dv = a dt dv A = (3t + 2) dt L v A = 1.5t 2 + 2t t = 5 = 47.5 rad>s (47.5)(50) = v C (50) v C = 47.5 rad>s v B (75) = 47.5(50) v B = 31.7 rad>s Ans.

Transcription

1 16 9. When only two gears are in mesh, the driving gear and the driven gear will always turn in opposite directions. In order to get them to turn in the same direction an idler gear is used. In the case shown, determine the angular velocity of gear when t = 5 s, if gear starts from rest and has an angular acceleration of a = (3t + 2) rad>s 2, where t is in seconds. 75 mm 50 mm 50 mm a dv = a dt Idler gear Driving gear L0 v t dv = (3t + 2) dt L 0 v = 1.5t 2 + 2t t = 5 = 47.5 rad>s (47.5)(50) = v (50) v = 47.5 rad>s v (75) = 47.5(50) v = 31.7 rad>s

2 The disk is originally rotating at v 0 = 8 rad>s. If it is subjected to a constant angular acceleration of a = 6 rad>s 2, determine the magnitudes of the velocity and the n and t components of acceleration of point just after the wheel undergoes 2 revolutions. v 2 = v a c (u - u 0 ) v 2 = (8) 2 + 2(6)[2(2p) - 0] v = rad>s 1.5 ft V 0 8 rad/s 2 ft v = vr = 14.66(1.5) = 22.0 ft>s (a ) t = ar = 6(1.5) = 9.00 ft>s 2 (a ) n = v 2 r = (14.66) 2 (1.5) = 322 ft>s 2

3 * If bar has an angular velocity v = 4 rad>s, determine the velocity of the slider block at the instant shown. For link : Link rotates about a fixed point. Hence 200 mm 30 y = v r = 4(0.15) = 0.6 m>s For link v = {0.6 cos 30 i sin 30 j}m>s v = y i v = v k r > = {-0.2 sin 30 i cos 30 j} m 150 mm v 4 rad/s 60 v = v + v * r > y i = (0.6 cos 30 i sin 30 j) + (v k) * (-0.2 sin 30 i cos 30 j) y i = ( v )i - ( v )j Equating the i and j components yields: 0 = v v = -3 rad>s y = (-3) = 1.04 m>s :

4 If link has an angular velocity of v = 4 rad>s at the instant shown, determine the velocity of the slider block E at this instant.lso, identify the type of motion of each of the four links. E 1 ft D 30 2 ft Link rotates about the fixed point. Hence 2 ft 1 ft 30 v 4 rad/s y = v r = 4(2) = 8 ft>s For link D v = {-8 cos 60 i - 8 sin 60 j} ft>s v D = -y D i v D = v D k r D> = {1i} ft v D = v + v D * r D> -y D i = (-8 cos 60 i - 8 sin 60 j) + (v D k) * (1i) -y D i = -8 cos 60 i + (v D - 8 sin 60 )j : + -y D = -8 cos 60 y D = 4 ft>s (+ c) 0 = v D - 8 sin 60 v D = rad>s For Link DE v D = {-4i} ft>s v DE = v DE k v E = -y E i r E>D = {2 cos 30 i + 2 sin 30 j} ft v E = v D + v DE * r E>D -y E i = -4i + (v DE k) * (2 cos 30 i + 2 sin 30 j) -y E i = (-4-2 sin 30 v DE )i + 2 cos 30 v DE j : + 0 = 2 cos 30 v DE v DE = 0 + c -y E = -4-2 sin 30 (0) y E = 4ft>s ;

5 If end of the hydraulic cylinder is moving with a velocity of v = 3 m>s, determine the angular velocity of rod at the instant shown. 0.4 m 0.4 m Rotation bout a Fixed xis: Referring to Fig. a, v = v r = v (0.4) v 3 m/s 45 General Plane Motion: The location of the I for rod is indicated in Fig. b. From the geometry shown in this figure, we obtain r >I = 0.4 cos 45 r >I = m Thus, the angular velocity of rod can be determined from Then, v = r >I = 0.4 tan 45 = 0.4 m v r >I = 3 = rad>s v = v r >I v (0.4) = 5.303(0.4) v = 5.30 rad>s

6 * t the given instant member has the angular motions shown. Determine the velocity and acceleration of the slider block at this instant. 7 in. 3 rad/s 2 rad/s 2 v = 3(7) = 21 in.>s ; 5 in. v = v + v * r > -v a 4 5 bi - v a 3 bj = -21i + vk * (-5i - 12j) 5 : v = v (+ c) -0.6v = -5v Solving: v = rad>s in. v = in.>s = 9.38 in.>s (a ) n = (3) 2 (7) = 63 in.>s 2 T (a ) t = (2)(7) = 14 in.>s 2 ; a = a + a * r > - v 2 r > -a a 4 5 bi - a a 3 5 bj = -14i - 63j + (ak) * (-5i - 12j) - (1.125)2 (-5i - 12j) : a = a (+ c) -0.6a = -63-5a a = 54.7 in.>s 2 a = rad>s 2

7 t the instant shown rod has an angular velocity v = 4 rad>s and an angular acceleration a = 2 rad>s 2. Determine the angular velocity and angular acceleration of rod D at this instant. The collar at is pin connected to D and slides freely along. v 4 rad/s a 2 rad/s m oordinate xes: The origin of both the fixed and moving frames of reference are located at point. The x, y, z moving frame is attached to and rotate with rod since collar slides along rod m D Kinematic Equation: pplying Eqs and 16 27, we have a = a +Æ # v = v +Æ*r > + (v > ) xyz * r > +Æ*(Æ *r > ) + 2Æ *(v > ) xyz + (a >) xyz [1] [2] Motion of moving reference v = 0 a = 0 Æ=4k rad>s Æ # = 2k rad>s 2 Motion of with respect to moving reference r > = 50.75i6m (v > ) xyz = (y > ) xyz i (a > ) xyz = (a > ) xyz i The velocity and acceleration of collar can be determined using Eqs and with r >D = {-0.5 cos 30 i sin 30 j }m = { i j} m. v = v D * r >D = -v D k * ( i j) = v D i v D j a = a D * r >D - v 2 D r >D = -a D k * ( i j) - v 2 D( i j) = v 2 D a D i a D v 2 Dj Substitute the above data into Eq.[1] yields v = v +Æ*r > + (v > ) xyz v D i v D j = 0 + 4k * 0.75i + (y > ) xyz i v D i v D j = (y > ) xyz i j Equating i and j components and solve, we have (y > ) xyz = m>s v D = rad>s = 6.93 rad>s Substitute the above data into Eq.[2] yields a = a +Æ # * r > +Æ*(Æ *r > ) + 2Æ *(v > ) xyz + (a > ) xyz a D Di a D Dj = 0 + 2k * 0.75i + 4k * (4k * 0.75i) + 2 (4k) * (-1.732i) + (a > ) xyz i ( a D )i + ( a D + 12)j = (a > ) xyz Di j Equating i and j components, we have (a > ) xyz = m>s 2 a D = rad>s 2 = 56.2 rad>s 2 d

8 The quick return mechanism consists of the crank D and the slotted arm. If the crank rotates with the angular velocity and angular acceleration at the instant shown, determine the angular velocity and angular acceleration of at this instant. 2 ft v D 6 rad/s a D 3 rad/s 2 D Reference Frame: The xyz rotating reference frame is attached to slotted arm and coincides with the XYZ fixed reference frame at the instant considered, Fig. a. Thus, the motion of the xyz reference frame with respect to the XYZ frame is v = a = 0 v = v k v # = a k ft For the motion of point D with respect to the xyz frame, we have r D> = [4i] ft (v rel ) xyz = (v rel ) xyz i (a rel ) xyz = (a rel ) xyz i Since the crank D rotates about a fixed axis, v D and a D with respect to the XYZ reference frame can be determined from v D = v D * r D = (6k) * (2 cos 30 i - 2 sin 30 j) = [6i j] ft>s a D = a D * r D - v D 2 r D = (3k) * (2 cos 30 i - 2 sin 30 j) (2 cos 30 i - 2 sin 30 j) = [-59.35i j] ft>s 2 Velocity: pplying the relative velocity equation, v D = v + v * r D> + (v rel ) xyz 6i j = 0 + (v k) * (4i) + (v rel ) xyz i 6i j = (v rel ) xyz i + 4v j Equating the i and j components yields (v rel ) xyz = 6 ft>s = 4v v = rad>s = 2.60 rad>s cceleration: pplying the relative acceleration equation, a D = a + v # * r D> + v * (v * r ) + 2v * (v rel ) xyz + (a rel ) xyz i j = 0 + (a k) * 4i k * [(2.598k) * (4i)] + 2(2.598k) * (6i) + (a rel ) xyz i i j = c(a rel ) xyz - 27 di + (4a )j Equating the i and j components yields = 4a a = 2.50 rad>s 2