91962_06_s16_p0513-0640
6/8/09
3:27 PM
Page 633
© 2010 Pearson Education, Education, Inc., Upper Saddle River, River, NJ. All rights reserved. reserved. This material material is protected under all copyright laws as they currently currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
shown, car B travels with a speed *16–152. At the instant shown, of 15 m s , which is increasing at a constant rate of 2 m s2 , while car C travels travels with a speed of 15 m s , which is increasing increasing at a constant constant rate of 3 m s2 . Determine Determine the velocity and acceleration of car B with respect to car C .
>
>
>
>
45
250 m
15 m/ s 2 m/ s2
C
B 200 m
Reference Frame: The xyz rotating reference frame is attached to C and coincides
A
with the XYZ fixed reference frame at the instant considered, Fig. a. Since Since B and C
2 m/ s2
move along the circular circular road, their normal normal components components of acceleration acceleration are vB 2 vC 2 152 152 2 2 (aB)n = and (aC)n = = = 0.9 m s = = 0.9 m s . Thus, the r r 250 250 motion of cars B and C with respect to the XYZ frame are
>
>
>
vB
=
[ - 15i] m s
vC
=
[ - 15 cos 45°i
aB
=
[ - 2i
aC
=
( - 0.9 cos 45° - 3 cos 45°)i
15 sin 45° j]
-
[ - 10.607i
=
>
10.607 j] m s
-
>
0.9 j] m s2
+
(0.9 sin 45° - 3 sin 45°) j
+
=
[ - 2.758i
-
>
1.485 j] m s2
Also, the angular velocity and angular acceleration of the xyz reference frame with respect to the XYZ reference frame are v =
#
v =
vC r
(aC)t r
>
=
15 250
=
0.06 rad s
=
3 250
=
0.012 rad s2
v =
>
#
v =
>
[ - 0.06k] rad s
>
[ - 0.012k] rad s2
From the geometry shown in Fig. a,
>
= - 250
rB C
sin 45°i
-
(250
-
250 cos 45°) j
[ - 176.78i
=
73.22 j] m
-
Velocity: Applying the relative velocity equation, vB
=
vC
- 15i =
+ v *
( - 10.607i
- 15i = - 15i +
(vrel)xyz
>
rB C
=
-
(v rel)xyz
+
10.607 j)
( - 0.06k)
+
*
( - 176.78i
-
73.22 j)
+
(vrel)xyz
(vrel)xyz
0
Ans.
Acceleration: Applying the relative acceleration equation, aB
=
aC
- 2i +
# + v *
0.9 j
=
>
rB C
+ v *
( - 2.758i
-
(v
*
1.485 j)
>
rB C) +
0.9 j
(a rel)xyz
=
= - 3i +
0.9 j
+
2v
*
( - 0.012k)
+ ( - 0.06k) * - 2i +
+
(vrel)xyz
*
[( - 0.06k)
(a rel)xyz
+
( - 176.78i
*
-
( - 176.78i
73.22 j)
-
73.22 j)]
+
2( - 0.06k)
(arel)xyz
>
[1i] m s2
Ans.
633
25 m/ s
*
0
+
(a rel)xyz
15 m/ s 3 m/ s2
91962_06_s16_p0513-0640
6/8/09
3:28 PM
Page 634
© 2010 Pearson Education, Inc., Upper Saddle River,NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•16–153. At the instant shown, boat A travels with a speed of 15 m s, which is decreasing at 3 m s2 , while boat B travels with a speed of 10 m s , which is increasing at 2 m s2 . Determine the velocity and acceleration of boat A with respect to boat B at this instant.
>
>
>
>
30 m
15 m/ s 50 m
A
B
50 m 10 m/ s 2 m/ s2
3 m/ s2
Reference Frame: The xyz rotating reference frame is attached to boat B and
coincides with the XYZ fixed reference frame at the instant considered, Fig.a. Since boats A and B move along the circular paths, their normal components of vA 2 vB 2 152 102 2 2 acceleration are (aA)n = = = 4.5 m s and (aB)n = = = 2m s . r r 50 50 Thus, the motion of boats A and B with respect to the XYZ frame are
>
>
vA
=
[15 j] m s
aA
=
[ - 4.5i
>
=
vB
>
3 j] m s2
-
=
aB
>
[ - 10 j] m s [2i
-
>
2 j] m s2
Also, the angular velocity and angular acceleration of the xyz reference frame with respect to the XYZ reference frame are vB
v =
#
v =
r
10 50
=
(aB)t
2 50
=
r
>
0.2 rad s
=
=
>
[0.2k] rad s
v =
>
#
0.04 rad s2
v =
>
[0.04k] rad s2
And the position of boat A with respect to B is
>
=
rA B
[ - 20i] m
Velocity: Applying the relative velocity equation, vA
=
+ v *
vB
>
rA B
+
(vrel)xyz
*
( - 20i)
15 j
= - 10 j +
(0.2k)
15 j
= - 14 j +
(vrel)xyz
(vrel)xyz
=
+
(vrel)xyz
>
[29 j] m s
Ans.
Acceleration: Applying the relative acceleration equation, aA
=
# + v *
aB
( - 4.5i
-
3 j)
- 4.5i -
3 j
(arel)xyz
=
=
>
rA B
(2i
-
= - 8.8i -
[4.3i
-
(v
+ v *
2 j)
+
>
rA B )
(0.04k)
+
2.8 j
*
*
2v
*
(vrel)xyz
( - 20i)
+
(0.2k)
+
*
+
(arel)xyz
C (0.2k)
*
(arel)xyz
>
0.2 j] m s2
Ans.
634
( - 20i) D
+
2(0.2k)
*
(29 j)
+
(arel)xyz
91962_06_s16_p0513-0640
6/8/09
3:28 PM
Page 635
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–154. At the instant shown, boat A travels with a speed of 15 m s, which is decreasing at 3 m s2 , while boat B travels with a speed of 10 m s , which is increasing at 2 m s2 . Determine the velocity and acceleration of boat B with respect to boat A at this instant.
>
>
>
>
30 m
15 m/ s 50 m
A
B
50 m 10 m/ s 2 m/ s2
3 m/ s2
Reference Frame: The xyz rotating reference frame is attached to boat A and
coincides with the XYZ fixed reference frame at the instant considered, Fig.a. Since boats A and B move along the circular paths, their normal components of vA 2 vB 2 152 102 2 2 acceleration are (aA)n = = = 4.5 m s and (aB)n = = = 2m s . r r 50 50 Thus, the motion of boats A and B with respect to the XYZ frame are
>
>
vA
=
[15 j] m s
aA
=
[ - 4.5i
-
>
>
3 j] m s2
>
vB
=
[ - 10 j] m s
aB
=
[2i
-
>
2 j] m s2
Also, the angular velocity and angular acceleration of the xyz reference frame with respect to the XYZ reference frame are v =
#
v =
vA r
=
(aA)t r
15 50
=
=
3 50
>
0.3 rad s
=
v =
>
#
0.06 rad s2
v =
>
[0.3k] rad s
>
[ - 0.06k] rad s2
And the position of boat B with respect to boat A is
>
=
rB A
[20i] m
Velocity: Applying the relative velocity equation, vB
=
+ v *
vA
>
rB A
+
(vrel)xyz
*
(20i)
- 10 j =
15 j
+
(0.3k)
- 10 j =
21 j
+
(vrel)xyz
(vrel)xyz
=
+
(vrel)xyz
>
[ - 31 j] m s
Ans.
Acceleration: Applying the relative acceleration equation, # + v *
aB
=
aA
(2i
-
2 j)
2i
=
>
rB A
( - 4.5i
2 j
=
12.3i
(arel)xyz
=
[ - 10.3i
-
-
-
+ v(v *
3 j)
4.2 j +
+
+
>
rB A)
( - 0.06k)
+ *
2v
*
(20i)
(vrel)xyz +
+
(0.3k)
(arel)xyz *
C (0.3k)
*
(arel)xyz
>
2.2 j] m s2
Ans.
635
(20i) D
+
2(0.3k)
*
( - 31 j)
+
(arel)xyz
91962_06_s16_p0513-0640
6/8/09
3:28 PM
Page 636
© 2010 Pearson Education, Inc., Upper Saddle River,NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–155. Water leaves the impeller of the centrifugal p ump with a velocity of 25 m s and acceleration of 30 m s2 , both measured relative to the impeller along the blade line AB. Determine the velocity and acceleration of a water particle at A as it leaves the impeller at the instant shown. The impeller rotates with a constant angular velocity of v = 15 rad s.
>
y
>
B 30
A
>
x v
Reference Frame: The xyz rotating reference frame is attached to the impeller and coincides with the XYZ fixed reference frame at the instant considered, Fig.a.Thus, the motion of the xyz frame with respect to the XYZ frame is =
vO
=
aO
v =
0
>
#
[ - 15k] rad s
v =
0.3 m
0
The motion of point A with respect to the xyz frame is
>
rA O
=
[0.3 j] m
>
(vrel)xyz
=
( - 25 cos 30° i
+
25 sin 30° j)
=
[ - 21.65i
+
12.5 j] m s
(arel)xyz
=
( - 30 cos 30° i
+
30 sin 30° j)
=
[ - 25.98i
+
15 j] m s2
+
12.5 j)
>
Velocity: Applying the relative velocity equation. vA
=
vO
+ v *
=
0
=
[ - 17.2i
+
>
rA O
( - 15k) +
*
+
(vrel)xyz
(0.3 j)
+
( - 21.65i
>
12.5 j] m s
Ans.
Acceleration: Applying the relative acceleration equation, aA
# + v *
=
aO
=
0
=
[349i
+
( - 15k) +
>
+ v *
(v
*
[( - 15k)
*
rA O
*
>
rA O )
(0.3 j)]
+
+
2v
*
(vrel)xyz
2( - 15k)
*
+
(arel)xyz
( - 21.65i
>
597 j] m s2
+
12.5 j)
+
Ans.
636
( - 25.98i
+
15 j)
15 rad/ s
91962_06_s16_p0513-0640
6/8/09
3:29 PM
Page 637
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*16–156. A ride in an amusement park consists of a rotating arm AB having a constant angular velocity vAB = 2 rad s about point A and a car mounted at the end of the arm which has a constant angular velocity V ¿ = - 0.5k rad s, measured relative to the arm.At the instant shown, determine the velocity and acceleration of the passenger at C .
5
>
rB A vB aB
> >
= vAB *
rB A
= aAB * =
Æ =
vC
(10 cos 30° i
=
=
0
-
(2 vB
-
rB A
{ - 7.00i
5 j)
+
=
{ - 10.0i
v AB
+
>
>
20 j} ft s2
+
0
>
1.5k
+
( - 2 j)
*
>
>
+ Æ *
rC B
20 j
Ans.
0
+
(Æ
>
rC B )
*
(1.5k)
+
*
+
(1.5k)
2Æ
>
(vC B)xyz
*
( - 2 j)
*
+
0
+
>
(aC B)xyz
0
+
>
Ans.
>
5
=
(10 cos 30°i
> >
6 >
10 sin 30° j)
+
2k
5 j)
+
0
rB A
2 - vAB rB A
# Æ =
(1
vC
=
vB
-
0.6)k
+ Æ *
= - 10.0i + =
{ - 7.00i
+
= - 39.64i =
{ - 38.8i
-
5 j)
+
v AB
>
A
17.32 j} ft s
-
=
{ - 39.64i
-
=
>
11.34 j} ft s2
0.4k
>
rC B
>
(vC B)xyz
+
1.5k
+
*
( - 2 j)
>
17.3 j} ft s
>
rC B
Ans.
(Æ
*
rC B )
>
+
2Æ
(0.4k)
*
( - 2 j)
+
(1.5k)
+ Æ *
11.34 j
+
*
>
(vC B)xyz *
(1.5k)
+ *
>
(aC B)xyz ( - 2 j)
>
6.84 j} ft s2
+
0
+
0 Ans.
637
2 ft
60
1.5k
17.32 j
# aC = aB + Æ *
y
2 rad/ s
30
+
0.5 rad/ s
B
>
(8.66i =
+
10 ft
5 j} ft
+
= aAB *
(2 - 0.5)k
+
(2)2(8.66i
aB
Æ =
{8.66i
{ - 10.0i
=
v¿
6 >
=
rB A
*
=
5
5 j)
= vAB *
(1k)
*
(8.66i
vB
=
x
(vC B)xyz
+
>
>
C
>
-
•16–157. A ride in an amusement park consists of a rotating arm AB that has an angular acceleration of 2 aAB = 1 rad s when vAB = 2 rad s at the instant shown. Also at this instant the car mounted at the end of the arm has an angular acceleration of A ¿ = - 0.6k rad s2 and angular velocity of V ¿ = - 0.5k rad s, measured relative to the arm. Determine the velocity and acceleration of the passenger C at this instant. rB A
2 rad/ s
30
15.5 j} ft s2
-
17.32 j} ft s
17.3 j} ft s
= - 34.64i -
2 ft
60
A
{ - 34.64i
=
10 ft
y
5 j} ft
+
0.5 rad/ s
1.5k
17.32 j
+
{ - 34.6i
{8.66i
>
5 j)
+
rC B
# aC = aB + Æ *
=
=
B
- vAB rB A
=
+ Æ *
(8.66i
*
2
(2) (8.66i
= - 10.0i + =
2k
=
2
0.5)k
10 sin 30° j)
+
v¿
> 6 >
C
x
91962_06_s16_p0513-0640
6/8/09
3:29 PM
Page 638
© 2010 Pearson Education, Inc., Upper Saddle River,NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–158. The “quick-return” mechanism consists of a crank AB, slider block B, and slotted link CD. If the crank has the
D
angular motion shown, determine the angular motion of the slotted link at this instant.
=
vB
(aB)t
3(0.1)
=
vB
=
vC
vB
>
C =
vCD =
=
=
+ Æ *
>
+
+ (0.866k) * - 0.3294i +
>
+
9 rad/ s2
vC D, aC D
>
(vB C)xyz
>
=
(vCDk)
+
0
*
(0.3i)
+
vB
>
Ci
d
>
rB C
Ans.
+ Æ *
0.9 cos 30°i (0.866k
1.2294 j
C = - 0.104
aCD =
>
0.9 m s2
rB C
0.866 rad s
# aB = aC + Æ *
aB
>
0.3 sin 60° j
>
-
3 rad/ s
30
0.15 m s
0.9 cos 60°i
a AB
0.9 m s2
(3)2 (0.1)
0.3 cos 60°i
v AB
0.3 m s
=
9(0.1)
=
(aB )n
>
100 mm
=
>
*
+
0.3i)
(Æ
*
>
rB C)
0.9 sin 60° j +
0.3aCD j
2(0.866k -
0.225i
+ + * +
2Æ
*
>
(vB C)xyz
0.9 sin 30° j 0.15i) 0.2598 j
+ +
=
0
+ +
>
(aB C)xyz (aCD k)
*
(0.3i)
>
aB C i aB
>
C
i
m s2
>
3.23 rad s2 d
Ans.
638
C
30
A
300 mm
B
91962_06_s16_p0513-0640
6/8/09
3:29 PM
Page 639
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–159. The quick return mechanism consists of the crank CD and the slotted arm AB. If the crank rotates with the angular velocity and angular acceleration at the instant shown, determine the angular velocity and angular acceleration of AB at this instant.
B 2 ft
D
6 rad/ s 2 aC D 3 rad/ s
vC D
60
Reference Frame: The xyz rotating reference frame is attached to slotted arm AB 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 =
vA
aA
=
#
vAB = vABk
0
vAB = aAB
C 4 ft
k
30
For the motion of point D with respect to the xyz frame, we have
>
rD A
=
[4i] ft
(vrel)xyz
=
(vrel)xyzi
(arel)xyz
=
(arel)xyz i A
Since the crank CD rotates about a fixed axis, vD and aD with respect to the XYZ reference frame can be determined from vD
aD
= vCD * =
(6k)
=
[6i
rD
(2 cos 30° i
*
2 sin 30° j)
-
>
10.39 j] ft s
+
= aCD *
rD
=
(3k)
=
[ - 59.35i
*
2
- vCD
rD
(2 cos 30° i
2 sin 30° j)
-
62(2 cos 30° i
-
2 sin 30° j)
-
>
41.20 j] ft s2
+
Velocity: Applying the relative velocity equation, vD
=
vA
>
+ vAB *
6i
+
10.39 j
6i
+
10.39 j
= =
0
rD A
+
(vrel)xyz
+
(vABk)
(vrel)xyz i
+
*
(4i)
+
(vrel)xyz i
4vAB j
Equating the i and j components yields
>
(vrel)xyz
=
10.39
4vAB
=
6 ft s
>
2.598 rad s
vAB =
>
2.60 rad s Ans.
=
Acceleration: Applying the relative acceleration equation, aD
=
aA
# + vAB *
>
rD A
- 59.35i +
41.20 j
=
0
- 59.35i +
41.20 j
=
c(
+ vAB *
+
(aABk)
arel)xyz
-
(vAB *
4i
d
27 i
*
+ +
rAB)
2.598k (4aAB
2vAB
+ *
+
*
(vrel)xyz
[(2.598k)
*
(4i)]
31.18) j
Equating the i and j components yields 41.20
=
aAB =
4aAB
+
31.18
>
2.50 rad s2
Ans.
639
+
(arel)xyz +
2(2.598k)
*
(6i)
+
(arel)xyz i
91962_06_s16_p0513-0640
6/8/09
3:29 PM
Page 640
© 2010 Pearson Education, Inc., Upper Saddle River,NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*16–160. The Geneva mechanism is used in a packaging system to convert constant angular motion into intermittent angular motion. The star wheel A makes one sixth of a revolution for each full revolution of the driving wheel B and the attached guide C .To do this,pin P , which is attached to B, slides into one of the radial slots of A, thereby turning wheel A, and then exits the slot. If B has a constant angular velocity of vB = 4 rad s , determine V A and AA of wheel A at the instant shown.
vB
B
4 rad/ s
C
P
>
4 in.
A u
The circular path of motion of P has a radius of =
rP
4 tan 30°
2.309 in.
=
Thus, vP
= - 4(2.309) j = - 9.238 j
2 = - (4) (2.309)i = - 36.95i
aP
Thus, vP
=
vA
- 9.238 j =
+ Æ * +
0
>
(vA k)
>
(vP A)xyz
+
rP A *
(4 j)
-
>
vP A j
Solving, vA =
vP # aP = aA + Æ *
>
rP A
- 36.95i =
0
>
Ans.
>
9.238 in. s
A =
+ Æ * +
0
(aAk)
>
(Æ
*
rP A)
*
(4 j)
+
0
+
2Æ
+
0
-
*
>
(vP A)xyz
+
>
(aP A)xyz
>
aP A j
Solving, - 36.95 = - 4aA aA =
>
9.24 rad s2 d aP
>
A =
Ans.
0
640
30