC = 0. We also know that ωCB is constant,therefore εCB = 0. Substituting these quantities and rewriting (6.38) and (6.39) with unitvectors, we get

aA = −εOArOAex − ω2OArOAey , (6.40)

aB = −ω2CBrCBex . (6.41)

The acceleration of point A with respect to point B is

aA = aB + εBA × rBA − ω2BArBA . (6.42)

Substituting (6.41) into (6.42) and rewriting (6.42) with unit vectors, yields

aA = −ω2CBrCBex+εBAez×((rBA)xex)+εBAez×((rBA)yey)−ω2

BA(rBA)xex−ω2BA(rBA)yey .

(6.43)

Substituting (6.40) into (6.43) and equating separately the coefficients of the ex-componentsand the coefficients of the ey-components, yields

−εOArOA = −ω2CBrCB − εBA(rBA)y + ω2

BA(rBA)x , (6.44)

−ω2OArOA = εBA(rBA)x − ω2

BA(rBA)y . (6.45)

Plugging in the known quantities in (6.45), we get

εBA = −0.105 rad/s2 ,

εBA = εBAez = −0.105ez rad/s2 . (6.46)

Substituting the known quantities and (6.46) into (6.44), we get

εOA = −4.34 rad/s2 ,

εOA = εOAez = −4.34ez rad/s2 . (6.47)

Since the unit vector ez points in the positive z-direction, we see that the angular accel-erations of BA and OA are both clockwise.


6.6 Homework If the slider block A is moving tothe right at vA = 8 m/s, determine the velocity ofblocks B and C at the instant shown. Member CD ishinged to member ADB.

Solution

0. NotationWe use the following notations for vectors:

rOA =

(rOA)x(rOA)y(rOA)z

= (rOA)xex + (rOA)yey + (rOA)zez

If only one vector component is different from 0, we use the following notation:

rOA =

0(rOA)y

0

= (rOA)yey = rOAey

I. Choose the reference systemSet the right-handed reference system as shown on the figure below.

II. Express the variables with respect to the chosen reference system

The position vectors are expressed as

rAB = (rAB)xex + (rAB)yey,rAD = (rAD)xex + (rAD)yey,


rCD = (rCD)xex + (rCD)yey,

respectively. The velocities are

vA = vAex,vB = vBey,vC = vCey,vD = (vD)xex + (vD)yey,

respectively. The angular velocities of the members ABD and CD are denoted by

ωADB = ωADBez,ωCD = ωCDez.

III. Velocity transfer formula

The velocity transfer formula for point B with respect to point A is written as

vB = vA + ωADB × rAB , (6.48)

which can be expressed in components

vBey = vAex+ωADBez×((rAB)xex+(rAB)yey) = vAex+ωADB(rAB)xey−ωADB(rAB)yex .

(6.49)

Equating ex and ey components and substituting the known values, we get

ωADB = 2.828 rad/s

and

vB = 8 m/s.

In the same fashion, we can write the velocity transfer formula for point D with respectto point A

vD = vA + ωADB × rAD (6.50)

(vD)xex + (vD)yey = vAex + ωADBez × ((rAD)xex + (rAD)yey) (6.51)

(vD)xex + (vD)yey = vAex + ωADB(rAD)xey − ωADB(rAD)yex (6.52)

Equating ex and ey components and substituting the known quantities, we obtain

(vD)x = 4 m/s and (vD)y = 4 m/s.

In order to get velocity of C, we use velocity transfer formula for point C with respect topoint D.

vC = vD + ωCD × rCD (6.53)


vCey = (vD)xex + (vD)yey + ωCD(rCD)xey − ωCD(rCD)yex (6.54)

Equating ex and ey components and substituting the known quantities, we get

ωCD = 4.00 rad/s.

and

vC = −2.93 m/s.

.


6.7 Homework Consider the cardan joint intro-duced during the lecture. Determine the angularspeed ω2 of the driven shaft in terms of ω1, α andψ.

Solution

The angular velocity of the driving shaft can be written as

ω1 =

00

ψ

and the angular velocity of the driven shaft as

ω2 =

0ω2 sinαω2 cosα

.

Since angular velocities are additive, we can write:

ωcross = ω1 + φ = ω2 + γ. (6.55)

Observe that: γ ⊥ φ and ω2 ⊥ γ.Based on these bservations, we can determine the direction of γ.

eγ =φ× ω2

|φ× ω2|=

∣∣∣∣∣∣ex ey ez

cosψ sinψ 00 sinα cosα

∣∣∣∣∣∣ =

sinψ cosα− cosψ cosαcosψ sinα

1√cos2 α + cos2 ψ sin2 α

.

(6.56)

Then γ can be expressed as

γ = γ

sinψ cosα− cosψ cosαcosψ sinα

1√cos2 α + cos2 ψ sin2 α

. (6.57)

For simplicity we introduce the following notation:

γ1 = γ1√

cos2 α + cos2 ψ sin2 α. (6.58)


Now we can write (6.55) component-wise

φ cosψ = γ1 sinψ cosα. (6.59)

φ sinψ = ω2 sinα− γ1 cosψ cosα. (6.60)

ω1 = ω2 cosα + γ1 cosψ sinα. (6.61)

We have now a system of equations with 3 equations and 3 unknowns. From (6.59) wecan express φ as

φ = γ1sinψ

cosψcosα. (6.62)

Substituting (6.62) into (6.60), we get

γ1sin2 ψ

cosψcosα = ω2 sinα− γ1 cosψ cosα, (6.63)

which can be written as

γ1

(sin2 ψ

cosψcosα + cosψ cosα

)= ω2 sinα, (6.64)

γ1

(sin2 ψ + cos2 ψ

cosψcosα

)= ω2 sinα, (6.65)

γ1

(cosα

cosψ

)= ω2 sinα. (6.66)

Inserting (6.65) into (6.61), we finally obtain the ratio between the angular velocity of thedriving and the driven shaft

ω2 = ω1

(cosα

cos2 α + sin2 α cos2 ψ

). (6.67)

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6 Kinematics of planar rigid bodies - ETH Z