Chapter_16_Planar_Kinematics_of_Rigid_Bo.ppt

CHAPTER-16
PLANAR KINEMATICS OF A
RIGID BODY

PLANAR KINEMATICS OF A RIGID BODY
• Planar Rigid Body Motion
• Translation
• Rotation About a Fixed Axis
• General Plane Motion
• Velocity Analysis of a Rigid Body
• Relative Velocity Method
• Instantaneous Center (IC) Method
• Acceleration Analysis of a Rigid Body

RIGID BODY MOTION
• When all the particles of a rigid body move along paths which are
equidistant from a fixed plane, the body is said to undergo planar motion.
•
• There are three types of planar motion:
• Translation
• Rotation about a fixed axis
• General Plane Motion

RIGID BODY MOTION
1. TRANSLATION:
• Translation occurs if any line segment on the body remains parallel to its
original direction during the motion
• When the paths of motion for any two particles of the body are along
equidistant straight lines, the motion is called rectilinear translation
• If the paths of motion are along curved lines which are equidistant the
motion is called curvilinear translation

RIGID BODY MOTION
2. ROTATION ABOUT A FIXED AXIS:
• When a rigid body rotates about a fixed axis, all the particles of the body,
except those which lie on the axis of rotation, moves along circular paths.

RIGID BODY MOTION
3. GENERAL PLANE MOTION ( COMPLEX MOTION):
• When a body is subjected to general plane motion, it undergoes a
combination of translation and rotation
• The translation occurs within a reference plane, and the rotation occurs
about an axis perpendicular to the reference plane

TRANSLATION
• Consider a rigid body which is subjected to either rectilinear or curvilinear
translation in the xy plane
Position:
• The location of points A and B are defined from the fixed xy reference frame by
using position vectors rA and rB
• The position of B with respect to A is denoted by the relative position vector rB/A
• By vector addition:
rB = rA + rB/A

TRANSLATION
Velocity:
• A relationship between the instantaneous velocities of A and B is obtained by taking
the time derivative of the position equation, which yields:
vB = vA + ( d rB/A / dt )
• Here vB and vA denote absolute velocities
• The term ( d rB/A / dt ) = 0, since the magnitude of rB/A is constant by definition of a
rigid body, and because the body is translating the direction of rB/A is constant.
• Therefore:
vB = vA

TRANSLATION
• Acceleration:
• Taking the time derivative of the acceleration equation yields:
aB = aA
• The above two equations indicate that all points in a rigid body subjected to either
curvilinear or rectilinear translation move with the same velocity and acceleration
• So the kinematics of particle motion may be used to specify the kinematics of points
located in a translating rigid body

ROTATION ABOUT A FIXED AXIS
• When a body is rotating about a fixed axis, any point P
located in the body travels along a circular path.
Angular Motion:
• Since a point is without dimension, it has no angular
motion
• Only lines and bodies undergo angular motion
• Consider the body shown in the figure and the angular
motion of a radial line r.
Angular Position:
• At any instant, the angular position of r is defined by the
angle θ measured between a fixed reference line and r.
Angular Displacement:
• The change in angular position, which can be measured as
a differential dθ, is called the angular displacement.

Angular Velocity:
• The time rate of change in the angular position is called
the angular velocity ω
• Its magnitude is:
ω = dθ / dt -----------------------------(1)
Angular Acceleration:
• The angular acceleration α measures the time rate of
change of the angular velocity.
• Its magnitude is:
α = dω / dt -------------------------------(2)
or α = d2θ / dt2 ------------------------------(3)
• The line of action of α is same as that for ω, however, its
sense of direction depends on whether ω is increasing or
decreasing with time.

• By eliminating dt from equations (1) and (2), we get:
α dθ = ω dω ----------------------------(4)
• Similarity between the differential relations for angular motion and those
developed for rectilinear motion of a particle
Constant angular acceleration:
• If the angular acceleration of the body is constant, α = αc, then eq (1), (2) and (4)
when integrated, yield a set of formulas which relate the body’s angular velocity,
angular position and time. The results are:
ω = ω0 + αct -----------------------------------(5)
θ = θ0 + ω0t + (1/2) αct2 ----------------------(6)
ω2 = ω0
2 + 2 αc (θ – θ0 ) ------------------------(7)

Motion of Point P:
• As the rigid body rotates, point P travels along a circular path of radius r
and center at point O.
Position:
• The position of P is defined by the position vector r, which extends from
O to P.

Velocity:
• The velocity of P has a magnitude which can be
determined from the circular motion of P using its
polar coordinate components vr = r0 and vθ = rθ0
• Since r is constant, the radial component vr = 0,
so that v = vθ = rθ0
• Because ω = θ0, therefore:
v = ωr
• The direction of v is tangent to the circular path.
• Both the magnitude and direction of v can also be
found by using the cross product of w and r:
v = ω x r

Acceleration:
• The acceleration of P can be expressed in terms of its
normal and tangential components
• Using at = dv /dt, an = v2 / ζ, noting that ζ = r, v = ω r,
and α = dω /dt, we have:
at = α r
an = ω2 r
• The tangential component of acceleration represents
the time rate of change in the velocity’s magnitude. If
the speed of P is increasing, then at acts in the same
direction as v; if the speed is decreasing, at acts in the
opposite direction of v; and if the speed is constant
at = 0
• The normal component of acceleration represents the
time rate of change in the velocity’s direction. The
direction of an is always towards O, the center of the
circular path

Acceleration:
• Like the velocity, the acceleration of point P may be
expressed in terms of the vector cross product.
• Since dω/dt = α and dr/dt = v = ω x r, so the above
equation becomes:
a = α x r + ω x (ω x r )
• In terms of the components, a can be expressed as:
a = at + an
a = α x r – ω2 r

Examples:
16.1, 16.2
Fundamental Problems:
F16.1, F16.3, F16.5
Practice Problems:
16.3, 16.5, 16.18, 16.21,
16.29, 16.30

EXAMPLE 16-1
A cord is wrapped around a wheel which is initially at rest as shown.
If a force is applied to the cord and gives it an acceleration a = (4t)
m/s2, where ‘t’ is in seconds, determine as a function of time (a) the
angular velocity of the wheel (b) the angular position of the line OP
in radians

EXAMPLE 16-2
The motor shown in the photo is used to
turn a wheel and attached blower
contained within the housing. The
details of the design are shown in Fig. If
the pulley A connected to the motor
begins to rotate from rest with a
constant angular acceleration of αA = 2
rad/s2, determine the magnitudes of the
velocity and acceleration of point P on
the wheel, after the pulley A has turned
two revolutions. Assume the
transmission belt does not slip on the
pulley and wheel.

PROBLEM 16-5
The operation of reverse gear in an automotive transmission is shown.
If the engine turns shaft A at ωA = 40 rad/s, determine the angular
velocity of the drive shaft, ωB The radius of each gear is listed in the
figure.

PROBLEM 16-18
Gear A is in mesh with gear B as shown. If A starts from rest and has
a constant angular acceleration of αA = 2 rad/s2, determine the time
needed for B to attain an angular velocity of ωB = 50 rad/s

PROBLEM 16-29
At the instant shown, gear A is rotating with a constant angular
velocity of ωA=6 rad/s, determine the largest angular velocity of gear
B and the maximum speed of point C.

PROBLEM 16-30
If the operator initially drives the pedals at 20 rev/min, and then begins
an angular acceleration of 30 rev /min2, determine the angular velocity
of the flywheel F when t = 3 s. Note that the pedal arm is fixed
connected to the chain wheel A, which in turn drives the sheave B using
the fixed connected clutch gear D. The belt wraps around the sheave
then drives the pulley E and fixed connected flywheel.

RELATIVE MOTION ANALYSIS: VELOCITY
• General plane motion consists of a combination of translation and rotation
• So it is often convenient to view these components separately using a relative-
motion analysis.
• Two set of coordinate axes will be used to do this
• The x, y coordinate system is fixed and will be used to measure the absolute
positions, velocities and accelerations of two points A and B on the body
• The origin of the x’, y’ coordinate system will be attached to the selected “base
point” A, which generally has a known motion
• The axes of this coordinate system are not fixed to the body, rather they will be
allowed to translate with respect to the fixed frame

POSITION:
• The position vector rA (above figure) specifies the location of the base point A,
whereas the relative position vector rB/A locates point B on the body with respect to
A.
• By vector addition, the position of B is then:
rB = rA + rB/A

DISPLACEMENT:
• During an instant dt, points A and B undergo displacements drA and drB as shown.
• The entire body first translates by an amount drA so that A, the base point, moves to its final
position and B moves to B’.
• The body is then rotated by an amount dθ about A, so that B’ undergoes a relative displacement
drB/A and thus moves to its final position B.
• Since the body is rigid, rB/A has a fixed magnitude, and thus drB/A accounts only for a change in
the direction of rB/A
• Due to rotation about A the magnitude of the relative displacement is drB/A = rB/A dθ, and the
displacement of B is thus:
drB = drA + drB/A

VELOCITY:
• To determine the relationship between the velocities of points A and B, it is necessary
to take the time derivative of position equation, i.e.
(drB / dt) = (drA /dt) + (drB/A /dt)
• Since (drB / dt) = vB, and (drA /dt) = vA,
• The magnitude of the third term is (rB/A dθ /dt) = rB/A θ0 = rB/A ω
• We therefore have:
vB = vA + vB/A
vB = vA + ω x rB/A

Examples:
16.6, 16.7, 16.8, 16.9
F16.7. F16.9, F16.10
Practice Problems:
16.62, 16.68, 16.70, 16.73, 16.81

EXAMPLE 16-6
The link shown is guided by two blocks A and B which move in the
fixed slots. If the velocity of A is 2 m/s downward, determine the
velocity of B at the instant θ = 45º

EXAMPLE 16-8
The collar C is moving downward with a velocity of 2 m/s. Determine
the angular velocities of CB and AB at this instant.

EXAMPLE 16-9
The bar AB shown has a clockwise angular velocity of 30 rad/s when
θ=60º. Determine the angular velocities of member BC and the wheel
at this instant.

PROBLEM 16-73
If link AB has an angular velocity of ωAB = 4 rad/s at the instant shown,
determine the velocity of the slider block E at this instant. Also, identify
the type of motion of each of the four links.

PROBLEM 16-81
If the slider block A is moving to the right at vA= 8 ft/s, determine the
velocity of blocks B and C at the instant shown. Member CD is pin
connected to member ADB.

INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY
• The velocity of any point B located on a rigid body can be obtained in a simple way if
one chooses the base point A to be a point that has zero velocity at the instant
considered.
• In this case vA = 0, and therefore the velocity equation, vB = vA + ω x rB/A, becomes
vB = ω x rB/A
• For a body having general plane motion, point A so chosen is called the instantaneous
center of zero velocity (IC) and it lies on the instantaneous axis of zero velocity.
• This axis is always perpendicular to the plane used to represent the motion, and the
intersection of the axis with this plane defines the location of the IC.

• Since point A is coincident with the IC, then vB = ω x rB/IC and so point B moves
momentarily about the IC in a circular path; in other words, the body appears to rotate
about the instantaneous axis.
• The magnitude of vB is simply vB = ωrB/IC, where ω is the angular velocity of the
body.
• Due to circular motion, the direction of vB must always be perpendicular to rB/IC.
• When a body is subjected to general plane motion, the point determines as the
instantaneous center of zero velocity for the body can only be used for an instant of
time.
• Since the body changes its position from one instant to the next, then for each position
of the body a unique instantaneous center must be determined.

LOCATION OF INSTANTANEOUS CENTER (IC)
• If the location of the IC is unknown, it may be determined by using the fact that the
relative position vector extending from the IC to a point is always perpendicular to the
velocity of the point. Folowing possibilities exist:
1. Given the velocity vA of a point A on the body and the angular velocity ω of
the body:
• In this case, the IC is located along the line drawn perpendicular to vA at A, such that
the distance from A to the IC is rA/IC = vA / ω.
• Note that the IC lies up and to the right of A since vA must cause a clockwise angular
ω about the IC.

2. Given the lines of action of two non-parallel
velocities vA and vB
• Consider the body shown where the lines of
action of the velocities vA and vB are known
• From each of these lines construct at points A and
B line segments that are perpendicular to vA and
vB and therefore define the lines of action of rA/IC
and rB/IC respectively.
• Extending these perpendiculars to their point of
intersection locates the IC at the instant
considered.
• The magnitudes of rA/IC and rB/IC are generally
determined from the geometry of the body and
trigonometry.
• Furthermore, if the magnitude and sense of vA are
known, then the angular velocity of the body is
determined from ω = vA / rA/IC.
• Once computed, ω can then be used to determine
vB = ωrB/IC

3. Given the magnitude and direction of two parallel velocities vA and vB
• When the velocities of points A and B are parallel and have known magnitudes vA
and vB then the location of the IC is determined by proportional triangles.
• Examples are shown in figures (a) and (b). In both cases rA/IC = vA/ω and rB/IC =
vB/ω
• If d is a known distance between points A and B, then from figure (a) rA/IC + rB/IC
= d, and from figure (b) rB/IC – rA/IC = d
• As a special case, note that if the body is translating, vA = vB and the IC would be
located at infinity, in which case rA/IC = rB/IC --- ∞.
• This being the case, ω = vA/∞ = vB/∞ = 0

Examples:
16.11, 16.12
F16.13, F16.18
Practice Problems:
16.89, 16.93, 16-91, 16.101,
16.102, 16.104, 16-105

EXAMPLE 16-11
Block D moves with a speed of 3 m/s. Determine the angular
velocities of links BD and AD, at the instant shown.

EXAMPLE 16-13
The crankshaft AB turns with a clockwise angular velocity of 10 rad/s.
Determine the velocity of the piston at the instant shown.

PROBLEM 16-91
If the center O of the gear is given a velocity of vO = 10 m/s, determine
the velocity of the slider block B at the instant shown.

PROBLEM 16-101
If rod AB is rotating with an angular velocity ωAB = 3 rad/s, determine
the angular velocity of rod CD at the instant shown.

PROBLEM 16-102
The mechanism used in a marine engine consists of a crank AB and two
connecting rods BC and BD. Determine the velocity of the piston at C
the instant the crank is in the position shown and has an angular
velocity of 5 rad/s.

PROBLEM 16-105
If crank AB is rotating with an angular velocity of ωAB = 6 rad/s,
determine the velocity of the center ) of the gear at the instant shown.

RELATIVE MOTION ANALYSIS: ACCELERATION
• The acceleration of two points on a rigid body subjected to general plane motion may
be determined by differentiating the velocity equation vB = vA + vB/A
(dvB / dt) = (dvA /dt) + (dvB/A /dt)
• Since (dvB / dt) = aB, (dvA /dt) = aA, and (dvB/A /dt) = aB/A
• We therefore have:
aB = aA + aB/A
aB = aA + (aB/A)t + (aB/A)n
aB = aA + (α x rB/A) – ω2 rB/A

RELATIVE MOTION ANALYSIS: ACCELERATION
Examples:
16.13, 16.17, 16.18
F16.19, F16.23, F16.24
Practice Problems:
16.113, 16.116, 16.120,
16.121, 16.125

EXAMPLE 16-13
The rod AB is confined to move along the inclined planes at A and B.
If point A has an acceleration of 3 m/s2 and a velocity of 2 m/s, both
directed down the plane at the instant the rod becomes horizontal,
determine the angular acceleration of the rod at this instant.

EXAMPLE 16-17
The collar C is moving downward with an acceleration of 1 m/s2. At
this instant it has a speed of 2 m/s which gives links CB and AB an
angular velocity ωAB= ωCB= 10 rad/s. Determine the angular
accelerations of CB and AB at this instant.

EXAMPLE 16-18
The crankshaft AB of an engine turns with a clockwise angular
acceleration of 20 rad/s2. Determine the acceleration of the piston at
the instant AB is in the position shown. At this instant ωAB=10 rad/s
and ωBC=2.43 rad/s.

PROBLEM 16-120
The slider block moves with a velocity of vB = 5 ft/s and an acceleration of
aB = 3 ft/s2. Determine the acceleration of A at the instant shown.

PROBLEM 16-121
Crank AB rotates with an angular velocity of ωAB = 6 rad/s and an
angular acceleration of αAB = 2 rad/s2. Determine the acceleration of C
and the angular acceleration of BC at the instant shown.

Chapter_16_Planar_Kinematics_of_Rigid_Bo.ppt

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