dynamics 17-92 17-93 solution

8/12/2019 Dynamics 17-92 17-93 Solution

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EAS 208 Solution for 17-92 and 17-93 2014 D Joseph Mook

Figure 1: 17.92, 17.93

17.92: GivenvP = 2i= constant, = 45

Find OQ andPQ

17.93: Given = 50 andOQ =1 radsFindaP

The problem consists of two rotating rigid bodies, connected by a pin so that their motions are related to each other.The general approach to such problems (which arise very frequently!) is to use the point of connection to obtainequations that can be solved for unknown components of the motion of one body or the other. The steps to solutionmay be visualized as

1. Use a specific set of coordinates, selected for convenience, to describe the kinematics of each body by itself.

Here, as in almost all situations, polar coordinates will be the most convenient. Thus, we normally have adifferent set of coordinates for each body in the problem. Some of our problems in class may involve 3-4 bodies;in real-world practice, the number may be much larger, but the procedure is the same and the individual stepsare not any more difficult, there are just more of them.

2. Using the perspective and coordinate system within each individual body, write velocity and accelerationequations for point(s) common to two or more bodies - in this case, point Q. Although these equations maybe written independently, Q has exactly one velocity and exactly one acceleration, and we use that fact to setour velocity and acceleration equations equal to each other.

3. Although we may set our velocity and acceleration equations from each bodys perspective equal to each otherfor any common point (Q, in this problem), since each bodys equations are written in its own convenientcoordinate system, we will then need to use geometry to convert to a single common coordinate system for

solution.

Solution of 17-92

1. For body OQ, well use a polar coordinate system with origin at O, with eROQ pointing from O to Q, eOQperpendicular to eROQ positive defined by the right-hand rule. For body P Q, use a polar coordinate systemwith origin at P and eRPQ pointing from P to Q, etc.

2. The velocity ofQ may be easily written in the polar coordinates of body OQ:

vQ= vO+ vQ/O= vO+ (rOQeROQ+rOQOQ eOQ)

We may substitute the known information that vO = 0, rOQ = 1.2 m, and rOQ = 0 to simplify

vQ= 1.2OQeOQ (1)

The exact same procedure is now applied using body P Q:

vQ= vP+ vQ/P =vP+ (rPQeRPQ+rPQPQePQ)

Again, we may substitute the known information, in this case: vP = 2i, rPQ= 1.2, and rPQ= 0:

vQ= 2i+ 1.2PQePQ (2)

Now, point Q only has one velocity, so the right-hand sides of Eq. (1) and Eq. (2) must be equal:

1.2OQ eOQ = 2i+ 1.2PQePQ (3)

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Eq. (3) is a vector equation in two dimensions (i.e., in the plane of the motion), so it contains two independentscalar equations. There are two unknowns, OQ and PQ, so there is enough information to solve for them.The key steps, to repeat, are (1) write velocity equations of any common points, using convenient coordinatesystems for each body the point is part of; and then (2) since any point may have only one actual velocity,we may require that all velocity expressions for any given point must be equal, even though they appear indifferent coordinate systems.

3. Eq. (3) currently contains three different coordinate systems, all defined in the same 2-D plane of the motion.

In order to solve, we need to convert to a single 2-D coordinate system using simple geometry. The choiceof conversion is yours, but as discussed often in class, the algebra involved may sometimes be much easier(or harder) depending on the choice. Also, you must always avoid choosing two parallel directions as yourcoordinates, since parallel coordinates do not provide 2-D, only 1-D.

In this case, note that since = 45 and rOQ = rPQ, we know that angle OQP is 90; thus, eOQis perpendicular to ePQ, so Im going to use those two as my common coordinates. Thus, I need only convert

i into eOQ and ePQ:

i= 12

eOQ 12

ePQ

Thus, Eq. (3) may be written

1.2OQ eOQ = 2( 1

2 eOQ 1

2 ePQ) + 1.2PQePQ

from which we may solve each coordinate independently to find

OQ = 21.2

2

=1.179 rads

, PQ= 2

1.2

2= 1.179

rad

s

Lets do a quick reality check by visualizing the motion. It should be clear that ifvP= (+)i (i.e., moving tothe right), then P Q is rotating counterclockwise ( = (+)) and OQ is rotating clockwise ( = ()), whichour math has found. Moreover, since the lengths of the two bars are equal, it makes sense that their angularvelocities must be the same speed (opposite directions).

Now repeat the process for accelerations:

aQ= aO+ aQ/O=aO+{(rOQr2OQ)eROQ+ (2rOQOQ+rOQOQ)eOQ}SubstitutingaO = 0, rOQ = 1.2, rOQ = 0, rOQ = 0, and OQ =1.179, we have

aQ= (1.21.1792)eROQ+ (1.2OQ)eOQ =1.667eROQ+ 1.2OQeOQ (4)

Similarly,aQ= aP+ aQ/P =aP+{(rPQr2PQ)eRPQ+ (2rPQPQ+rPQPQ)ePQ}

SubstitutingaP = 0, rPQ= 1.2, rPQ= 0, rPQ= 0, and PQ= 1.179, we have

aQ= (1.21.1792)eRPQ+ (1.2PQ)ePQ=1.667eRPQ+ 1.2PQePQ (5)Since Q has only one acceleration, the right-hand sides of Eq. (4) and Eq. (5) are equal:

1.667eROQ+ 1.2OQeOQ =1.667eRPQ+ 1.2PQePQ (6)There are two coordinate systems in Eq. (6), so we need to convert to a common one in order to solve. Forthis particular problem, since OPQ forms a 90 angle, we see that

eROQ=ePQ , eRPQ= eOQThus, in Eq. (6),

OQ =1.6671.2

=1.39 rads2

, PQ=1.667

1.2 = 1.39

rad

s2

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Solution of 17-93:The velocity ofQ is again written in the polar coordinates of body OQ:

vQ= vO+ vQ/O=vO+ (rOQeROQ+rOQOQeOQ)

We substitute the known information that vO = 0, rOQ = 1.2 m, and OQ =1rads to obtain

vQ=1.2eOQAs in 17.92, we write the velocity vQ using the coordinates of body P Q:

vQ= vP+ vQ/P =vP+ (rPQeRPQ+rPQPQePQ)

Again, we may substitute the known information, in this case: vP =vPi(i.e., direction is constrained to i),rPQ= 1.2,and rPQ= 0:

vQ= vPi+ 1.2PQePQ

We set vQ equal to vQ to obtain:

vQ=1.2eOQ = vPi+ 1.2PQePQWe now have available two scalar equations to find the two unknowns vP and PQ, but we do need to convert to a

common set of coordinates to solve. Unlike 17.92, = 50 here, so I find it simpler to convert to iandj in this case:

eOQ =0.766i+ 0.663j , ePQ=0.766i0.663j

Thus,1.2eOQ =1.2(0.766i+ 0.663j) = vPi+ 1.2PQ(0.766i0.663j)

from which we may solve each coordinate independently to find

PQ=1.20.6631.20.663= 1

rad

s , vP =1.20.766 + (1.2)(0.766) = 1.84 m

s

Now repeat the process for accelerations:

aQ= aO+ aQ/O = aO+{(rOQr2OQ)eROQ+ (2rOQOQ+rOQOQ)eOQ}SubstitutingaO = 0, rOQ = 1.2, rOQ = 0, rOQ = 0, OQ =1, and OQ = 0, we have

aQ= (1.212)eROQ =1.2eROQIn terms of body P Q,

aQ= aP+ aQ/P =aP+{(rPQr2PQ)eRPQ+ (2rPQPQ+rPQPQ)ePQ}

SubstitutingaQ=1.2eROQ,aP =aPi, rPQ= 1.2, rPQ= 0, rPQ= 0, and PQ= 1, we have

aQ= (1.2)eROQ = aPi1.2eRPQ+ 1.2PQePQWe have two scalar equations to solve for aP and PQ. We convert to a common coordinate system, say, i andj,using

eROQ = 0.663i+ 0.766j , eRPQ=0.663i+ 0.766j , ePQ=0.766i0.663jThus,

1.2(0.663i+ 0.766j) = aPi1.2(0.663i+ 0.766j) + 1.2PQ(0.766i0.663j)Solving, we find

PQ=(1.20.766) + (1.20.766)

(1.20.663) = 0 rad

s2 , aP = (1.2)(0.663)(1.2)(0.663) = 1.59 m

s2

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dynamics 17-92 17-93 solution

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