0.5 rad r C 20 mm. 30 deg r s 50 mm. r A. 200 mm. Solution: v C 0.01 m s. v C. r s. 0.2 rad. v A v E s r A

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1 The mechanim for a car window winder i hown in the figure. Here the handle turn the mall cog C, which rotate the pur gear S, thereby rotating the fixed-connected lever which raie track D in which the window ret. The window i free to lide on the track. If the handle i wound with angular velocity c, determine the peed of point and E and the peed v w of the window at the intant. Given: c.5 rad r C 2 mm 3 deg r 5 mm r 2 mm Solution: v C c r C v C.1 m v C r.2 rad v v E r v r v 4 mm Point and E move along circular path. The vertical component cloe the window. v w v co v w 34.6 mm

2 Due to the crew at E, the actuator provide linear motion to the arm at F when the motor turn the gear at. If the gear have the radii lited in the figure, and the crew at E ha a pitch p = 2 mm, determine the peed at F when the motor turn at v = 2 rad>. Hint: The crew pitch indicate the amount of advance of the crew for each full revolution. D E F v r = v r C r 1 mm r 5 mm r C 15 mm r D 6 mm v C r C = v D r D Thu, v D = r r r C r D v = = 1 rad> 6 v F = 1 rad > 2p rad 1 rev (2 mm) =.318 mm >

3 16 4. The mechanim i ued to convert the contant circular motion v of rod into tranlating motion of rod CD and the attached vertical lot. Determine the velocity and acceleration of CD for any angle u of. D u l V C x x = l co u x # = v x = -l in uu # x # = a x = -l(in uu $ + co uu # 2 ) Here v,, and u #, u $ x = v CD a x = a CD = v = a =. v CD = -l in u(v) = -vl in u a CD = -lcin u() + co u(v) 2 D = - v 2 l co u Negative ign indicate that both v CD and a CD are directed oppoite to poitive x.

4 Peg mounted on hydraulic cylinder D lide freely along the lot in link C. If the hydraulic cylinder extend at a contant rate of.5 m>, determine the angular velocity and angular acceleration of the link at the intant u = 45. C.5 m/ u Poition Coordinate: From the geometry hown in Fig. a, y C =.6 tan u m Time Derivative: Taking the time derivative, D.6 m v C = y # C = (.6 ec 2 uu # ) m> (1) Here, v C =.5 m> ince v C act in the poitive ene of y C. When u = 45, Eq. (1) give.5 =.6 ec 2 45 u # v = u # =.4167 rad> =.417 rad> The time derivative of Eq. (1) give a C = y $ C =.6ec 2 uu $ + 2 ec u ec u tan uu # 2 a C =.6 ec 2 uu $ + 2 tan uu # 2 (2) Since i contant, a C =. Thu, Eq. (2) give v C = u $ + 2 tan a = u $ = rad> 2 =.347 rad> 2 d The negative ign indicate that a act counterclockwie.

5 rm ha an angular velocity of V and an angular acceleration of. If no lipping occur between the dik and the fixed curved urface, determine the angular velocity and angular acceleration of the dik. ω ', α' C r d = (R - r) du = - rdf ω, α R (R - r) du df = -r dt dt (R - r) v v =- r (R - r) a a =- r

6 The Geneva wheel provide intermittent rotary motion v for continuou motion v D = 2 rad> of dik D. y chooing d = 122 mm, the wheel ha zero angular velocity at the intant pin enter or leave one of the four lot. Determine the magnitude of the angular velocity V of the Geneva wheel at any angle u for which pin i in contact with the lot. 1 mm v D u d 1 u 2 mm f v f D 1 mm tan f =.1 in u co u = in u 22 - co u ec 2 ff # = 22 - co u(co uu# ) - in u(in uuª ) 22 - co u 2 = 22 co u co u 2 u# (1) From the geometry: r 2 = (.1 in u) 2 + [ co u] 2 = co u ec 2 f = r co u = = 2 [ co u] [ cou] 2 (3-222 co u) 22 - co u 2 From Eq. (1) co u 22 co u - 1 fª= co u 22 - co u u# 2 f # = 22 co u co u u# Here f = v and u # = v D = 2 rad> 22 co u - 1 v = 2a co u b

7 If rod CD i rotating with an angular velocity DC, determine the angular velocitie of rod and C at the intant hown. Given: DC 8 rad r 15 mm 1 45 deg r C 4 mm 2 3 deg r CD 2 mm Solution: Guee 3 2 deg 1 rad C 1 rad Given.DC r in 1 r C in 3 r CD in 2 r.cd co.2 r.cd in.2.c r.c co.3 r.c in.3 r. co.1 r. in.1 =. 3 C Find 3 C deg C Poitive mean CCW Negative mean CW rad

8 t the intant hown, the truck i traveling to the right at peed v, while the pipe i rolling counterclockwie at angular velocity without lipping at. Determine the velocity of the pipe center G. Given: v 3 m 8 rad r 1.5 m Solution: v G v v G v G v r v G 9. m

9 If bar ha an angular velocity, determine the velocity of the lider block C at the intant hown. Given: 6 rad 45 deg 3 deg r r C 2 mm 5 mm Solution: Guee C 2 rad v C 4 m Given r co r in C r C co r C in v C C v C rad m Find C v C C 1.96 v C 1.34

10 If the lider block i moving downward at v = 4 m>, determine the velocity of block C at the intant hown. 25 mm mm E C 4 mm 3 D 3 mm v 4 m/ General Plane Motion: pplying the relative velocity equation by referring to the kinematic diagram of link hown in Fig. a, v = v + V * r > v i = -4j + (-v k) * c -.55a 4 5 b i +.55a 3 5 bj d v i =.33v i + (.44v - 4) j Equating j component, =.44v - 4 v = 9.91 rad>b Uing the reult of, Uing the reult of v v D = v + V * r D> = -4j + (-9.91k) * c -.3a 4 5 bi +.3a 3 5 bj d = {1.636i j} m> v D to conider the motion of link CDE,Fig.b, v C = v D + V CD * r C>D v C i = (1.636i j) + (-v CD k) * (-.4 co 3 i -.4 in 3 j) v C i = ( v CD )i + (.3464v CD )j Equating j and i component, =.3464v CD v CD = rad>b v C = (5.249) =.587 m> :