Chapter 16

RIGID BODY MOTION: TRANSLATION &
ROTATION (Sections 16.1-16.3)
Objectives :
To analyze the kinematics of a
rigid body undergoing planar
translation or rotation about a
fixed axis.
Methodology
 Understand 1D translational motion
 Use analogy:


To relate rotation with translation
To relate general motion of rigid body with
relative motion between two particles using
vector (tensor) notations.
APPLICATIONS
Passengers ride are subjected to
curvilinear translation.
Given the angular motion of the
rotating arms, determine v and a
experienced by the passengers?
Does each passenger feel the
same acceleration?
How can we relate the angular motions
of contacting bodies that rotate about
different fixed axes?
PLANAR KINEMATICS OF A RIGID BODY
There are cases where an object cannot be treated as
a particle. In these cases the size or shape of the
body must be considered. Also, rotation of the body
about its center of mass requires a different approach.
For example, in the design of gears and links in
machinery or mechanisms, rotation of the body is an
important aspect in the analysis of motion.
PLANAR RIGID BODY MOTION
There are three types of planar rigid body motion.
Translation: when every line segment on the body remains
parallel to its original direction during motion. When all
points move along straight lines: motion is rectilinear, when
the paths are curved: motion is curvilinear translation.
PLANAR RIGID BODY MOTION (continued)
Rotation about a fixed axis: all particles of
the body, except those on the axis of
rotation, move along circular paths in planes
perpendicular to the axis of rotation.
General plane motion: the body
undergoes both translation and rotation.
Translation occurs within a plane and
rotation occurs about an axis
perpendicular to this plane.
PLANAR RIGID BODY MOTION (continued)
Example of body undergoing
the three types of motion:
The wheel and crank (A and
B) undergo rotation about a fixed axis. Both axes of
rotation are at the location of the pins and perpendicular
to the plane of the figure.
The piston (C) undergoes rectilinear translation. The
connecting rod (D) undergoes curvilinear translation, since it
will remain horizontal as it moves along a circular path.
The connecting rod (E) undergoes general plane motion,
as it will both translate and rotate.
RIGID-BODY MOTION: TRANSLATION
The positions of two points
A and B on a translating
body can be related by
rB = rA + rB/A
The velocity at B is vB = vA+ drB/A/dt .
Now drB/A/dt = 0 since rB/A is constant. So, vB = vA, also: aB = aA.
Note, all points in a rigid body subjected to translation
move with the same velocity and acceleration.
RIGID-BODY MOTION:
ROTATION ABOUT A FIXED AXIS
When a body rotates, point P travels along a circular
path. The angular position of P is defined by .
d is the angular displacement, with units of either
radians or revolutions (1 revol. = 2 radians)
Angular velocity, , is the time derivative of
angular displacement:
 = d/dt (rad/s) +
Similarly, angular acceleration is  = d2/dt2 =
d/dt or  = (d/d) rad/s2
Time issues:
 Clockwise versus counterclockwise
 Design of stairs
 Time of days, months
 Hour time?!
RIGID-BODY MOTION:
ROTATION ABOUT A FIXED AXIS (continued)
If  is constant (C) by analogy:
 = O + Ct
 = O + Ot + 0.5Ct2
2 = (O)2 + 2C ( – O)
O and O are initial angular position and
angular velocity.
RIGID-BODY ROTATION:
VELOCITY OF POINT P
The velocity of P: v = r, its direction is
tangent to the circular path of P.
In vector formulation, the magnitude and
direction of v can be determined from the
cross product of  and rp . rp is a vector from
any point on the axis of rotation to P.
v =  x rp =  x r
The direction of v is determined by the righthand rule.
RIGID-BODY ROTATION:
ACCELERATION OF POINT P
The acceleration of P is expressed in terms of
its normal (an) and tangential (at) components.
In scalar form, these are at =  r and an = 2 r.
The tangential component, at, represents the
time rate of change in the velocity's
magnitude. It is directed tangent to the path of
motion.
The normal component, an, represents the time
rate of change in the velocity’s direction. It is
directed toward the center of the circular path.
RIGID-BODY ROTATION:
ACCELERATION OF POINT P (continued)
Using the vector formulation, the acceleration of
P is a = dv/dt = d/dt x rP +  x drP/dt
=  x r P +  x (  x r P)
It can be shown that this equation reduces to
a =  x r – 2r = at + an
The magnitude of the acceleration vector is a = (at)2 + (an)2
EXAMPLE
Given:The motor M begins
rotating at
 = 4(1 – e-t) rad/s, where
t is in seconds. The radii
of the motor, fan pulleys,
and fan blades are 2.5cm,
10cm, and 40cm,
respectively.
Find: The magnitudes of the velocity and acceleration at point P
on the fan blade when t = 0.5 s.
EXAMPLE (Solution:)
1) The angular velocity is given as a function of time, m =
4(1 – e-t), thus:
m = dm/dt = 4e-t rad/s2
When t = 0.5 s,
m = 4(1 – e-0.5) = 1.5739 rad/s, m = 4e-0.5 = 2.4261 rad/s2
2) Since the belt does not slip (and is assumed inextensible), it must
have the same speed and tangential component of acceleration.
Thus the pulleys must have the same speed and tangential
acceleration at their contact points with the belt. Therefore, the
angular velocities of the motor (m) and fan (f) are related as
v = m rm = f rf => (1.5739)(2.5) = f(10) => f = 0.3935 rad/s
EXAMPLE (continued)
3) Similarly, the tangential accelerations are related as
at = m rm = f rf => (2.4261)(2.5) = f(10) => f = 0.606rad/s2
4) The speed of point P on the fan, at a radius of 40cm, is now
determined as
vP = frP = (0.3935)(40) = 15.8 cm/s
The normal and tangential components of acceleration of P are
an = (f)2 rP = (0.3935)2 (40) = 6.2 cm/s2
at = f rP = (0.6065) (40) = 24.3 cm/s2
GROUP PROBLEM SOLVING
Given: Starting from rest when s = 0,
pulley A (rA = 50 mm) is given
a constant angular acceleration,
A = 6 rad/s2. Pulley C (rC =
150 mm) has an inner hub D
(rD = 75 mm) which is fixed to
C and turns with it.
Find: The speed of block B when it has risen s = 6 m.
GROUP PROBLEM SOLVING (Solution)
1) Assuming the belt is inextensible and does not slip,
C = 2 rad/s2
2) Assuming the cord attached to block B is inextensible and
does not slip, the speed and acceleration of B is
aB = 0.15 m/s2
GROUP PROBLEM SOLVING (continued)
3) Since A is constant,
vB = 1.34 m/s
ABSOLUTE MOTION ANALYSIS
(Section 16.4)
Objective:
To determine the velocity
and acceleration of a rigid
body undergoing general
plane motion using an
absolute motion analysis.
Note: to be read by student if
interested in geometry. In
class we only use vector
(tensor) methodology.
EXAMPLE 1
Given:Two slider blocks are connected
by a rod of length 2 m. Also,
vA = 8 m/s and aA = 0.
Find: Angular velocity, , and
angular acceleration, , of the
rod when  = 60°.
EXAMPLE 1 (Solution by geometry))
In class solution by vectors
By geometry, sA = 2 cos 
reference
By differentiating with respect to time,
vA = -2  sin 

A
sA
Using  = 60° and vA = 8 m/s and solving
for :  = 8/(-2 sin 60°) = - 4.62 rad/s
Differentiating vA and solving for ,
aA = -2 sin  – 22 cos  = 0
 = - 2/tan  = -12.32 rad/s2
EXAMPLE 2 (Solution by geometry))
In class solution by vectors
Given:Crank AB rotates at a
constant  = 150 rad/s
Find: Velocity of P when
 = 30°
Solution: xP = 60 cos  +
(225)2 – (60 sin )2
vP = -60 sin  + (0.5)[(225)2
– (60sin )2]-0.5(-2)(60sin )(60cos )
vP = -60 sin  – [0.5(60)2sin2 ] / (225)2 – (60 sin )2
At  = 30°,  = 150 rad/s and vP = -5550mm/s
GROUP PROBLEM SOLVING
How to deal with flexible bodies!
Given: The  and  of the disk of
the dimensions shown.
Find: The velocity and
acceleration of cylinder B
in terms of .
GROUP PROBLEM SOLVING (Solution)
Law of cosines:
aB =
RELATIVE MOTION ANALYSIS:
VELOCITY (Section 16.5)
Objectives:
a) Describe the velocity of a rigid
body in terms of translation
and rotation components.
b) Perform a relative-motion
velocity analysis of a point on
the body.
APPLICATIONS
As block A moves to the left with vA,
it causes the link CB to rotate
counterclockwise, thus vB is directed
tangent to its circular path. Which
link undergoes general plane motion?
How to find its angular velocity?
Gear systems are used in many
automobile automatic transmissions.
By locking or releasing different gears,
this system can operate the car at
different speeds.
RELATIVE MOTION ANALYSIS:
DISPLACEMENT
A body undergoes a combination of translation and rotation (general motion)
=
Point A is called the base point, has a known motion. The x’-y’ frame
translates with the body, but does not rotate. The displacement of B:
Disp. due to translation
drB = drA + drB/A
Disp. due to translation and rotation
Disp. due to rotation
RELATIVE MOTION ANALYSIS:
VELOCITY
=
+
The velocity at B is : (drB/dt) = (drA/dt) + (drB/A/dt) or
vB = vA + vB/A
Since the body is taken as rotating about A,
vB/A = drB/A/dt =  x rB/A
Here  will only have a k component since the axis of
rotation is perpendicular to the plane of translation.
RELATIVE MOTION ANALYSIS:
VELOCITY (continued)
vB = vA +  x rB/A
When using the relative velocity equation, points A and B
should generally be points on the body with a known motion.
Often these points are pin connections in linkages.
Both points A and B have circular
motion since the disk and link BC
move in circular paths. The
directions of vA and vB are tangent
to the circular path of motion.
RELATIVE MOTION ANALYSIS:
VELOCITY (continued)
vB = vA +  x rB/A
When a wheel rolls without slipping, point A has zero velocity.
Furthermore, point B at the center of the wheel moves along a
horizontal path. Thus, vB has a known direction, e.g., parallel
to the surface.
EXAMPLE 1
Given: Block A is moving down
at 2 m/s.
Find: The velocity of B at the
instant  = 45.
EXAMPLE 1 (Solution)
vB = vA + AB x rB/A
vB i = -2 j + ( k x (0.2 sin 45 i - 0.2 cos 45 j ))
vB i = -2 j + 0.2  sin 45 j + 0.2  cos 45 i
Equating the i and j components gives:
vB = 0.2  cos 45
0 = -2 + 0.2  sin 45
Solving:
 = 14.1 rad/s or AB = 14.1 rad/s k
vB = 2 m/s or vB = 2 m/s i
EXAMPLE 2
Given:Collar C is moving
downward with a velocity of
2 m/s.
Find: The angular velocities of CB
and AB at this instant.
EXAMPLE 2 (solution)
Link CB. Write the relative-velocity
equation:
vB = vC + CB x rB/C
vB i = -2 j + CB k x (0.2 i - 0.2 j )
vB i = -2 j + 0.2 CB j + 0.2 CB i
By comparing the i, j components:
i: vB = 0.2 CB
=> vB = 2 m/s i
j: 0 = -2 + 0.2 CB
=> CB = 10 rad/s k
EXAMPLE 2 (continued)
Link AB experiences only rotation about
A. Since vB is known, there is only one
equation with one unknown to be found.
vB = AB x rB/A
2 i = AB k x (-0.2 j )
2 i = 0.2 AB i
By comparing the i-components:
2 = 0.2 AB
So, AB = 10 rad/s k
GROUP PROBLEM SOLVING
Given: The crankshaft AB is rotating
at 500 rad/s about a fixed axis
passing through A.
Find: The speed of the piston P at
the instant it is in the position
shown.
GROUP PROBLEM SOLVING (solution)
1) First draw the kinematic diagram of link AB.
Applying the relative velocity equation
vB = -50 j m/s
GROUP PROBLEM SOLVING (continued)
2) Now consider link BC.
Applying the relative velocity equation:
vC = -50 j m/s
INSTANTANEOUS CENTER (IC)
OF ZERO VELOCITY (Section 16.6)
To be read by student if interested in geometry.
Note: every body has an
instantaneous center of zero
velocity. Think of motion of planets
and an instantaneous center for the
universe
RELATIVE MOTION ANALYSIS:
ACCELERATION (Section 16.7)
Objectives:
a) Resolve the acceleration of a
point on a body into
components of translation and
rotation.
b) Determine the acceleration of a
point on a body by using a
relative acceleration analysis.
APPLICATIONS
The forces delivered to the crankshaft, and
the angular acceleration of the crankshaft,
depend on the speed and acceleration of
the piston in an automotive engine.
How can we relate the accelerations of the
piston, connection rod, and crankshaft in
this engine?
RELATIVE MOTION ANALYSIS:
ACCELERATION
The equation relating the accelerations of two points on the
body is determined by:
dvB
dvA
dv
=
+ B/ A
dt
dt
dt
These are absolute accelerations
of points A and B. They are
measured from a set of fixed x,y
axes.
This term is the acceleration
of B wrt A. It will develop
tangential and normal
components.
The result is aB = aA + (aB/A)t + (aB/A)n
RELATIVE MOTION ANALYSIS:
ACCELERATION
Graphically:
aB = aA +
=
(aB/A)t + (aB/A)n
+
The relative tangential acceleration component (aB/A)t is ( x rB/A)
and perpendicular to rB/A.
The relative normal acceleration component (aB/A)n is (-2 rB/A)
and the direction is always from B towards A.
RELATIVE MOTION ANALYSIS:
ACCELERATION (continued)
Since the relative acceleration components can be expressed
as (aB/A)t =   rB/A and (aB/A)n = - 2 rB/A the relative
acceleration equation becomes
aB = aA +   rB/A - 2 rB/A
APPLICATION OF RELATIVE
ACCELERATION EQUATION
In applying the relative acceleration equation, the two points used in the
analysis (A and B) should generally be selected as points which have a
known motion, such as pin connections with other bodies.
In this mechanism, point B is known to travel along a circular path, so aB can
be expressed in terms of its normal and tangential components. Note that
point B on link BC will have the same acceleration as point B on link AB.
Point C, connecting link BC and the piston, moves along
a straight-line path. Hence, aC is directed horizontally.
EXAMPLE 1
Given:Point A on rod AB has an
acceleration of 3 m/s2 and
a velocity of 2 m/s at the
instant the rod becomes
horizontal.
Find: The angular acceleration
of the rod at this instant.
Solution: First, we need to find the angular velocity of the rod
at this instant. Locating the instant center (IC) for
rod AB (which lies above the midpoint of the rod),
we can determine :
A = vA/rA/IC = vA/(5/cos 45) = 0.283 rad/s
EXAMPLE 1 (continued)
Since points A and B both move
along straight-line paths,
aA = 3 (cos 45 i - sin 45 j) m/s2
aB = aB(cos 45 i + sin 45 j) m/s2
Applying the relative acceleration equation
aB = aA +  x rB/A – 2rB/A
(aB cos 45 i + aB sin 45 j) = (3 cos 45 i – 3 sin 45 j)
+ ( k x 10 i) – (0.283)2(10i)
EXAMPLE 1 (continued)
By comparing the i, j components;
aB cos 45 = 3 cos 45 – (0.283)2((10)
aB sin 45 = -3 sin 45 + (10)
Solving:
aB = 1.87 m/s2
 = 0.344 rad/s2
BODIES IN CONTACT
Consider two bodies in contact with one another without slipping,
where the points in contact move along different paths.
In this case, the tangential components of acceleration will be the
same, i. e.,
(aA)t = (aA’)t (which implies BrB = CrC ).
The normal components of acceleration will not be the same.
(aA)n  (aA’)n so aA  aA’
EXAMPLE: ROLLING MOTION
A common problem involves rolling motion without slip; e.g., a ball
or disk rolling along a flat surface without slipping. This problem
can be analyzed using relative velocity and acceleration equations.
As the cylinder rolls, point G (center) moves along a straight line, while
point A, on the rim of the cylinder, moves along a curved path called a
cycloid. If  and  are known, the relative velocity and acceleration
equations can be applied to these points at the instant A is in
contact with the ground:
EXAMPLE: ROLLING MOTION(continued)
• Velocity analysis. Since no slip occurs, v = 0. From the
A
kinematic diagram: vG = vA +  x rG/A
vG i = 0 + (- k) x (r j)
vG = r or vG = r i
• Acceleration. Since G moves along a straight-line path, aG is
horizontal. Just before A touches ground, its
velocity is directed downward, and just after
contact, its velocity is directed upward. Thus,
point A accelerates upward as it leaves the ground.
aG = aA +  x rG/A – 2rG/A => aG i = aA j + (- k) x (r j) – 2(r j)
Evaluating and equating i and j components:
aG = r and aA = 2r or aG = r i and aA = 2r j
ROLLING MOTION(continued):
different view
vG = v i and v=2πr/Δt
ω=2π/Δt=2π/(2πr/v)=v/r
vG =r ω i
vA = vG + vA/G =r ω i – ωkX-r j=0
EXAMPLE 2
Given:The ball rolls without
slipping.
Find: The accelerations of
points A and B at this
instant.
Solution: Since the ball is rolling without slip, aO is
directed to the left with a magnitude of
aO = r = (4 rad/s2)(0.15 m)=0.6 m/s2
EXAMPLE 2 (continued)
Apply the relative acceleration
equation between points O and B:
aB = aO +  x rB/O – 2rB/O
aB = -0.6i + (4k) x (0.15i) –
(6)2(0.15i)= (-6i + 0.6j) m/s2
Now do the same for point A.
aA = aO +  x rA/O – 2rA/O
aA = -0.6i + (4k) x (0.15j) –
(6)2(0.15j)= (-1.2i – 5.4j) m/s2
GROUP PROBLEM SOLVING
Given: The disk is rotating with
 = 3 rad/s,  = 8 rad/s2 at
this instant.
Find: The acceleration at point
B, and the angular
velocity and acceleration
of link AB.
Solution: …..
GROUP PROBLEM SOLVING (continued)
Draw the kinematic diagram and then
apply the relative-acceleration
equation:
aBi = (1.6 + 0.4 sin 30 AB)i + (-1.8 + 0.4 cos 30 AB)j
GROUP PROBLEM SOLVING (continued)
Solving:
aB = 2.64 m/s2
AB = 5.20 rad/s2
CHAPTER 17
MASS MOMENT OF INERTIA (Section 17.1)
Objectives:
To determine the mass
moment of inertia of a rigid
body or a system of rigid
bodies.
APPLICATIONS
The flywheel on the engine has a large
mass moment of inertia about its axis of
rotation. Once it is set into motion, it will
be difficult to stop. Does the mass
moment of inertia depend on the radius
of the wheel? Its thickness?
The crank undergoes rotation about a
fixed axis that is not at its mass center.
The crank develops a kinetic energy
directly related to its mass moment of
inertia. As the crank rotates, its kinetic
energy is converted to potential energy
and vice versa.
MOMENT OF INERTIA
The mass moment of inertia is a measure of an
object’s resistance to rotation:
I =  r2 dm =  r2r dV
r acts as the moment arm of the mass element
and r is the density of the body. Thus, the value
m axis about
of I differs for each
which it is
V
computed.
In Section 17.1, the focus is on obtaining the
mass moment of inertia via integration.
MOMENT OF INERTIA (continued)
The figures show the mass moment of inertia for two flat plate shapes
commonly used when working with 3D bodies. The shapes are often
used as the differential element being integrated over the entire body.
PROCEDURE FOR ANALYSIS
(review of statics/ not required)
Shell element
• If a shell element having a height z,
radius r = y, and thickness dy is
chosen for integration, then the
volume element is dV = (2πy)(z)dy
Disk element
If a disk element having a radius y and
a thickness dz is chosen for
integration, then the volume element
is dV = (πy2)dz
PARALLEL-AXIS THEOREM
The parallel axis theorem states: IA= IG + md2
where IG = mass moment of inertia about the body’s mass center
m = mass of the body
d = perpendicular distance between the parallel axes
PARALLEL-AXIS THEOREM (continued)
Radius of Gyration: has units of length and is a measure of the
distribution of the body’s mass about the axis at which the moment
of inertia is defined.
I = m k2 or k = (I/m)
Composite Bodies: If a body is constructed of a number of
shapes, the mass moment of inertia of the body about any axis is
the algebraic addition of all the mass moments of inertia, found
about the same axis, of the different shapes.
EXAMPLE 2
Given:Two rods assembled as shown,
with each rod weighing 10 N.
Find: The location of the center of
mass G and moment of inertia
about an axis passing through G
of the rod assembly.
Solution: The center of mass is located relative to the pin at O
at a distance y, where
EXAMPLE 2 (continued)
The mass moment of inertia of each rod about its center of mass is:
I = (1/12)ml2 = (1/12)(10/9.80)(2)2 = 0.340 kg·m2
The moment of inertia IG is calculated
using the parallel axis theorem.
IG = [I + m(y-1)2]OA + [I + m(2-
)2]BC
IG=[0.34+(10/9.8)(0.5)2]+[0.34+(10/9.80)(0.5)2]
= 1.19 kg·m2
PLANAR KINETIC EQUATIONS OF MOTION:
TRANSLATION (Sections 17.2-17.3)
Objectives:
a) Apply the three equations
of motion for a rigid body
in planar motion.
b) Analyze problems
involving translational
motion.
APPLICATIONS
The boat and trailer undergo
rectilinear motion. In order to
find the reactions at the trailer
wheels and the acceleration
of the boat center of mass,
we need to draw the FBD for
the boat and trailer.
=
How many equations of motion do we need to
solve this problem? What are they?
EQUATIONS OF TRANSLATIONAL
MOTION
• We will limit our study of planar kinetics to rigid bodies that are
symmetric with respect to a fixed reference plane.
• As discussed when a body is subjected to general plane motion, it
undergoes a combination of translation and rotation.
• First, a coordinate system with its origin at an arbitrary point P
is established. The x-y axes should not rotate and can either be
fixed or translate with constant velocity.
EQUATIONS OF TRANSLATIONAL MOTION
(continued)
• If a body undergoes translational motion, the equation of motion
is F = m aG . In scalar form:
 Fx = m(aG)x
and
 Fy = m(aG)y
• In words: the sum of all the external forces acting on the
body is equal to the body’s mass times the acceleration of it’s
mass center.
=
EQUATIONS OF ROTATIONAL MOTION
We need to determine the effects caused by the moments of the
external force system. The moment about point P is:
 (ri  Fi) +  Mi = rG  maG + IG
 Mp = ( Mk )p
where  Mp is the resultant moment about P due to all the external
forces. The term (Mk)p is called the kinetic moment about P.
=
EQUATIONS OF ROTATIONAL MOTION
(continued)
If point P coincides with the mass center G, this equation reduces to
 MG = IG .
Thus, three independent scalar equations of motion are used to
describe the general planar motion of a rigid body:
 Fx = m(aG)x
 Fy = m(aG)y
and  MG = IG or  Mp =  (Mk)p
EQUATIONS OF MOTION:
TRANSLATION ONLY
All the particles of the body have the same acceleration so aG = a
and  = 0. The equations of motion become:
 Fx = m(aG)x
 Fy = m(aG)y
 MG = 0
Note that the moment equation can be applied about other points
instead of the mass center, e.g.
MA = (m aG ) d .
EQUATIONS OF MOTION: TRANSLATION ONLY
(continued)
When a rigid body is subjected to
curvilinear translation, it is best to
use an n-t coordinate system. Then
apply the equations of motion, as
written below, for n-t coordinates.
 Fn = m(aG)n
 Ft = m(aG)t
 MG = 0 or
 MB = e[m(aG)t] – h[m(aG)n]
EXAMPLE
Given: A 50 kg crate rests on
a horizontal surface for
which the kinetic friction
coefficient k = 0.2.
Find: The acceleration of the
crate if P = 600 N.
Note that the load P can cause the crate either to
slide or to tip over. Let’s assume that the crate slides.
We will check this assumption later.
EXAMPLE (solution)
The coordinate system and FBD
are as shown. The weight of
(50)(9.81) N is applied at the
center of mass and the normal
force Nc acts at O. Point O is
some distance x from the crate’s
center line. The unknowns are
Nc, x, and aG .
Applying the equations of motion:
 Fy = m(aG)y: Nc – 490.5 = 0,
Nc = 490.5 N
 Fx = m(aG)x: 600 – 0.2 Nc = 50 aG,
aG = 10.0 m/s2
 MG = 0: -600(0.3) + Nc(x)-0.2 Nc (0.5) = 0,x
= 0.467 m
EXAMPLE (continued)
Since x = 0.467 m < 0.5
m, the crate slides as
originally assumed.
If x was greater than 0.5
m, the problem would
have to be reworked with
the assumption that
tipping occurred.
Example 2
Given: A 2kg bottle rests on
the checkout conveyor
at a grocery store. The
static friction coefficient
is s = 0.2.
Find: The largest acceleration
the conveyor can have
without causing the
bottle to slip or tip
Answer: 1.96m/s2 no slip
1.84m/s2 no tip
GROUP PROBLEM SOLVING
Given: A uniform rod BC
has a mass of 3 kg.
The crank is rotating
at a constant AB = 5
rad/s.
Find: The vertical
forces on rod BC
at points B and C
when  = 0 and
90 degrees.
GROUP PROBLEM SOLVING (continued)
Solution: Rod BC’s FBD at  = 0º:
Applying the equations of motion:
By = 7.215 N
Cy = 7.215 N
GROUP PROBLEM SOLVING (continued)
When  = 90º, the FBD is:
Applying the equations of motion:
By = 14.7 N
Cy = 14.7 N
EQUATIONS OF MOTION: ROTATION
ABOUT A FIXED AXIS (Section 17.4)
Objectives:
To analyze the planar kinetics
of a rigid body undergoing
rotational motion.
APPLICATIONS
The crank undergoes rotation
about a fixed axis, caused by the
driving torque M from a motor.
Pin at the center
of rotation.
As the crank turns, a dynamic
reaction is produced at the pin. It
is a function of angular velocity,
angular acceleration, and the
orientation of the crank.
EQUATIONS OF MOTION FOR
PURE ROTATION
When a rigid body rotates, the body’s
center of gravity G moves in a circular
path of radius rG. Thus, the acceleration
of G can be represented by (aG)t = rG 
and (aG)n = rG 2.
The body experiences an angular acceleration, its inertia creates a
moment IG equal to the moment of the external forces about G. Thus
 Fn = m (aG)n = m rG 2 , Ft = m (aG)t = m rG 
 MG = IG 
EQUATIONS OF MOTION
(continued)
MG moment equation may be replaced by a moment summation
about any arbitrary point. Summing the moment about the center
of rotation O yields
MO = IG + rG m (aG) t = (IG + m (rG)2 ) 
Since, IO = IG + m(rG)2, therefore the three equations of
motion for the body becomes:
Fn = m (aG) n = m rG 2
Ft = m (aG) t = m rG 
MO = IO 
EXAMPLE
Given:A rod with mass of 20
kg is rotating at 5rad/s
at the instant shown. A
moment of 60 N·m is
applied to the rod.
Find: The angular acceleration  and the reaction at pin O when
the rod is in the horizontal position.
EXAMPLE (continued)
FBD & Kinetic Diagram
Equations of motion:
+ Fn = man = mrG2
On = 20(1.5)(5)2 = 750 N
+ Ft = mat = mrG
-Ot + 20(9.81) = 20(1.5)
+ MO = IG  + m rG  (rG)
EXAMPLE (continued)

Using IG = (ml2)/12 and rG = (0.5)(l):
MO = [(ml2/12) + (ml2/4)] = (ml2/3) where (ml2/3) = IO.
After substituting:
60 + 20(9.81)(1.5) = 20(32/3)
Solving:  = 5.9 rad/s2
Ot = 19 N
GROUP PROBLEM SOLVING
Given: Wdisk = 15 N,
Wrod = 10 N,
ω = 8 rad/s at
this instant.
Find: The horizontal and vertical components of the
reaction at pin O when the rod is horizontal.
GROUP PROBLEM SOLVING (continued)
Solution: FBD & Kinetic Diagrams
Equations of motion:
Fx = m(aG)x:
Fy = m(aG)y:
MO = Ioα:
Therefore,
α = 2.85rad/s2, Oy = 4.29 N
Ox = 465N
EQUATIONS OF MOTION: GENERAL
PLANE MOTION (Section 17.5)
Objectives:
To analyze the planar
kinetics of a rigid body
undergoing general plane
motion.
APPLICATIONS
As the soil compactor accelerates
forward, the front roller experiences
general plane motion (both
translation and rotation).
What are the loads experienced
by the roller shaft or bearings?
=
The forces shown on the
roller’s FBD cause the
accelerations shown on
the kinetic diagram.
APPLICATIONS (continued)
During an impact, the center of
gravity of this crash dummy will
decelerate with the vehicle, but
also experience another
acceleration due to its rotation
about point A.
How can engineers use this
information to determine the
forces exerted by the seat belt on
a passenger during a crash?
EQUATIONS OF MOTION:
GENERAL PLANE MOTION
When a rigid body is subjected to
external forces and couple-moments,
it undergoes both translational and
rotational motion.
Using an x-y inertial coordinate
system, the equations of motions
about the center of mass, G, may be
written as
 Fx = m (aG)x
P
 Fy = m (aG)y
 MG = I G 
EQUATIONS OF MOTION: GENERAL
PLANE MOTION (continued)
Sometimes, it is convenient to write
the moment equation about some
point P other than G. Then the
equations of motion are:
 Fx = m (aG)x
 Fy = m (aG)y
 MP =  (Mk )P
P
Here,  (Mk )P is the sum of the moments
of IG α and maG about point P.
FRICTIONAL ROLLING PROBLEMS
When analyzing the rolling motion it may not be known if the
body rolls without slipping or if it slides as it rolls.
For example, consider a disk with mass m
and radius r, subjected to a known force P.
The equations of motion will be
 Fx = m(aG)x => P - F = maG
 Fy = m(aG)y => N - mg = 0
 MG = IGα
=> F r = IG α
There are 4 unknowns (F, N, α, and aG)
in these three equations.
FRICTIONAL ROLLING PROBLEMS
(continued)
Hence, we need an assumption to
provide another equation. The 4th
equation can be obtained from the slip or
non-slip condition of the disk.
Case 1: Assume no slipping and use aG =  r and DO NOT use
Ff = sN. After solving, you need to verify the assumption
was correct by checking Ff  sN.
Case 2: Assume slipping and use Ff = kN. In this case, aG  r.
EXAMPLE
Given: A spool has a mass of 8 kg
and a radius of gyration (kG)
of 0.35 m. Cords of
negligible mass are wrapped
around its inner hub and
outer rim. There is no
slipping.
Find: The angular acceleration ()
of the spool.
EXAMPLE (solution)
FBD
The moment of inertia of the spool is
IG = m (kG)2 = 8 (0.35)2 = 0.980 kg·m 2
Method I
Equations of motion:
Fy = m (aG)y
T + 100 -78.48 = 8 aG
MG = IG 
100 (0.2) – T(0.5) = 0.98 
3 unknowns, T, aG, . No slipping, the 3rd equation: aG = r=0.5
Solving, we find:  =10.3 rad/s2, aG = 5.16 m/s2, T = 19.8 N
EXAMPLE (continued)
FBD
Method II
A moment equation about A will
be used. This will eliminate the
unknown cord tension (T).
 MA=  (Mk)A: 100 (0.7) - 78.48(0.5) = 0.98  + (8 aG)(0.5)
No slip: aG = 0.5α, solving: α = 10.3 rad/s2, aG = 5.16 m/s2
GROUP PROBLEM SOLVING
Given:A 50 N wheel has a
radius of gyration kG =
0.7 m.
Find: The acceleration of the
mass center if M = 35
N.m is applied. s = 0.3,
k = 0.25.
Solution:
The moment of inertia of the wheel
GROUP PROBLEM SOLVING (continued)
FBD:
Equations of motion:
Do you need another equation before solving for the unknowns?
GROUP PROBLEM SOLVING (continued)
Solving for the unknowns yields:
NA = 50.0 N
FA = 0.25 NA = 12.5 N
 = 7.75 rad/s2
aG = 2.45 m/s2