Assume that the point O is fixed in an inertial reference frame and determine the absolute velocity and absolute acceleration of the slider.. If the horizontal translation of the pivot
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Solutions to Exercises
Chapter 1
Section 1.1
1.1 Choose any two contributors from Table 1.1, study their contributions, and write a
paragraph about each of them
Section 1.2.1 1.2 Consider the planar pendulum kinematics discussed in Example 1.1, start with
position vector r P O resolved in terms the unit vectors i and j, and verify the expressions
obtained for the acceleration and velocity given by Eq (f) of Example 1.1
Solution 1.2
The relationships among the unit vectors are
2
cos sin sin cos
1
Then, the position vector is
2 sin ( cos )
The velocity vector is
/
1
( sin ) ( cos )
P O
L
r
e
The acceleration vector is
/ /
2
2
( cos ) ( sin )
( sin cos ) (cos sin )
P O
v
1.3 Consider the kinematics of the rolling disc considered in Example 1.2, and verify that
the instantaneous acceleration of the point of contact is not zero
Trang 2Solution 1.3
Making use of Eq (1.7) that relates accelerations of two arbitrary points on a rigid body, we have
a a r r
From Example 1.2, we have
/ /
C G
G O
r
r
r j
a i
k k
and, hence,
/
2 2
C O
r
i j i j
Thus, it is verified that the instantaneous acceleration of the point of contact is not zero
1.4 Show that the acceleration of the particle in the rotating frame of Example 1.3 is
(x p 2y p x p y p) (y p 2x p y p x p)
where is the magnitude of the angular acceleration of the rotating frame about the z
axis
Solution 1.4
From Eq (b) of Example 1.3
d
Then the acceleration is
v
where = d/dt and we have used Eq (1.8) We note that = k; therefore,
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1.5 In Figure E1.5, a slider of mass M r is located on a bar whose angular displacement in the plane is described by the coordinate The motion of the slider from the pivot point
is measured by the coordinate r1 The acceleration due to gravity acts in a direction
normal to the plane of motion Assume that the point O is fixed in an inertial reference
frame and determine the absolute velocity and absolute acceleration of the slider
Solution 1.5
Choose unit vectors e1 and e2 fixed to the slider as shown in the figure Then, the position vector from
point O to the slider is
1 1
m r
r e
The absolute velocity is then given by
m
O
2
r1
Noting that k , where k is the unit vector that points out of the plane, we find that
1 1 1 1 1 1 1 2
m r r r r
The absolute acceleration is found from
1 1 1 2 1 1 1 1 2 1 2 1
2
m
1.6 A pendulum of mass m is attached to a moving pivot of mass M as shown in Figure
E1.6 Assume that the pivot point cannot translate in the vertical direction If the
horizontal translation of the pivot point from the fixed point O is measured by the coordinate x and the angle is used to describe the angular displacement of the pendulum from the vertical, determine the absolute velocity of the pendulum
Trang 4Solution 1.6
The position vector of the location of the mass with
respect to the fixed-point O is given by
2
m
Then the velocity is
1
m m
d
dt
r
V i e i ω e i k e
i e i i j
l
M
m
x
O
j
i
1
ˆe
2
ˆe
O
We could have also obtained the solution as follows:
sin cos sin cos
Then the velocity is
Section 1.2.2 1.7 Determine the number of degrees of freedom for the systems shown in Figure E1.7
Assume that the length L of the pendulum shown in Figure E1.7a is constant and that the length between each pair of particles in Figure E1.7b is constant Hint: For Figure E1.7c,
the rigid body can be thought of as a system of particles where the length between each pair of particles is constant
Solution 1.7
a) Three coordinates (n = 3) and one inextensible constraint (m = 1): number of
degrees of freedom is 3 1 = 2
b) Three particles, each with 3 coordinates and three inextensible constraints: number
of degrees of freedom is 9 3 = 6
c) Rigid body: number of degrees of freedom is 6; 3 translation and 3 rotational The system of three particles shown in Figure E1.7b is an example of a rigid body, since the particles are a constant distance apart from each other In three-dimensional space, for a
system of N particles representing a rigid body, the number of degrees of freedom is 3N
(3N 6), where 3N is the total number of coordinates associated with the N particles and 3N 6 is the total number of inextensional constraints that ensures that the particles are always at a constant distance apart By extension to the case shown in Figure E1.7c, the number of degrees of freedom is 6
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1.8 Draw free-body diagrams for each of the masses shown in Figure E1.6 and obtain the
equations of motion along the horizontal direction by using Eq (1.15)
Solution 1.8
The free-body diagrams for each of the masses in
figure E1.14 are as shown to the right The force N is shown in the free-body diagram of mass M to account
for the constraint that the pivot mass cannot move in
the j-direction Summing forces along the i-direction
for each of the two masses, we find for the pivot mass from Eqs (1.15) that
sin
and for the pendulum mass that
Mg
N
i
j
M
m
mg
T
1.9 Draw the free-body diagram for the whole system shown in Figure E1.6, obtain the
system equation of motion by using Eq (1.14) along the horizontal direction, and verify that this equation can be obtained from Eq (1.15)
Solution 1.9
The free-body diagram is shown to the right Note that the
internal force T does not appear in this diagram Making use of
Eqs (1.14) for the motion along the horizontal direction, we find that
0Mxm xl sin l cos
which can be obtained by adding the two equations obtained to the solution to Exercise 1.8
Mg
N
M
m
mg
1.10 Determine the linear momentum for the system shown in Figure E1.5 and discuss if
it is conserved Assume that the mass of the bar is M bar and the distance from the point O
to the center of the bar is L bar
Solution 1.10
The linear momentum of the system is given by
bar slider
where
2
bar M bar L bar
Trang 6Making use of the slider velocity V m determined in the solution to Exercise 1.5, we arrive
at
1 1 1 2
slider M r r r
Thus,
Since there are no external forces acting on the system, by virtue of Eq (1.14), the system’s linear momentum is conserved
1.11 Determine the angular momentum of the system shown in Figure 1.6 about the point
O and discuss if it is conserved
Solution 1.11
The angular momentum of the system about the point O is given by
pendulum disc
k r p
where k is the unit vector normal to the plane, the position vector r m runs from point O to
the pendulum, and p is the linear momentum of the pendulum Making use of the
solution to Example 1.4, we have
m
m
and
O
which can be written as
O
If no external moments act on the system, since O is a fixed point, from Eq (1.17) it is
clear that the angular momentum of the system is conserved
1.12 A rigid body is suspended from the ceiling by two elastic cables that are attached to
the body at the points O and O, as shown in Figure E1.12 Point G is the center of mass
of the body Which of these points would you choose to carry out an angular-momentum balance based on Eq (1.17)?
Solution 1.12
Since point G is the center of mass of the body, this point would be used to carry out
an angular momentum balance based on Eq (1.17) One cannot use Eq (1.17) with
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inertia J G about the center of mass G It is suspended from a point O on the ceiling by using an elastic suspension The point of attachment O is at a distance l from the center
of mass G of this body M(t) is an external moment applied to the system along an axis normal to the plane of the body Use the generalized coordinates x, which describes the
up and down motions of point O from point O, and , which describes the angular oscillations about an axis normal to the plane of the rigid body For the system shown in Figure E1.13, use the principle of angular-momentum balance given by Eq (1.17) and obtain an equation of motion for the system Assume that gravity loading is present
Solution 1.13
The principle of angular-momentum balance given by Eq (1.17) is applied with respect to the center of mass of the system Thus,
where k is the unit vector normal to the plane of the system Then we have
( ) G
If, instead, the principle of angular-momentum balance given by Eq (1.17) is applied
with respect to the fixed point O of the system, the result is
G
d
dt d
dt
where, we have substituted the velocity of the center of mass from the solution to Exercise 1.13 Carrying out the different cross-product operations, we obtain
M t( ) mglcos J G m d l x lsin sin l x l cos cos
dt
which leads to
dt
Section 1.2.4 1.14 For the system shown in Figure E1.13, construct the system kinetic energy
Trang 8Solution 1.14
We make use of Eq (1.23) to determine the system kinetic energy Note that the velocity of
the center of mass G can be obtained as in the
solution to Exercise 1.6; that is,
cos sin
Then, from Eq (1.23), the kinetic energy is l
G
m, J G
O
i
2 cos
G
2
V V
1.15 Determine the kinetic energy of the planar pendulum of Example 1.1
Solution 1.15
From Eqs (b) and (e) of Example 1.1, we find that
Then, from Eq (1.22), the kinetic energy is
1.16 Consider the disc rolling along a line in Figure E1.16 The disc has a mass m and a
rotary inertia J G about the center of mass G Answer the following: (a) How many
degrees of freedom does this system have? and (b) Determine the kinetic energy for this system
Solution 1.16
a) One degree of freedom, since one independent coordinate (x or ) is needed to
describe the motion Due to the non-slip constraint, x = r, and x or can always be expressed in terms of the other coordinate
b) Making use of Eq (1.24), we have
v v
Noting from Example 1.2 that
G O
we arrive at
T m r r J 2 mr J 2
i i
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the pendulum is r, and the rotary inertia of the disc about the point O is J O, determine the system kinetic energy
Solution 1.17
The kinetic energy of the system can be written as
pendulum disc
where
1
2 1 2
V V
Noting from the solution to Example 1.4 that
sin 1 cos 2
we determine that
2 2
2
O
2 2
1.18 Referring to Figure E1.6 and assuming that the bar to which the pendulum mass m is
connected is massless, determine the kinetic energy for the system
Solution 1.18
The kinetic energy of the system is given by
M pendulum
T T T
where
1 2
M
V V
i i
Making use of the solution for the velocity in Exercise 1.6, we find that
1
2 1
2
pendulum
Trang 10Thus,
2 cos
1.19 Determine the kinetic energy of the system shown in Figure E1.5
Solution 1.19
The kinetic energy of the system is given by
bar slider
where
1
2 1 2
V V
Making use of the velocity determined in the solution of Exercise 1.5, we find that
T M r ere r ere M r r
and, therefore,
2 2 2 2 2 2