122
Chapter 3. Kinetics of Particles
The geometry of the bases cfw_Ex , Ey , Ez and cfw_er , e , ez is shown in Fig. 3-15 from
which we obtain
Ex
= cos er sin e
(3.420)
Ey
= sin er + cos e
(3.421)
Ey
e
er
Ez
Figure 3-15
Ex
Geometry of Question .
The pos
150
Chapter 4. Kinetics of a System of Particles
Now we have that
A
d F
vP /O
dt
F A
F vP /O
= le
(4.22)
= ez le = l 2 er
(4.23)
Adding Eq. (4.22) and Eq. (4.23) gives
F
aP /O = l 2 er + le
(4.24)
Then, adding Eq. (4.7) and Eq. (4.24), we obtain the acce
Chapter 4
Kinetics of a System of Particles
Question 41
A particle of mass m is connected to a block of mass M via a rigid massless
rod of length l as shown in Fig. P4-1. The rod is free to pivot about a hinge
attached to the block at point O. Furthermore
148
Chapter 4. Kinetics of a System of Particles
Solution to Question 41
Kinematics
Let F be a reference frame xed to the ground. Then, choose the following
coordinate system xed in reference frame F :
Ex
Ez
Ey
Origin at O at t = 0
=
To the Right
=
Into P
149
The velocity of point P in reference frame F is then given as
F
vP =
F
F
d
d
rP /O = F vO + F vP /O
(rO ) +
dt
dt
(4.10)
Now we already have F vO from Eq. (4.6). Next, since rP /O is expressed in the basis cfw_er , e , ez and cfw_er , e , ez rotates
144
Chapter 3. Kinetics of Particles
Finally, the force of gravity is given as
mg = mgey
(3.584)
Then, adding Eq. (3.576), Eq. (3.583), and Eq. (3.584), the resultant force acting
on the particle is then obtained as
(2x/a)ex + ey
+ Nb ez K(x x0 )ex mgey
(
143
Therefore,
en = eb et = ez
ex + (2x/a)ey
2x
1+
a
2
=
(2x/a)ex + ey
2x
1+
a
2
(3.573)
Suppose now that we dene
=
Then we can write
en =
2x
a
1+
2
(2x/a)ex + ey
(3.574)
(3.575)
The reaction force exerted by the parabola on the particle is then given as
151
Now we have that
F = F Ex
(4.28)
R = Rer
(4.29)
N = NEy
(4.30)
Mg = MgEy
(4.31)
Then the resultant force acting on the block is given as
FO = F + R + N + Mg = F Ex + NEy + Rer + MgEy
(4.32)
Using the expression for er from Eq. (4.3), we have that
FO =
153
Eq. (4.49) simplies to
ml 2 cos + ml sin mg R cos = 0
(4.50)
Then, multiplying Eq. (4.50) by sin , we obtain
ml 2 cos sin + ml sin2 mg sin R sin = 0
(4.51)
Next, multiplying Eq. (4.34) by cos , we obtain
F cos + R sin cos = M x cos
(4.52)
Rearranging
160
Chapter 4. Kinetics of a System of Particles
Equating the energy of the block at times t0 and t1 we have that
1
2
mv1 (t1 ) mgh
2
0=
(4.103)
Solving Eq. (4.103) for v1 (t1 ), we obtain
v1 (t1 ) = 2gh
(4.104)
From Eq. (4.104), the velocity of the block
152
Chapter 4. Kinetics of a System of Particles
Therefore, the resultant force acting on the block-particle system is given as
FT = F + N + (M + m)g = F Ex + NEy + (M + m)gEy
(4.39)
Eq. (4.39) simplies to
FT = F Ex + N + (M + m)g Ey
(4.40)
a
Setting FT e
159
The velocity of the center of mass of the block-plate system in reference frame
F is then given as
F
v=
mv1 + Mv2
m F v1 + M F v2
=
Ex
m+M
m+M
(4.93)
Kinetics
Phase 1: During Descent of Block
The only force acting on the block during its descent is th
158
Chapter 4. Kinetics of a System of Particles
Question 43
A block of mass m is dropped from a height h above a plate of mass M as shown
in Fig. P4-3. The plate is supported by three linear springs, each with spring
constant K, and is initially in stati
155
Rearranging, we have
(mv0 cos + Fx )Ex + (Fy mv0 sin )Ey = mvmx Ex + mvmy Ey
(4.66)
We then obtain the following two scalar equations:
mv0 cos + Fx
Fy mv0 sin
= mvmx
(4.67)
= mvmy
(4.68)
Application of Linear Impulse and Linear Momentum to the Entire
157
Simplifying this last expression, we have that
Fx = mv0 cos
M
m+M
(4.85)
Also, from Eq. (4.68) we have that
Fy mv0 sin = mvmy = mvy = 0
(4.86)
Solving this last equation for Fy , we obtain
Fy = mv0 sin
(4.87)
Consequently, the impulse exerted by the
156
Chapter 4. Kinetics of a System of Particles
Rearranging Eq. (4.74), we obtain
mv0 cos Ex + (P mv0 sin )Ey = (mvmx + MvMx )Ex + (mvmy + MvMy )Ey
(4.75)
We then obtain the following two scalar equations:
mv0 cos
= mvmx + MvMx
(4.76)
P mv0 sin
= mvmy
154
Chapter 4. Kinetics of a System of Particles
Solution to Question 42
First, let F be a xed reference frame. Then, choose the following coordinate
system xed in reference frame F :
Origin at Location of Block
Ex To The Left
Ez Into Page
Ey = E z E x
(4
142
Chapter 3. Kinetics of Particles
Kinetics
The free body diagram of the particle is shown in Fig. 3-22. Using Fig. 3-22, it is
Nn
Fs
Nb
mg
Figure 3-22
Free Body Diagram of Particle for Question 325.
seen that the following forces act on the particle:
=
141
choose the following coordinate system:
Origin at Point O
=
=
=
Ex
Ey
Ez
Along OP When t = 0
Along Oy When t = 0
Ex Ey
Furthermore, corresponding to reference frame A, we choose the following coordinate system:
Origin at Point O
=
Along OP
ex
=
Along
133
Fs
mg
Figure 3-19
Free Body Diagram of Particle for Question 322.
Kinetics
The free body diagram of the particle is shown in Fig. 3-19.
Using Fig. 3-19. we see that the following forces act on the particle:
= Force of Spring
Fs
mg = Force of Gravity
F
130
Chapter 3. Kinetics of Particles
Question 322
A particle of mass m is attached to a linear spring with spring constant K and
unstretched length r0 as shown in Fig. P3-22. The spring is attached at its other
end to a massless collar where the collar sl
129
Conservation of Angular Momentum Relative to Point O
First, it is important to observe that O is xed in the inertial reference frame F .
Then, using the denition of the angular momentum of a particle relative to an
inertially xed point O, we have that
128
Chapter 3. Kinetics of Particles
N
F
mg
Figure 3-17
Free Body Diagram for Question 320.
Now we have that
N = NEz
(3.468)
mg = mgEz
(3.469)
F = F er
(3.470)
It is noted that, because the rope exerts a known force on the particle, it is
necessary that t
124
Chapter 3. Kinetics of Particles
where A is the attachment point of the spring. It is seen from the geometry that
the spring is attached a distance 2a from the center of the circle. In terms of
the direction Ex , we have that
rA = 2REx
(3.436)
Then, u
125
(b) Reaction Force of Track on Particle As a Function of
Solving for the reaction force using Eq. (3.444), we obtain
N = mR 2 + KR(1 + 2 cos )
(3.448)
It is seen from Eq. (3.448) that, in order to obtain N as a function of , it is
necessary to nd 2 a
127
Then the position of the particle is given as
r = r er
(3.458)
Furthermore, the angular velocity of reference frame A in reference frame F is
given as
F
A = ez
(3.459)
The velocity of the particle in reference frame F is then obtained from the rate
of
123
Now we have that
F
d F
v
= R e
dt
F
A F v = ez (R e ) = R 2 er
(3.429)
(3.430)
Adding the expressions in Eq. (3.429) and Eq. (3.430), we obtain the acceleration
of the collar in reference frame F as
F
v = R 2 er + R e
(3.431)
Kinetics
The dierential e
132
Chapter 3. Kinetics of Particles
Furthermore, since rP /Q is expressed in terms of the basis cfw_er , e , ez and cfw_er , e , ez
is xed in reference frame A, we can obtain F vP /Q using the rate of change
transport theorem between reference frame A