2. 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
3. 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
4. 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
5. 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.
6. 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
7. 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
8. 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
9. 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
10. 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.
11. ROTATION ABOUT A FIXED AXIS
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.
12. ROTATION ABOUT A FIXED AXIS
• 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)
13. ROTATION ABOUT A FIXED AXIS
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.
14. ROTATION ABOUT A FIXED AXIS
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
15. ROTATION ABOUT A FIXED AXIS
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
16. ROTATION ABOUT A FIXED AXIS
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
17. ROTATION ABOUT A FIXED AXIS
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
18. 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
19. 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.
20. 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.
21. 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
22. 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.
23. 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.
24. 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
25. RELATIVE MOTION ANALYSIS: VELOCITY
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
26. RELATIVE MOTION ANALYSIS: VELOCITY
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
27. RELATIVE MOTION ANALYSIS: VELOCITY
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
29. 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
30. 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.
31. 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.
32. 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.
33. 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.
34. 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.
35. INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY
• 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.
36. 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.
37. LOCATION OF INSTANTANEOUS CENTER (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
38. LOCATION OF INSTANTANEOUS CENTER (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
39. INSTANTANEOUS CENTER (IC) OF ZERO VELOCITY
Examples:
16.11, 16.12
Fundamental Problems:
F16.13, F16.18
Practice Problems:
16.89, 16.93, 16-91, 16.101,
16.102, 16.104, 16-105
40. 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.
41. 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.
42. 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.
43. 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.
44. 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.
45. 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.
46. 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
48. 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.
49. 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.
50. 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.
51. 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.
52. 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.
