Problem 6.98
One way to convert rotational motion into linear motion and vice versa
is by the use of a mechanism called the Scotch yoke, which consists of
a crank
C
that is connected to a slider
B
by a pin
A
. The pin rotates
with the crank while sliding within the yoke, which, in turn, rigidly
translates with the slider. This mechanism has been used, for example,
to control the opening and closing of valves in pipelines. Letting the
radius of the crank be
RD25 cm
, determine the angular velocity
E!C
and the angular acceleration
E˛C
of the crank at the instant shown if
D25ı
and the slider is moving to the right with a constant speed
vBD40 m=s.
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1281
Problem 6.99
Collar
C
moves along a circular guide with radius
RD2ft
with
a constant speed
vCD18 ft=s
. At the instant shown, the bars
AB
and
BC
are vertical and horizontal, respectively. Letting
LD4ft
and
HD5ft
, determine the angular accelerations of the bars
AB
and BC at this instant.
the solution of which is
C
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permission of McGraw-Hill, is prohibited.
Problem 6.100
A sphere
S
of radius
RSD5in:
is rolling without slip inside a
stationary spherical bowl
B
of radius
RBD17 in:
Assume that
the motion of the sphere is planar. The center of the sphere is at
C
and the point of contact between the sphere and the bowl is at P.
If, at the instant shown, the center of the sphere
C
is traveling
counterclockwise with a speed
vCD32 ft=s
and, such that
PvCD
0
, determine the acceleration of
C
, as well as the acceleration of
point
P
, which is the point on the sphere that is in contact with the
bowl at this instant.
12 ft
12 ft
acceleration of Cto that of P, we obtain
EaCD EaPC E˛S ErC =P !2
SErC=P
C
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1283
Therefore, the acceleration of Pis
C
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permission of McGraw-Hill, is prohibited.
Problem 6.101
A sphere
S
of radius
RSD5in:
is rolling without slip inside a
stationary spherical bowl
B
of radius
RBD17 in:
Assume that
the motion of the sphere is planar. The center of the sphere is at
C
and the point of contact between the sphere and the bowl is at P.
If, at the instant shown, the center of the sphere
C
is travel-
ing counterclockwise with a speed
vCD32 ft=s
and, such that
PvCD24 ft=s2
, determine the acceleration of
C
, as well as the
acceleration of
P
, which is the point on the sphere that is in contact
with the bowl at this instant.
12 ft
12 ft
relating the acceleration of Cto that of P, we obtain
EaCD EaPC E˛S ErC =P !2
SErC=P
C
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1285
Therefore, the acceleration of Pis
C
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Problem 6.102
A truck on an exit ramp is moving in such a way that, at the instant shown,
jEaAj D 6m=s2
and
D13ı
.
Let the distance between points Aand Bbe dAB D4m. If, at this instant, the truck is turning clockwise,
D59ı
,
aBx D6:3 m=s2
, and
aBy D 2:6 m=s2
, determine the angular velocity and angular acceleration
of the truck.
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permission of McGraw-Hill, is prohibited.
of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the
permission of McGraw-Hill, is prohibited.
Problem 6.103
As the circular cam whose center is at
A
rotates, it causes the follower
B
to move back and forth. The cam angle is
, the radius of the cam is
R
,
the angular speed of the cam is
P
D!OA
, and the angular acceleration
of the cam is
R
D˛OA
. The cam is pin-connected to the fixed point
O
.
Using the given
x
coordinate, determine the acceleration of the
follower at the instant
D30ı
if
RD1:5 in:
,
P
D!OA D1000 rpm
,
and R
D˛OA D25 rad=s2.
Solution
The position of the follower can be described by the variable x, which is shown in the given figure. Further,
we can write xin terms of the variable , for which we know its time derivatives, as
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 1289
Problem 6.104
As the circular cam whose center is at
A
rotates, it causes the follower
B
to move back and forth. The cam angle is
, the radius of the cam is
R
,
the angular speed of the cam is
P
D!OA
, and the angular acceleration
of the cam is
R
D˛OA
. The cam is pin-connected to the fixed point
O
.
Using the given
x
coordinate, determine the acceleration of the
follower as a function of the cam angle
, the radius of the cam
R
, the
given angular speed of the cam
!OA
, and the given angular acceleration
of the cam ˛OA.
Solution
The position of the follower can be described by the variable x, which is shown in the given figure. Further,
we can write xin terms of the variable , for which we know its time derivatives, to obtain
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permission of McGraw-Hill, is prohibited.