Solution 6(1)
Solution 6. Motion relative to a body attached rotating reference
Purpose and Requirement
In this tutorial the application of acceleration combination principle are practiced. The physical
meanings of relative, entrained and Coriolis accelerations should be made clear to the students,
so that they can find them by their physical meanings not by just memorizing the mathematical
expressions. Through the exercises the basic skill of vector calculation should be enhanced.
1. Two rotating rods are connected by slider block P. The rod
attached at A rotates with a constant clockwise angular
velocity A. Assuming b = 200 mm, A = 6 rad/s,
determine for the position shown (a) the angular velocity of
the rod attached at B, (b) the relative velocity of slider
block P with respect to the rod on which it slides. (BJ
P
E
A
60
B
20
b
15.115)
Solution:
Using the given geometry of triangle PAB, we have
rPB
rPA
200
 
 
sin 40
sin 120
sin 20 

v P'

vP
E
rPA = 106.42 (mm)
rPB = 269.46 (mm)

Given  A = 6 k (cw), so
P
50
40

v P/ f
f
60
20
A
 

v P =  A  rPA = 638.530 (mm/s)

Attach frame f on rod BE. Since the direction lines of v P' and

v P / f are known, we can directly have

v P' = 638.5 sin50 70 = 489.12 70,
489.12 

 B = 
k = 1.82 k (rad/s) (cw) (ans)
269.46

v P / f = 638.5 cos50 20 = 410.42 20 mm/s) (ans)
O
B
Solution 6(2)
y
B
2m

x
O
3m
2. Block B moves along the slot in the platform with a
constant speed of 2 m/s, measured relative to the platform in
the direction shown. If the platform is rotating at a constant
rate of  =5 rad/s counterclockwise, determine the velocity of
the block (a) at the instant  =90, (b) at the instant  =60.
(H.16-148, p337)
Solution:
Set up O-xy and O-x”y”, which are coincide at the instant
y,y”

v B / f =  2 i ,
(a) Given

v B/ f

v B'

rB' = 2 j ,
B

 = 5 k .
x, x”
O
So we have
 

v B' =   rB' =  5(2) i =  10 i ,

v B/ f
Ov

vB

vB

v B'



v B  v B'  v B/ f =  12 i (ans)
Hence

v B'
y,y”
(b) Give

v B / f =  2 i ,

rB' =1.1547 i + 2 j ,

v B/ f
2m
x, x”
60

 = 5 k
So we have
 

v B' =   rB' =5 k (1.1547 i + 2 j )
= 10 i + 5.77 j = 11.55 150

v B/ f

vB



Hence v B  v B'  v B/ f =  12 i i + 5.77 j

v B'
= 13.32 154.32
Ov
(ans)
Solution 6(3)
3. At the instant shown the length of the boom is being
decreased at the constant rate of 150 mm/s and the boom is
being lowered at the constant rate of 0.075 rad/s. Knowing
that  = 30, determine (a) the velocity of point B, (b) the
acceleration of point B. (BJ 15.123)
B
6m
30o
A
Solution:
(a) Given

rBA = 630 (m)

v B / f = 0.15150 (m/s),

 = 0.075 k
60
18.4
 

v B' =   rBA = 0.4560 (m/s)

B

v B/ f

v B'

f
30


vB
A



v B = v B / f + v B'
= 0.47478.43 (m/s) (ans)
(b)


a B' =   2 rBA = 0.0752 (6) 150

a BC

aB
= 0.03375150 (m/s2)
120
56.31

aB/ f = 0

aB'
 

a BC = 2   v B / f = 2(0.075)(0.15) 120
f
30
= 0.0225120

(m/s2)


 
a B = a B' + a B / f + a BC
= 0.04056(56.31+120)
= 0.04056176.31
(m/s2) (ans)
A


B
150
Solution 6(4)
A
D
C
B
188 mm
E
4. The hydraulic cylinder CD is welded to an arm, which
rotates clockwise about A at the constant rate  = 2.4
rad/s. Knowing that in the position shown rod BE is
being moved to the right at the constant rate of 375 mm/s
with respect to the cylinder, determine (a) the velocity of
point B, (b) the acceleration of point B. (BJ 15.129)
250 mm
Solution:
Give
O

v B'



 = 2.4(  k ) (rad/s)

v B / f =375 i (mm/s)

rBO = 312.8(36.94180)
188
B
36.94
(a)

v B/ f
250
 

v B' =   rBO = 2.4(312.8) (36.94+90)
= 750.72126.94
=  451.2 i + 600.02 j (mm/s)

vB

v B'



v B = v B / f + v B'
= 76.17 i + 600.02 j
Ov

v B/ f



a B'
B
= 604.897.23 (mm/s) (ans)



(b) a B' = a Bt'  a Bn'

since a Bt'  0,
v B2 '
n

36.94 = 1801.73 36.94
a B' = a B' =
rOB

a BC
= 1440.06 i + 1082.8 j

a B / f =0

a B'
Oa

aB

a BC
 

a BC =2   v B / f = 2(2.4)(375)(  j ) =  1800 j




 a B = a B' + a B / f + a BC = 1440.06 i 717.2 j = 160926.5
(ans)
Solution 6(5)
5. The platform (f) is rotating about a fixed point O with a
constant angular velocity 0.3 rad/s counterclockwise. The
particle P is traveling clockwise along a circular track with
a constant speed 10 mm/s with respect to the platform. For
the instant when the platform and the particle are at the
positions shown, determine
(a) the absolute velocity of particle P;
(b) the absolute acceleration of particle P.
Oy
Answer: (a) 14.42-13.9o

(b) 3.28102.2o
20 mm

vP'
Solution: Attach O”-x”y” on the plate
P
y”



(a) v P  v P / f  v P '
P’
40 mm
 

v P"    rPO =10,3930
O”
30

v P / f  10-60
x”

vP / f
f



v P  v P / f  v P ' =14.42-13.9 mm/s
 



(b) a P  a P / f  a P '  2  v P / f

a P '  3.12120
 

a PC  2  v P / f =6 30 
Oy

20 mm

a P / f  5210 
y”
P
40 mm

a P  3.28 102.2  mm/s2
O”
30
x”
f

vP / f