SOLUTION If link AB is rotating at v AB = 6 rad>s, determine the angular velocities of links BC and CD at the instant u = 60.

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1 If link i rotating at v = 6 rad>, determine the angular velocitie of link C and CD at the intant u = mm 3 3 mm ω = 6 rad/ C 4 mm r IC - =.3co 3 =.2598 m r IC - C =.3co 6 =.15 m θ D v C = 1.5 = = 5.77 rad>.2598 v C = 5.774(.15) =.8661 m> v CD = = 2.17 rad

2 Determine the angular velocity of the double-tooth gear and the velocity of point C on the gear. v 4 m/.3 m.15 m C v 6 m/ General Plane Motion: The location of the IC can be found uing the imilar triangle hown in Fig. a. r >IC 4 =.45 - r >IC 6 r >IC =.18 m Then, y =.3 - r >IC = =.12 m and r C>IC = =.3231 m f = tan - 1 a.12.3 b = 21.8 Thu, the angular velocity of the gear can be determined from v = v = 4 = rad> = 22.2 rad> r >IC.18 Then v C = vr C>IC = 22.2(.3231) = 7.18 m> nd it direction i f = 9 - f = = 68.2

3 16 1. The imilar link and CD rotate about the fixed pin at and C. If ha an angular velocity v = 8 rad>, determine the angular velocity of DP and the velocity of point P. 3 mm 3 mm 3 mm 6 6 D 3 mm =8rad/ ω C 7 mm Kinematic Diagram: Since link and CD i rotating about fixed point and C. then v and v D are alway directed perpendicular to link and CD repectively. The magnitude of v and v D are v = v r = 8(.3) = 2.4 m> and v. t the intant hown. v and v D are D = v CD r CD =.3 v CD directed at 3 with the horizontal. P Intantaneou Center: The intantaneou center of zero velocity of link DP at the intant hown i located at the interection point of extended line drawn perpendicular from v and v D. From the geometry. r >IC = The angular velocity of link DP i given by.3 co 6 =.6 m r P>IC =.3 tan = 1.22 m v DP = v = 2.4 = 4. rad r >IC.6 Thu, the velocity of point P i given by v P = v DP r P IC = 4.(1.22) = 4.88 m ;

4 The flywheel rotate with angular velocity and angular acceleration Determine the angular acceleration of link and C at the intant hown. Given: 2 rad a.4 m 6 rad 2 b.5 m r.3 m e 3 Solution: d 4 d b r 1 r r 2 e r e 2 d 2 3 a k 1 Guee 1 rad C 1 rad 1 rad 2 C 1 rad 2 Given k r 1 k r 2 C k r 3 k r 1 k k r 1 k r 2 k k r 2 C k r 3 C k C k r 3 C C Find C C C C rad rad 2

5 The center O of the gear and the gear rack P move with the velocitie and acceleration hown. Determine the angular acceleration of the gear and the acceleration of point located at the rim of the gear at the intant hown. 15 mm v O 3 m/ a O 6 m/ 2 v P 2 m/ a P 3 m/ 2 O P ngular Velocity: The location of the IC i indicated in Fig. a. Uing imilar triangle, 3 r O>IC = r O>IC r O>IC =.9 m Thu, v = v O r O>IC = 3.9 = rad> cceleration and ngular cceleration: pplying the relative acceleration equation to point O and and referring to Fig. b, a = a O + a * r >O - v 2 r >O -3i + (a ) n j = 6i + (-ak) * (-.15j) (-.15j) -3i + (a ) n j = (6 -.15a)i j Equating the i component, -3 = a a = 6 rad> 2 Uing thi reult, the relative acceleration equation i applied to point O and, Fig. b, which give a = a O + a * r >O - v 2 r >O (a ) t i - (a ) n j = 6i + (-6k) * (.15j) (.15j) (a ) t i - (a ) n j = 15i j Equating the i and j component, (a ) t = 15 m> 2 (a ) n = m> 2 Thu, the magnitude of a i a = 3(a ) t 2 + (a ) n 2 = = 167 m> 2 and it direction i u = tan -1 c (a ) n d = tan -1 a b = 84.9 c (a ) t 15

6 The hoop i cat on the rough urface uch that it ha angular velocity and angular acceleration. lo, it center ha a velocity v and a deceleration a. Determine the acceleration of point at thi intant. Given: 4 rad a 2 m 2 5 rad 2 r.3 m v 5 m 45 deg Solution: a a rco rin rco rin 4.33 m a a 6.21 m 2

7 t the intant hown, rod ha an angular velocity v and an angular acceleration a = 5 rad> 2 = 3 rad>. Determine the angular velocity and angular acceleration of rod CD at thi intant. The collar at C i pin-connected to CD and lide over. v 3 rad/ a 5 rad/ m C.5 m r C> = (.75 in 6 )i - (.75 co 6 )j r C> = {.6495i -.375j} m D v C = v CD * r C>D = (v CD k) * (.5j) = {-.5v CD i}m> a C = a CD * r CD - v 2 CD r CD = (a CD k) * (.5j) - v 2 CD(.5j) a C = {-.5 a CD i - v 2 CD(.5)j} m> 2 v C = v +Æ*r C> + (v C> ) xyz -.5v CD i = + (3k) * (.6495i -.375j) + v C> in 6 i - v C> co 6 j -.5v CD = v C> = v C> v C> = m> v CD = -9. rad> = 9. rad>b a C = a +Æ # * r C> +Æ*(Æ *r C> ) + 2Æ *(v C> ) xyz + (a C> ) xyz a C = + (5k) * (.6495i -.375j) + (3k) * [(3k) * (.6495i -.375j)] + 2(3k) * [3.897(.866)i -.5(3.897)j] +.866a C> i -.5a C> j.5a CD i - (-9.) 2 (.5)j = i j i j i j +.866a C> i -.5a C> j.5a CD = a C> -4.5 = a C> a C> = m> 2 a CD = 249 rad> 2 b

8 The quick-return mechanim conit of a crank, lider block, and lotted link CD. If the crank ha the angular motion hown, determine the angular motion of the lotted link at thi intant. 1 mm D v 3 rad/ a 9 rad/ 2 3 v = 3(.1) =.3 m> (a ) t = 9(.1) =.9 m> mm (a ) n = (3) 2 (.1) =.9 m> 2 v = v C +Æ*r >C + (v >C ) xyz.3 co 6 i +.3in 6 j = + (v CD k) * (.3i) + v >C i v >C =.15 m> v CD, a CD C v CD =.866 rad> d a = a C +Æ # * r >C +Æ*(Æ *r >C ) + 2Æ *(v >C ) xyz + (a >C ) xyz.9 co 6 i -.9 co 3 i +.9 in 6 j +.9 in 3 j = + (a CD k) * (.3i) +(.866k) * (.866k *.3i) + 2(.866k *.15i) + a >C i i j =.3a CD j -.225i j + a >C i a >C = -.14 m> 2 a CD = 3.23 rad> 2 d

9 t the intant hown, boat travel with a peed of 15 m>, which i decreaing at 3m> 2, while boat travel with a peed of 1 m>, which i increaing at 2m> 2. Determine the velocity and acceleration of boat with repect to boat at thi intant. 3 m 15 m/ 5 m 5 m 3m/ 2 1 m/ 2m/ 2 Reference Frame: The xyz rotating reference frame i attached to boat and coincide with the XYZ fixed reference frame at the intant conidered, Fig. a. Since boat and move along the circular path, their normal component of acceleration are and (a ) n = v 2 (a. r = 12 ) n = v 2 r = 152 = 4.5 m>2 5 5 = 2m>2 Thu, the motion of boat and with repect to the XYZ frame are v = [15j] m> v = [-1j] m> a = [-4.5i - 3j] m> 2 a = [2i - 2j] m> 2 lo, the angular velocity and angular acceleration of the xyz reference frame with repect to the XYZ reference frame are v = v r = 15 =.3 rad> v = [.3k] rad> 5 v # = (a ) t r = 3 5 =.6 rad>2 v # = [-.6k] rad> 2 nd the poition of boat with repect to boat i r > = [2i] m Velocity:pplying the relative velocity equation, v = v + v * r > + (v rel ) xyz -1j = 15j + (.3k) * (2i) + (v rel ) xyz -1j = 21j + (v rel ) xyz (v rel ) xyz = [-31j] m> cceleration: pplying the relative acceleration equation, a = a + v # * r > + v(v * r > ) + 2v * (v rel ) xyz + (a rel ) xyz (2i - 2j) = (-4.5i - 3j) + (-.6k) * (2i) + (.3k) * C(.3k) * (2i)D + 2(.3k) * (-31j) + (a rel ) xyz 2i - 2j = 12.3i - 4.2j + (a rel ) xyz (a rel ) xyz = [-1.3i + 2.2j] m> 2

10 The dik rotate with the angular motion hown. Determine the angular velocity and angular acceleration of the lotted link C at thi intant. The peg at i fixed to the dik. Given: 6 rad 1 rad 2 l.75 m 3 deg 3 deg r.3 m Solution: u 1 in co co u 2 in k r 1 lu 1 r 2 ru 2 1 Guee C 1 rad C 1 rad 2 v rel 1 m a rel 1 m 2 Given k r 2 C k r 1 v rel u 1 k r 2 2 r 2 C k r 1 2 C r1 a rel u 1 2 C k v rel u 1 C C v rel a rel m Find C C v rel a rel v rel 1.8 a rel 3 m 2 C rad rad.c