An Introduction to Three-Dimensional, Rigid Body Dynamics. James W. Kamman, PhD. Volume I: Kinematics. Unit 4

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1 Summary n Introduction to Three-imensional, igid ody ynamics James W. Kamman, h Volume I: Kinematics Unit 4 Kinematics of a oint Moving on a igid ody This unit continues the development of the concepts of velocity and acceleration vectors and shows how to calculate them using formulae for points moving on non-stationary rigid bodies. s before, this technique can be applied to complex mechanical systems; however, at this point the focus will remain on systems in which components are connected by simple revolute (pin) joints. This technique will be generalized in Unit 7 to apply to systems with more complex connecting joints. age Count Examples Suggested Exercises Copyright James W. Kamman, 06

2 Kinematics of a oint Moving on a igid ody General Concept The kinematic analysis is now extended to include systems where interconnected bodies may rotate and translate relative to each other. In this case, the kinematics of points that are moving on (or relative to) rotating bodies is needed. To analyze this motion, consider the figure shown at the right. Here, : is a fixed reference frame, : is a rigid body, : is a point moving on, and ˆ : is a point fixed on that coincides with at this instant of time The velocity and acceleration of may be written as v v v ˆ a a a v ˆ where each of the terms are defined as follows v ˆ, a : the velocity and acceleration of on, assuming is fixed v, a : the velocity and acceleration of ˆ in (recall that ˆ is fixed on ) ˆ v : Coriolis acceleration of in Note: The velocity and acceleration of ˆ can be determined by using the two-point formulae for points that are fixed on rigid bodies as discussed in Unit 3. erivation The results shown above can be easily shown by using the derivative rule discussed in Unit. Consider the rigid body shown in the diagram at the right. Here, : is a fixed reference frame, : is a rigid body, : is a point moving on ˆ : is a point fixed on that coincides with Q : is a point fixed on not coincident with Copyright James W. Kamman, 06 Volume I Unit 4: page /0

3 The velocity of may be found as follows d p d d q d r v q r dt dt dt dt dr dt v v r vq r v v ˆ v Q The acceleration of may be found as follows Q d a v v r dt d d d vq v r dt dt dt (as r 0 ) d d r a Q v v r r dt dt a a v r v r Q Now, letting r 0 gives the desired result. a a a v ˆ Example : The system shown consists of a hoop H which is affixed-to and rotates with the vertical shaft. s the shaft rotates about its axis a ball slides freely in the hoop. The shaft and hoop are rotating at a rate of (rad/s), and the position of the ball in the hoop is given by the angle. The radius of the hoop is a. In the solution that follows, the following reference frames will be useful. Find: : i, j, k (fixed frame) H : e, e, k (rotates with the hoop) e k e : e, er, e (rotates with the hoop and ball) er e e v the velocity of point in a the acceleration of point in Copyright James W. Kamman, 06 Volume I Unit 4: page /0

4 Solution: s hoop H rotates about the vertical axis, the ball slides within the hoop. To find the velocity and acceleration of, use the formulae for points moving on non-stationary bodies. where v v v ( ˆ is a point fixed on H that coincides with at the instant shown) H ˆ H v a S e ( ˆ has circular motion around the vertical axis; radius of the circle is as ) ˆ v a e a C e S k (circular motion of on H in the ( e, k ) plane) dding these two results, gives the velocity of. v a S e ac e a S k The formula for the acceleration of a point moving on a body is a a a v ˆ H H H Here, ˆ has the same meaning as before and the last term is the Coriolis acceleration. a a S e a S e (circular motion of ˆ in ) ˆ H a a e a er (circular motion of on H) H H v k a C e S k a C e (Coriolis acceleration) dding these three results, gives the acceleration of. a as a C e a C a S a S e a S a C k Example : The system shown consists of two connected bodies the frame F and the disk. Frame F rotates at a rate of (rad/s) about the fixed vertical direction (annotated by the unit vector k ). isk is affixed to and rotates relative to F at a rate of (rad/s) about the horizontal arm of F (annotated by the rotating unit vector e ). eference frames: :( i, j, k ) (fixed frame) F :( e, e, k ) (rotating frame) Copyright James W. Kamman, 06 Volume I Unit 4: page 3/0

5 Find: (express the results using unit vectors fixed in F) a) v the velocity of point in using the formula for points moving on bodies b) a the acceleration of point in using the formula for points moving on bodies Solution: This problem has been solved in previous Units using other methods. Here the formulae for points moving on rigid bodies are used. To do this, point is treated as a point that is moving on frame F (rather than in previous notes where was defined to be a point fixed on the disk). a) Using this approach, the velocity of may be written as v v v (here, ˆ is fixed on F and coincides with at the instant shown) F ˆ The points ˆ and O are both fixed on F, so the two-point formula for velocity can be used to find v v v r k ac e e a S k v ac e e ˆ O ˆ / O F ˆ /O ˆ The velocity of on F is found by holding the frame F fixed. ( has circular motion on F) v v v r e ac e a S k v ac k a S e F F F F F Q / Q / Q dding these two results gives the velocity of. v a S e a C e a C k b) Using this same approach, the acceleration of may be written as a a a v (here again, ˆ is fixed on F and coincides with ) ˆ F F F where, using the two-point formula for acceleration a a a r v ˆ O ˆ / O F ˆ /O F ˆ / O k a C e e a S k k a C e e from part (a) a C e e a C e e a ˆ a C e a C e v. ˆ Copyright James W. Kamman, 06 Volume I Unit 4: page 4/0

6 a r v F F F F / Q / Q e ac e a S k e ac k a S e ac k a S e ac e a S k F a a S a C e a C a S k ( has circular motion on F) F F v k ac k a S e a S e (Coriolis acceleration) dding these three results gives the final result. ac a S k a a S a C e a S a C e Example 3: Slider on a otating ar The figure shows the top and side views of a system which consists of a disk, bar, and a slider. The disk is affixed-to and rotates with a shaft about a vertical axis (direction annotated by unit vector e ). The angle of rotation is given by the variable. The bar is pinned to the front edge of the disk and swings freely about the axis of the pin (along the e 3 direction). The angle of the bar relative to the vertical is. The slider slides along the bar. It s position relative to the pivot point Q is given by the variable x. s shown, could slide off the end of the bar, but it could easily be held on the bar using a spring. eference frames: : ( e, e, e ) 3 : ( e, e, e ) r 3 Find: a) v the velocity of in a fixed frame b) a the acceleration of in Solution: a) To find the velocity of as it moves on the bar, the angular velocity of the bar is needed. Using the summation rule for angular velocities, the angular velocity of relative to the ground frame may be written as e e 3 Copyright James W. Kamman, 06 Volume I Unit 4: page 5/0

7 Using the formula for points moving on bodies, the velocity of may now be written as ˆ v v v ( moves on ; ˆ is fixed on and coincides with ) where, using the two-point formula for velocity v ˆ v Q v ˆ/ Q Q ˆ / a e r 0 e e e 3 ae x S xc 0 v ( a xc ) e ( xs ) e ( x S ) e ˆ 3 v x e x ( S e C e ) ( has rectilinear motion on ) r dding the two previous equations gives the final result. v ( a xc x S ) e ( xs xc ) e ( x S ) e 3 b) To find the acceleration of as it moves on the bar, the angular acceleration of the bar is also needed. ecall that the angular acceleration is found by differentiating the angular velocity vector. d de de dt dt dt 3 e e3 e e3 e e3 e e e 3 Using the formulae for points moving on bodies, the acceleration of may now be written as where a a ˆ a v Q aq a ˆ / a Q v 3 a a e a e (Q has circular motion about O in ) a ˆ / r Q ˆ / Q v ˆ / Q ( ) ( ) e e e e e e x S xc 0 xc xs x S xc x S xc e3 a x C x S x S e x S x C e Q ˆ / ( ˆ and Q are both fixed on ) Copyright James W. Kamman, 06 Volume I Unit 4: page 6/0

8 r a x e x S e C e ( has rectilinear motion on ) e e e 3 v 0 x C e x S e x S e 3 x S xc 0 (Coriolis acceleration) dding the four terms gives the final result. ( ) a x S x C x S e3 a a x C x S x S x C e x S x C xc x S e Note that in the solution presented for this example, the velocity and acceleration of point ˆ were calculated using the two-point formulae for points fixed on bodies. s shown in Example, the formulae for points moving on bodies could also be used. See Example 4 for a solution that extends the application of this latter approach. Example 4: Slider on a otating ar evisited In this example the velocity and acceleration of the slider in Example 3 are calculated by recursively applying the formulae for points moving on bodies. Solution: a) Using the formula for points moving on bodies, the velocity of may now be written as where ˆ ( moves on ; ˆ is fixed on and moves on ; ˆ is fixed on ) (points, ˆ, and ˆ all coincide) v v ˆ O r ˆ e x S e xc e a e 3 a e x S e 3 ˆ/ O (circular motion) ˆ v x e x C e S e (circular motion of ˆ around Q) r v x e x S e C e (rectilinear motion of on ) dding the three previous results gives the final result v v v v ( a xc x S ) e ( xs xc ) e ( x S ) e ˆ ˆ v v v 3 b) Using the formula for points moving on bodies, the acceleration of may now be written as Copyright James W. Kamman, 06 Volume I Unit 4: page 7/0

9 a a ˆ a v ˆ ˆ ˆ a a v a v (points, ˆ, and ˆ as previously defined) where / a a ˆ O r ˆ v O ˆ e x S e xc e a e e a e x S e 3 3 a a x S e x S a e ˆ 3 r a ˆ x C x S e x S x C e a x e x e x C e S e x S e C e ˆ v e x C e S e x e (Coriolis acceleration) r ˆ 3 a x e x S e C e ( has rectilinear motion on ) ˆ has circular motion in ˆ has circular motion on e e e3 v 0 x C e x S e x S e 3 x S xc 0 (Coriolis acceleration) dding the previous five terms gives the final result. ( ) a x S x C x S e3 a a x C x S x S x C e x S x C xc x S e Compare this approach to that taken in Example 3. Notes. The formulae for points fixed on bodies together with the formulae for points moving on bodies may be used recursively to compute the velocities and accelerations of remote points in a multibody system.. Using this approach, calculations of velocities and accelerations do not require differentiation. It requires only vector multiplication and addition. esults may be calculated that are valid for all time or only a specific instant of time. ecall that when using direct differentiation, the method requires the results be valid for all time. s always, differentiation is still required to compute angular acceleration vectors. Copyright James W. Kamman, 06 Volume I Unit 4: page 8/0

10 Exercises: 4. The antenna system shown has two components, the base and the antenna dish. ase rotates relative to the ground about the fixed z-axis, and dish rotates relative to about the rotating x-axis. t any instant, the angle between the y-axis ( e ) and the fixed Y-axis is, and the angle between line O and the y-axis is. Given values for,, and their time derivatives, find v and a the velocity and acceleration of point in a fixed frame using the formulae for points moving on bodies. nswers: v LC e S e C k (results expressed in :( e, e, k ) ) a L C S e S C C e C S k 4. The system shown has three components, a vertical column C, a horizontal arm M, and a disk. The disk rotates relative to the arm at a rate 3 (rad/sec) about the n 3 direction (normal to ), the arm rotates relative to the column at a rate of (rad/sec) about the e direction, and the column rotates relative to the ground at a rate of (rad/sec) about the fixed e direction. The unit vector set C :( e, e, e 3) is fixed in the column, and the unit vector set M :( e, n, n 3) is fixed in arm M. Given values for,, 3, and their time derivatives, find v and a the velocity and acceleration of point in a fixed frame using the formulae for points moving on bodies. nswers: (expressed in M :( e, n, n 3) ) v r( S S S ) e ( r C ( L rc ) S ) n ( r S ( L rc ) C ) n a ( r S S r S r C ( L rc ) r S C r S C ) e ( r C ( L rc ) S r S r S S r S r S S ) n ( r S ( L rc ) C r S S C r S C r C ) n Hint: Here is the angle between the plane of the disk and the (, ) shows the position where 0. e e plane. The diagram Copyright James W. Kamman, 06 Volume I Unit 4: page 9/0

11 4.3 The system shown consists of two bodies and a collar. The bent bar O is connected to the ground by a simple revolute joint at which allows rotation about the fixed e direction only. ody C is connected to O with a simple revolute joint which allows rotation of C relative to O only about the e 3 direction. The collar at C is traveling along body C at a speed of c u. Calculate v and a the velocity and acceleration of C relative to the ground at the instant shown. C C nswers: vc u b e a c e be3 C b a c e3 eferences: (at the instant shown) a c b u e a c b b u e. T.. Kane,.W. Likins, and.. Levinson, Spacecraft ynamics, McGraw-Hill, 983. T.. Kane and.. Levinson, ynamics: Theory and pplication, McGraw-Hill, L. Huston, Multibody ynamics, utterworth-heinemann, H. aruh, nalytical ynamics, McGraw-Hill, H. Josephs and.l. Huston, ynamics of Mechanical Systems, CC ress, C. Hibbeler, Engineering Mechanics: ynamics, 3 th Ed., earson rentice Hall, 03. (at the instant shown) 7. F.. eer and E.. Johnston, Jr., Vector Mechanics for Engineers: ynamics, 4 th Ed., McGraw-Hill, 984. Copyright James W. Kamman, 06 Volume I Unit 4: page 0/0