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50 Solutions Manual
where, referring to the problem statement,
x.t/ Dx0⇥sin.2⇡!t/ 3sin.⇡!t /⇤:(5)
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.
Dynamics 2e 51
Problem 2.21
The position of point
P
as a function of time
t
,
t0
and expressed
in seconds, is
Er.t / D2:0 Œ0:5 Csin.!t/çO{C⇥9:5 C10:5 sin.!t/ C4:0 sin2.!t /⇤O|;
where !D1:3 rad=s and the position is measured in meters.
Find the trajectory of Pin Cartesian components and then, using
the
x
component of
Er.t /
, find the maximum and minimum values of
x
reached by
P
. The equation for the trajectory is valid for all values
of
x
, yet the maximum and minimum values of
x
as given by the
x
component of Er.t/ are finite. What is the origin of this discrepancy?
Solution
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permission of McGraw-Hill, is prohibited.
52 Solutions Manual
Problem 2.22
The position of point
P
as a function of time
t
,
t0
and expressed
in seconds, is
Er.t / D2:0 Œ0:5 Csin.!t/çO{C⇥9:5 C10:5 sin.!t/ C4:0 sin2.!t /⇤O|;
where !D1:3 rad=s and the position is measured in meters.
(a)
Plot the trajectory of
P
for
0t0:6
s,
0t1:4
s,
0t
2:3 s, and 0t5s.
(b) Plot the y.x/ trajectory for 10 mx10 m.
(c)
You will notice that the trajectory found in (b) does not agree with
any of those found in (a). Explain this discrepancy by analytically
determining the minimum and maximum values of
x
reached by
P
. As you look at this sequence of plots, why does the trajectory
change between some times and not others?
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 53
Part (b).
In this part of the problem we first need to write the trajectory in the form
yDy.x/
. To do so,
we start with solving the first of Eq. (1) for sin.1:3t/ as a function of x. This gives
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.
54 Solutions Manual
Problem 2.23
A bicycle is moving to the right at a speed
v0D20 mph
on a
horizontal and straight road. The radius of the bicycle’s wheels is
RD1:15 ft
. Let
P
be a point on the periphery of the front wheel.
One can show that the
x
and
y
coordinates of
P
are described by
the following functions of time:
x.t/ Dv0tCRsin.v0t=R/ and y.t/ DR⇥1Ccos.v0t=R/⇤:
Determine the expressions for the velocity, speed, and accelera-
tion of Pas functions of time.
Solution
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 55
Problem 2.24
A bicycle is moving to the right at a speed
v0D20 mph
on a
horizontal and straight road. The radius of the bicycle’s wheels is
RD1:15 ft
. Let
P
be a point on the periphery of the front wheel.
One can show that the
x
and
y
coordinates of
P
are described by
the following functions of time:
x.t/ Dv0tCRsin.v0t=R/ and y.t/ DR⇥1Ccos.v0t=R/⇤:
Determine the maximum and minimum speed achieved by
P
,
as well as the
y
coordinate of
P
when the maximum and minimum
speeds are achieved. Finally, compute the acceleration of
P
when
Pachieves its maximum and minimum speeds.
Solution
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permission of McGraw-Hill, is prohibited.
56 Solutions Manual
Problem 2.25
A bicycle is moving to the right at a speed
v0D20 mph
on a
horizontal and straight road. The radius of the bicycle’s wheels is
RD1:15 ft
. Let
P
be a point on the periphery of the front wheel.
One can show that the
x
and
y
coordinates of
P
are described by
the following functions of time:
x.t/ Dv0tCRsin.v0t=R/ and y.t/ DR⇥1Ccos.v0t=R/⇤:
Plot the trajectory of
P
for
0t1
s. For the same time inter-
val, plot the speed as a function of time, as well as the components
of the velocity and acceleration of P.
Solution
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permission of McGraw-Hill, is prohibited.
Dynamics 2e 57
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.
58 Solutions Manual
Acceleration Components
The components of the acceleration of
P
can be plotted for
0t1
s using
the expressions in Eq. (4) with
v0D20 mph D29:33 ft=s
and
RD1:15 ft
. The plot shown below was
generated using Mathematica with the following code:
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.
Dynamics 2e 59
Problem 2.26
Find the
x
and
y
components of the acceleration in Example 2.3 (except for the plots) by simply differenti-
ating Eqs. (4) and (5) with respect to time. Verify that you get the results given in Example 2.3.
Solution
Referring to Eqs. (4) and (5) of Example 2.3 on p. 38 of the textbook, we recall that the
x
and
y
components
of the velocity are, respectively,
To determine Rx, we differentiate Pxwith respect to time with the help of the chain rule:
dy
Differentiating the first of Eqs. (1) with respect to
y
and substituting the result along with the second of
Substituting the second of Eqs. (1) into Eq. (4) and simplifying, we have
0a2y
Our results match those in Example 2.3.
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.
