197
Determination of Dierential Equation of Motion
The dierential equation of motion can be obtained using Eulers law about the
point of contact of the sphere with the incline (i.e. about point Q), i.e. we can
apply
F
d F
r
HQ
(5.121)
MQ ( rQ ) mF aQ =
dt

198
Chapter 5. Kinetics of Rigid Bodies
Therefore,
MQ = (r Ex r Ey ) F Ex + (r Ex r Ey ) P Ey
+ (r Ey ) (mg sin Ex + mg cos Ey )
(5.130)
This last expression simplies to
MQ = r P Ez r P Ez = r P (1 )Ez mgr sin Ez
(5.131)
Furthermore, the angular momentum

166
Chapter 4. Kinetics of a System of Particles
The position of the center of mass of the system is then given as
=
r
mA rA + mB rB
mA + m B
(4.151)
Substituting the expressions for rA and rB from Eq. (4.149) and Eq. (4.150), respectively, into Eq. (4.15

168
Chapter 4. Kinetics of a System of Particles
where er is the direction from mA to mB . Now we have
er = sin Ex + cos Ey
(4.166)
Also, dening e = Ez er , we obtain
e = Ez er = Ez ( sin Ex + cos Ey ) = cos Ex sin Ey
(4.167)
Then the resultant force appl

193
Question 53
A bulldozer pushes a boulder of mass m with a known force P up a hill inclined
at a constant inclination angle as shown in Fig. P5-3. For simplicity, the boulder is modeled as a uniform sphere of mass m and radius r . Assuming that
the bou

170
Chapter 4. Kinetics of a System of Particles
Kinematics
First, let F be a xed reference frame. Then, choose the following coordinate
system xed in reference frame F :
Ex
Ez
Ey
Origin at Point O
=
=
=
Along OC When = 0
Out of Page
Ez Ex
Next, let A be

169
Question 417
A dumbbell consists of two particles A and B each of mass m connected by a
rigid massless rod of length 2l. Each end of the dumbbell slides without friction
along a xed circular track of radius R as shown in Fig. P4-17. Knowing that
is t

202
Chapter 5. Kinetics of Rigid Bodies
The angular velocity of the rod the instant after the impulse is applied is then
F
given as
6F
F R
ez
=
(5.157)
ml
Now, after has been applied, the rod starts to rotate. Therefore, the only
F
forces and torques acti

165
Question 48
A particle of mass mA slides without friction along a xed vertical rigid rod. The
particle is attached via a rigid massless arm to a particle of mass mB where mB
slides without friction along a xed horizontal rigid rod. Assuming that is th

164
Chapter 4. Kinetics of a System of Particles
For simplicity, let
a=
2(m + M)g
3K
(4.144)
a a2 + 4ah
2
(4.145)
Then Eq. (4.143) can be written as
xmax =
Eq. (4.145) can be rewritten as
xmax =
a a 1 + 4h/a
2
=
a
1 1 + 4h/a
2
(4.146)
Now since h and a ar

192
Chapter 5. Kinetics of Rigid Bodies
Now substitute the expression for Rr sin R cos from Eq. (5.69) into this
Eq. (5.87). This gives
mx = M x
Ml
Ml 2
sin +
cos
2
2
(5.88)
Rearranging this last equation, we obtain the second dierential equation as

181
and
rP rO = r Ey
Consequently,
F
(5.6)
vP F vO = Ez r Ey = r Ex
(5.7)
Furthermore,
rO = xEx
which implies that
F
We then have that
F
vO =
F
(5.8)
d
(rO ) = xEx
dt
vP = F vO r Ex = (x r )Ex
(5.9)
(5.10)
Now, since the paper is suddenly pulled to the ri

171
The velocity of each particle can then be obtained using the transport theorem
as
F
F
A
drA F A
+ rA
dt
A
drB F A
+ rB
=
dt
=
vA
vB
(4.181)
(4.182)
Now since a and l are constant, we have that
A
drA
dt
A
drB
dt
= 0
(4.183)
= 0
(4.184)
Furthermore,
F
F

186
Chapter 5. Kinetics of Rigid Bodies
Kinematics
First, let F be a xed reference frame. Then, choose the following coordinate
system xed in F :
Ex
Ez
Ey
Origin at Collar When x = 0
=
To The Right
=
Out of Page
=
Ez Ex
Next, let R be a reference frame xe

Chapter 5
Kinetics of Rigid Bodies
Question 51
A homogeneous circular cylinder of mass m and radius r is at rest atop a thin
sheet of paper as shown in Fig. P5-1. The paper lies at on a horizontal surface.
Suddenly, the paper is pulled with a very large v

191
Consequently,
l
rR = er
r
2
(5.78)
l
M = er (Rr er R e )
2
(5.79)
Then,
This gives
M=
Equating M and
obtain
F
l
R Ez
2
(5.80)
F
d H/dt using the expression for
F
F
d H/dt from Eq. (5.62), we
l
Ml2
= R
12
2
(5.81)
This gives
R =
Ml
6
(5.82)
System o

175
Now since rO = 0, Eq. (4.224) can be written as
MO = rA NA + rB NB + rg 2mg
(4.225)
Now from Eq. (4.218) and Eq. (4.219) we have that the forces NA and NB lie in
the direction of rA and rB , respectively. Consequently, we have that
rA NA
= 0
(4.226)
r

172
Chapter 4. Kinetics of a System of Particles
Using the accelerations obtained in Eq. (4.195) and Eq. (4.196), the angular
momentum of the system relative to the (inertially xed) point O is obtained as
F
HO = (rA rO ) mA F vA + (rB rO ) mB F vB
(4.197)

199
Question 54
A uniform slender rod of mass m and length l pivots about its center at the
xed point O as shown in Fig. P5-4. A torsional spring with spring constant
K is attached to the rod at the pivot point. The rod is initially at rest and the
spring

201
R
F
Figure 5-7
54.
Free Body Diagram of Rod During Application of for Problem
F
because is assumed to be applied instantaneously and, thus, the orientation
F
of the rod does not change during the application of . Next, we see that the
F
, is inconseq

194
Chapter 5. Kinetics of Rigid Bodies
Ff
P
mg
Q
P
R
Figure 5-5
N
Free Body Diagram of Sphere for Question 53.
Using Fig. 5-5, the forces acting on the sphere are given as
N
R
P
mg
Ff
=
=
=
=
=
Force
Force
Force
Force
Force
of
of
of
of
of
Incline on Sphe

185
Question 52
A collar of mass m1 is attached to a rod of mass m2 and length l as shown in
Fig. P5-2. The collar slides without friction along a horizontal track while the
rod is free to rotate about the pivot point Q located at the collar. Knowing that

167
Now we know that NA and NB act orthogonal to the horizontal and vertical
tracks, respectively, i.e.,
NA
= NA Ey
(4.157)
NB
= NB Ex
(4.158)
Furthermore, the force of gravity acts vertically downward, i.e.,
(mA + mB )g = (mA + mB )gEx
(4.159)
Then the t

200
Chapter 5. Kinetics of Rigid Bodies
Next, let R be a reference frame xed to the rod. Then, choose the following
coordinate system xed in reference frame R:
er
ez
e = ez er
Origin at Point O
=
=
Along Rod (Down)
Ez
Then, since the rod rotates about the

195
Ex
Ey
uv
Figure 5-6
where
F vplate
P
F vR
P
Unit Vector in Vertically Downward Direction for Question 53.
is the velocity of point P on the sphere in reference frame F and
is the of point P on the plate velocity of the plate in reference frame F
(we n

180
Chapter 5. Kinetics of Rigid Bodies
Next, in order to solve this problem, we need to apply linear impulse and momentum to the center of mass of the cylinder and angular impulse and momentum about the center of mass of the cylinder. In order to apply t

162
Chapter 4. Kinetics of a System of Particles
+
Solving for v(t1 ) gives
m
(4.124)
2ghEx
m+M
Therefore, the post-impact velocities of the block and plate in reference frame
F are given as
+
v(t1 ) =
F
+
v1 (t1 ) =
F
+
v2 (t1 ) =
m
2ghEx
m+M
m
2ghEx
m+M

182
Chapter 5. Kinetics of Rigid Bodies
Furthermore, using the result of the force resolution from above, we have that
F = N + mg + Ff = NEy + mgEy + N Ex = (N + mg)Ey + N Ex
(5.19)
Equating F and mF aO , we obtain
(N + mg)Ey + N Ex = mxEx
(5.20)
which yi

184
Chapter 5. Kinetics of Rigid Bodies
Therefore,
r mgt =
mr 2
(t)
2
(5.43)
Solving for (t), we obtain
(t) =
2g
t
r
(5.44)
The angular velocity of the disk at the instant that the paper is pulled out is
then given as
2g
F R
tEz
(t) =
(5.45)
r