PROBLEM 15.148*
A wheel of radius r rolls without slipping along the inside of a fixed cylinder
of radius R with a constant angular velocity
.ω
Denoting by P the point of the
wheel in contact with the cylinder at
0,t=
derive expressions for the
horizontal and vertical components of the velocity of P at any time t. (The
curve described by Point P is a hypocycloid.)
PROBLEM 15.149*
In Problem 15.148, show that the path of P is a vertical straight line when
/2.rR=
Derive expressions for the corresponding velocity and acceleration of
P at any time t.
PROBLEM 15.150
Pin P is attached to the collar shown; the motion of the pin is guided by a
slot cut in rod BD and by the collar that slides on rod AE. Knowing that
at the instant considered the rods rotate clockwise with constant angular
velocities, determine for the given data the velocity of pin P.
8
AE
ω
=
rad/s,
3
BD
ω
=
rad/s
SOLUTION
0.5
500 mm 0.5 m, 0.5 tan 30 , cos30
AB AP BP= = = °= °
8 rad/s
AE
=ω
,
3 rad/s
BD
=ω
Let
P′
be the coinciding point on AE and
1
u
be the outward velocity of the collar along the rod AE.
/
[( )
P P P AE AE
AP
ω
′
=+=vvv
1
][u+
]
Let
P′′
be the coinciding point on BD and
2
u
be the outward speed along the slot in rod BD.
/
[( )
P P P BD BD
BP
ω
′′
=+=vv v
2
30 ] [u°+
60 ]°
Equate the two expressions for
P
v
and resolve into components.
:
12
0.5 (3)(cos 30 ) cos 60
cos30
uu

= °+ °

°

or
12
1.5 0.5uu= +
(1)
:
2
0.5
(0.5 tan 30 )(8) (3)sin 30 sin 60
cos30 u

− ° =− °+ °

°

2
1[1.5 tan 30 4 tan 30 ] 1.66667 m/s
sin 60
u= °− ° =−
°
From (1),
1
1.5 (0.5)( 1.66667) 0.66667 m/su=+− =
[(0.5 tan 30 )(8)
P
= °v
]
[0.66667+
]
[2.3094 m/s=
]
[0.66667 m/s+
]
22
2.3094 0.66667 2.4037 m/s
P
v=−+=
2.3094
tan 73.9
0.66667
ββ
= = °
2.40 m/s
P
=v
73.9 °
PROBLEM 15.151
Pin P is attached to the collar shown; the motion of the pin is guided by
a slot cut in rod BD and by the collar that slides on rod AE. Knowing that
at the instant considered the rods rotate clockwise with constant angular
velocities, determine for the given data the velocity of pin P.
7
AE
ω
=
rad/s,
4.8
BD
ω
=
rad/s
SOLUTION
0.5
500 mm 0.5 m, 0.5 tan 30 , cos30
AB AP BP= = = °=°
7 rad/s
AE
=ω
,
4.8 rad/s
BD
=ω
Let
P′
be the coinciding point on AE and
1
u
be the outward velocity of the collar along the rod AE.
/
[( )
P P P AE AE
AP
ω
′
=+=vvv
1
][u+
]
Let
P′′
be the coinciding point on BD and
2
u
be the outward speed along the slot in rod BD.
/
[( )
P P P BD BD
BP
ω
′′
=+=vv v
2
30 ] [u°+
60 ]°
Equate the two expressions for
P
v
and resolve into components.
:
12
0.5 (4.8)(cos 30 ) cos 60
cos30
uu

= °+ °

°

or
12
2.4 0.5uu= +
(1)
:
2
0.5
(0.5 tan 30 )(7) (4.8)sin 30 sin 60
cos30 u

− ° =− °+ °

°

2
1[2.4 tan 30 3.5 tan 30 ] 0.73333 m/s
sin 60
u= °− ° =−
°
From (1),
1
2.4 (0.5)( 0.73333) 2.0333 m/su=+− =
[(0.5 tan 30 )(7)
P= °v
]
[2.0333+
]
[2.0207 m/s=
]
[2.0333 m/s+
]
22
(2.0333) (2.0207) 2.87 m/s
P
v= +=
2.0207
tan , 44.8
2.0333
ββ
=− =−°
2.87 m/s
P
=v
44.8 °
PROBLEM 15.152
Two rotating rods are connected by slider block P. The rod
attached at A rotates with a constant angular velocity
.
A
ω
For the
given data, 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.
b = 8 in.,
6 rad/s.
A
ω
=
SOLUTION
Dimensions:
Law of sines.
8 in.
sin 20 sin120 sin 40
AP BP
= =
° °°
/
P BE
PROBLEM 15.153
Two rotating rods are connected by slider block P. The rod
attached at A rotates with a constant angular velocity
.
A
ω
For
the given data, 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.
b = 300 mm,
10 rad/s.
A
ω
=
SOLUTION
Dimensions:
/
P AD
/
P AD
PROBLEM 15.154
Pin P is attached to the wheel shown and slides in a slot cut in bar
BD. The wheel rolls to the right without slipping with a constant
angular velocity of 20 rad/s. Knowing that x = 480 mm when
0,
θ
=
determine the angular velocity of the bar and the relative
velocity of pin P with respect to the rod for the given data.
(a)
0,
θ
=
(b)
90 .
θ
= °
/
PF
PROBLEM 15.154 (Continued)
Use
/P P PF
′
= +vvv
and resolve into components.
PROBLEM 15.155
Knowing that at the instant shown the angular velocity of bar
AB is 15 rad/s clockwise and the angular velocity of bar EF is
10 rad/s clockwise, determine (a) the angular velocity of rod
DE, (b) the relative velocity of collar B with respect to rod DE.
SOLUTION
10 rad/s
EF
ω
=
,
/E E F EF
r
ω
=
v
]
( )( )
15 10 150 in./s= =
[
/150 in./s
B E BE
′=+=v vv
]
[
20
DE
ω
+
]
/B ED
u=v
[
/
150 in./s
B B B ED
′
=+=vvv
] [
20
DE
ω
+
] [
u+
]
( )
B AB
AB
ω
=
v
( )
15
45 15
cos 45

° = 
°

[
45 225 in./s
°=
] [
225 in./s+
]
Equate the two expressions for
B
v
and resolve into components.
: 150 225, 75 in./suu+= =
: 20 225, 11.25 rad/s
DE DE
ωω
= =
(a)
11.25 rad/s
DE
ω
=
(b)
/75.0 in./s
B DE =v
PROBLEM 15.156
Knowing that at the instant shown the angular velocity of rod
DE is 10 rad/s clockwise and the angular velocity of bar EF is
15 rad/s counterclockwise, determine(a) the angular velocity of
bar AB, (b) the relative velocity of collar B with respect to rod
DE.
SOLUTION
15 rad/s
EF
ω
=
/E E F EF
r
ω
=
v
( )( )
15 15 225 in./s
= =
[
/
225 in./s
B E B ED
′
=+=v vv
] [
20
DE
ω
+
] [
225 in./s=
] [
200 in./s+
]
/B ED
u=v
[
/225 in./s
B B B ED
′
=+=vvv
] [
200 in./s+
] [
u+
]
( )
B AB
AB
ω
=
v
15
45 cos 45
AB
ω
°=
°
[
45 15
AB
ω
°=
] [
15
AB
ω
+
]
Equate the two expressions for
B
v
and resolve into components.
:
225 15 AB
u
ω
− +=−
(1)
:
200
200 15 13.333 rad/s
15
AB AB
ωω
= = =
(a)
13.33 rad/s
AB
ω
=
From (1),
( )( )
225 15 13.333 25 in./su=−=
(b)
/25.0 in./s
B ED =v
PROBLEM 15.157
The motion of pin P is guided by slots cut in rods AD and BE.
Knowing that bar AD has a constant angular velocity of 4 rad/s
clockwise and bar BE has an angular velocity of 5 rad/s
counterclockwise and is slowing down at a rate of 2 rad/s2,
determine the velocity of P for the position shown.
SOLUTION
Units: meters, m/s, m/s2
PROBLEM 15.157 (Continued)
Equating the two expressions for vP and resolving into components,
P
PROBLEM 15.158
Four pins slide in four separate slots cut in a circular plate as shown. When
the plate is at rest, each pin has a velocity directed as shown and of the
same constant magnitude u. If each pin maintains the same velocity relative
to the plate when the plate rotates about O with a constant
counterclockwise angular velocity
,
ω
determine the acceleration of each
pin.
SOLUTION
4
4
PROBLEM 15.159
Solve Problem 15.158, assuming that the plate rotates about O with a
constant clockwise angular velocity
.ω
PROBLEM 15.158 Four pins slide in four separate slots cut in a circular
plate as shown. When the plate is at rest, each pin has a velocity directed as
shown and of the same constant magnitude u. If each pin maintains the
same velocity relative to the plate when the plate rotates about O with a
constant counterclockwise angular velocity
,
ω
determine the acceleration
of each pin.
SOLUTION
PROBLEM 15.160
The cage of a mine elevator moves downward at a constant speed of 12.2 m/s. Determine the magnitude and
direction of the Coriolis acceleration of the cage if the elevator is located (a) at the equator, (b) at latitude
40°
north, (c) at latitude
40°
south.
SOLUTION
Earth makes one revolution
( )
2 radians
π
in 23.933 h (86160 s).
( )
6
272.926 10 rad/s
86160
π
−
= = ×jjΩ
Velocity relative to the Earth at latitude angle
.
f
( )
/earth
12.2 cos sin
P
ff
=−−v ij
Coriolis acceleration
.
c
a
( )
( )
( )
( )
3
1.7794 10 cos
f
−

= ×
k
(
a)
0 , cos 1.000
ff
=°=
(
b)
40 , cos 0.76604
ff
=°=
(
c)
40 , cos 0.76604
ff
=−° =
PROBLEM 15.161
Pin P is attached to the collar shown; the motion of the pin is guided by
a slot cut in bar BD and by the collar that slides on rod AE. Rod AE
rotates with a constant angular velocity of 6 rad/s clockwise and the
distance from A to P increases at a constant rate of 8 ft/s. Determine at
the instant shown (a) the angular acceleration of bar BD, (b) the relative
acceleration of pin P with respect to bar BD.
P AE AE
PROBLEM 15.161 (Continued)
/33.9 ft/s
P BD =a
PROBLEM 15.162
A rocket sled is tested on a straight track that is built along a meridian. Knowing that the track is located at
latitude
40°
north, determine the Coriolis acceleration of the sled when it is moving north at a speed of 900 km/h.
SOLUTION
C=a
PROBLEM 15.163
Solve the Geneva mechanism Sample Prob. 15.20 using vector
algebra.
SAMPLE PROBLEM 15.20 In the Geneva mechanism of Sample
Prob. 15.9, disk D rotates with a constant counterclockwise angular
velocity
D
ω
of 10 rad/s. At the instant when
150
f
= °
, determine the
angular acceleration of S.
SOLUTION
2
/S
SP
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PROBLEM 15.163 (Continued)
S
S=
Copyright © McGraw–Hill Education. Permission required for reproduction or display.