UNIT III. PLANE MOTION OF KINEMATICS OF RIGID BODIES
Overview
Within this unit we will analyze the planar kinematics of a rigid body. This study is
important for designing gears, cams, and mechanisms that are used in many mechanical
works. When the kinematics is completely understood, then we may apply the movement
equations, which link the forces on the body to the movement of the body.
Learning Objectives
At the end of the unit, I am able to:
1. Classify the various forms of rigid-body planar motion;
2. Analyze angular motion about a fixed axis and rigid-body translation; and
3. Test planar motion using measurement of absolute motion.
Topics
3.1.
3.2.
3.3.
3.4.
3.5.
Translation Motion
Plane Motion
Relative Motion Analysis: Velocity
Instantaneous Center of Zero Velocity
Absolute and Relative Reaction
1
Pre-Test
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. The angular velocity of the disk is defined by 𝜔 = (5𝑡 2 + 2) rad/s, where t is in seconds.
Assess the velocity and acceleration magnitudes of point A on the disk when t = 0.5 s.
2
2. The disk is originally rotating at 𝜔𝑂 = 12 rad/s. If it is subjected to a constant angular
acceleration of 𝛼 = 20 rad/s2, evaluate the velocity magnitudes and the acceleration n and t
components of point A at the moment t = 2s.
3
3. The disk is originally rotating at 𝜔𝑂 = 12 rad/s. If it is subjected to a constant angular
acceleration of 𝛼 = 20 rad/s2, evaluate the velocity magnitudes and the acceleration n and t
components of point A at the moment t = 2s.
4
Lesson Proper
PLANE MOTION OF KINEMATICS OF RIGID BODIES
Kinematics Planar Motion Equations
If all parts of the body move in parallel planes a rigid body performs plane motion.
For simplicity, we generally regard the motion plane as the plane that comprises the center
of mass, and we view the body as a thin slab whose motion is limited to the plane of the slab.
This idealization appropriately describes a very large category of rigid body movements
found in engineering
The plane motion of a rigid body may be divided into several categories, as
represented in Fig. 2-1. We note that in both of the two translation situations, the movement
of the body is entirely determined by the motion of any point in the body, as all points have
the same movement.
Translation is characterized as any movement in which every line in the body is
always parallel to its origin. In translation, no rotation of line in the body occurs. In
rectilinear translation, all body points travel in straight line. The curvilinear motion, by
the word itself, it moves in a curved direction.
Rotation about a fixed axis, it is the angular rotation of the axis. It says that every
object in a rigid body moves around the axis, circular in motion and all lines normal to the
axis of rotation rotate the same time around the same angle.
General plane motion is the combination of both translation and rotation.
5
Figure 2-1
Rotation
Figure 2-2 shows a rigid body that rotates in the direction of the image, as it
undergoes motion. The angular positions of any two lines 1 and 2 attached to the body are
specified by 𝜃1 and 𝜃2 measured from any convenient fixed reference direction. Because the
angle 𝛽 is constant, the relation 𝜃2 = 𝜃1 + 𝛽 upon differentiation with respect to time gives
𝜃2̇ = 𝜃1̇ or, during a finite interval, ∆𝜃2 = ∆𝜃1. Therefore, all lines on a rigid body have the
same angular displacement in its motion axis, the same angular velocity and the same angular
acceleration.
Figure 2-2
Note that a line 's angular motion depends only on its angular location with respect
to some arbitrary fixed reference, and on the displacement time derivatives. Angular motion
does not involve a fixed axis, natural to the motion plane on which the line and the body are
rotating.
6
Angular-Motion Relations
The angular velocity 𝜔 is the first derivative of the angular position 𝜃 and the angular
acceleration 𝛼 is the second derivative. These definitions give
Eqs. 2-1
For rotation with constant angular acceleration, the integrals of Eqs. 2-1 becomes
𝜔 = 𝜔𝑜 + 𝛼𝑡
𝜔2 = 𝜔𝑜 2 + 2𝛼(𝜃 − 𝜃𝑜 )
𝜃 = 𝜃𝑜 + 𝜔𝑜 𝑡 +
1 2
𝛼𝑡
2
Here, 𝜃𝑜 and 𝜔𝑜 are the values of the angular coordinate position and the angular
velocity at t = 0, respectively, and t is the length of the movement considered.
7
Rotation about a Fixed Axis
If a rigid body rotates around a fixed axis, all other points travel in concentric circles
around the axis. So, for the rigid body in Fig. 2-3 rotating around the normal fixed axis to the
plane of the figure through O, any point such as A is moving in the r-radius circle.
Figure 2-3
𝑣 = 𝑟𝜔
𝑎𝑛 = 𝑟𝜔2 =
𝑣2
= 𝑣𝜔
𝑟
𝑎𝑡 = 𝑟𝛼
Eqs. 2-2
Alternatively, those quantities can be expressed using the vector notation crossproduct relationship. The angular velocity can be expressed as 𝜔 as shown in Fig 2-4a, normal
to the rotational plane with a meaning regulated by the right-hand rule. From the definition
of the vector-cross product, we see that the vector v is obtained by crossing somewhere in r.
This cross product gives the right direction and magnitude for v and we write
𝑣= 𝜔𝑥𝑟
We must retain the order of the vectors to be crossed. The reverse order gives
𝑟 𝑥 𝜔 = −𝑣
8
Figure 2-4
The acceleration of point A is obtained by differentiating the cross-product
expression for v, which gives
Here 𝛼 = 𝜔̇ , stands for the angular acceleration of the body. Thus, the vector
equivalents to Eqs. 2-2 are
𝑣= 𝜔 × 𝑟
𝑎𝑛 = 𝜔 × (𝜔 × 𝑟)
𝑎𝑡 = 𝛼 × 𝑟
and are shown figure 2-4b.
Examples:
1. A flywheel rotating freely at 1800 𝑟𝑒𝑣 / 𝑚𝑖𝑛 in the clockwise direction is subjected to a
counterclockwise torque variable, which is first applied at t= 0. A counterclockwise angular
𝑟𝑎𝑑
acceleration is generated by the torque, 𝛼 = 4 𝑡 2 , where t is the time in seconds during
𝑠
which the torque is applied. Solve for (a) the time needed by the flywheel to reduce its angular
velocity to 900 𝑟𝑒𝑣 / 𝑚𝑖𝑛 in the clockwise direction, (b) the time needed by the flywheel to
reverse its rotational direction and (c) the overall number of revolutions, counterclockwise
plus clockwise, turned by the flywheel within the first 14 seconds of torque application.
Solution:
The counterclockwise path is arbitrarily acknowledged as positive.
We will integrate 𝛼 since 𝛼 is a known function of the time to obtain angular velocity.
With the initial angular velocity of -1800(2𝜋)/60 = -60𝜋 rad/s, we have
[𝑑𝜔 = 𝛼𝑑𝑡]
𝜔
𝑡
𝜔 = −60𝜋 + 2𝑡 2
∫−60𝜋 𝑑𝜔 = ∫0 4𝑡 𝑑𝑡
9
Substituting the clockwise angular speed of 900 rev/min or 𝜔 = −
𝑟𝑎𝑑
−30𝜋 𝑠
900(2𝜋)
60
=
gives
−30𝜋 = −60𝜋 + 2𝑡 2
𝑡 2 = 15𝜋
𝒕 = 𝟔. 𝟖𝟔 𝒔
The flywheel changes direction when its angular velocity is momentarily zero. Thus,
0 = −60𝜋 + 2𝑡 2
𝑡 2 = 30𝜋
𝒕 = 𝟗. 𝟕𝟏𝒔
The overall number of turns the wheel makes in 14 seconds is the number of
counterclockwise makes N1 in the first 9.71 seconds, plus the number of counterclockwise
turns N2 in the rest of the time.. Integrating the expression for 𝜔 in terms of 𝑡 gives us the
angular displacement in radians. Thus, for the first interval
𝜃1
[𝑑𝜃 = 𝜔𝑑𝑡]
0
𝜃1 = [−60𝜋𝑡 +
or 𝑁1 =
9.71
∫ 𝑑𝜃 = ∫
1220
2𝜋
(−60𝜋 + 2𝑡 2 ) 𝑑𝑡
0
2 3 9.71
𝑡 ]
= −1220 𝑟𝑎𝑑
3 0
= 194.2 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒.
For the second interval
𝜃2
14
∫ 𝑑𝜃 = ∫
0
(−60𝜋 + 2𝑡 2 )𝑑𝑡
9.71
𝜃2 = [−60𝜋𝑡 +
or 𝑁2 =
410
2𝜋
2 3 14
𝑡 ]
= 410 𝑟𝑎𝑑
3 9.71
= 65.3 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠 𝑐𝑜𝑢𝑛𝑡𝑒𝑟𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒.
The cumulative number of revolutions turned over in the 14 seconds is therefore
𝑁 = 𝑁1 + 𝑁2 = 194.2 + 65.3 = 𝟐𝟓𝟗 𝒓𝒆𝒗
2. The hoist motor pinion A drives gear B, which is fixed to the hoisting drum. The load L is
raised from its place of rest and with constant acceleration, it acquires an upward speed of 3
ft / sec in a vertical rise of 4 feet. Calculate (a) the acceleration of point C on the cable in
contact with the drum and (b) the angular velocity and angular acceleration of the pinion A.
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Solution:
The acceleration of the load L will generally be the same as the tangential velocity v
and the tangential acceleration at C. The n- and t-components of the acceleration C become
with continuous acceleration for the rectilinear motion of L.
[𝑣 2 = 2𝑎𝑠]
[𝑎𝑛 =
𝑣2
]
𝑟
[𝑎 = √𝑎𝑛 2 + 𝑎𝑡 2 ]
𝑎 = 𝑎𝑡 =
𝑎𝑛 =
𝑣2
32
=
= 1.125 𝑓𝑡/𝑠𝑒𝑐 2
2𝑠
2(4)
32
= 4.5 𝑓𝑡/𝑠𝑒𝑐 2
24
(12)
𝑎𝑐 = √(4.5)2 + (1.125)2 = 𝟒. 𝟔𝟒 𝒇𝒕/𝒔𝒆𝒄𝟐
The angular motion of gear A is determined by the velocity v1 and the tangential
acceleration a1 by their common point of contact from the angular motion of gear B. Firstly,
gear B angular movement is determined from point C motion on the attached drum. Thus,
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[𝑣 = 𝑟𝜔]
[𝑎𝑡 = 𝑟𝛼]
𝜔𝐵 =
𝑣
𝑟
𝛼𝐵 =
3
𝑟𝑎𝑑
= 1.5
24
𝑠𝑒𝑐
(12)
𝑎𝑡 1.125
=
= 0.562 𝑟𝑎𝑑/𝑠𝑒𝑐 2
24
𝑟
(12)
Then from 𝑣1 = 𝑟𝐴 𝜔𝐴 = 𝑟𝐵 𝜔𝐵 and 𝛼1 = 𝑟𝐴 𝛼𝐴 = 𝑟𝐵 𝛼𝐵 , we have
18
𝑟𝐵
𝜔𝐴 =
𝜔 = 12 (1.5) = 𝟒. 𝟓 𝒓𝒂𝒅/ 𝐬𝐞𝐜 𝑪𝑾
6
𝑟𝐴 𝐵
12
18
𝑟𝐵
𝛼𝐴 =
𝛼 = 12 (0.562) = 𝟏. 𝟔𝟖𝟖 𝒓𝒂𝒅/ 𝒔𝒆𝒄𝟐 𝑪𝑾
6
𝑟𝐴 𝐵
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Absolute and Relative Velocity in Plane Motion
The velocity and acceleration of a point P undergoing rectilinear motion can be
correlated with the angular velocity and angular acceleration of a line embedded within a
body using the method below.
Position Coordinate Equation
Use the position coordinate s to determine the location of point P on the body which
is measured according from its fixed origin and is directed along the straight-line path of
point P motion.
Measure the angular position θ of a line lying in the body from a fixed reference axis.
From the body dimensions, relate 𝑠 𝑡𝑜 𝜃, 𝑠 = 𝑓(𝜃) using geometry and/or
trigonometry.
Time Derivatives
Take the first derivative of s = f (𝜃) with respect to time to get a relation between
𝑣 𝑎𝑛𝑑 𝜔.
Take the second time derivative to get a relation between a and a.
For each case the chain rule of calculus must be used when the time derivatives of the
coordinate position equation are taken.
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Examples:
1. The end of the rod R shown in the figure maintains contact with the cam through a spring.
Solve the speed and acceleration of the rod when the cam is in an arbitrary position 𝜃, if the
cam rotates around an axis at point O with an angular acceleration of α and angular velocity
𝜔.
Solution:
In order to relate the rotational motion of the line segment OA on the cam to the
rectilinear translation of the rod, coordinates θ and x are chosen. These coordinates are
measured from the fixed-point O and can be related to each other using trigonometry. Since
OC = CB = r cos 𝜃, then
𝑥 = 2𝑟𝑐𝑜𝑠𝜃
With the use of chain rule of calculus, we have
𝑑𝑥
𝑑𝜃
= −2𝑟(𝑠𝑖𝑛𝜃)
𝑑𝑡
𝑑𝑡
𝒗 = −𝟐𝒓𝝎𝒔𝒊𝒏𝜽
𝑑𝑣
𝑑𝑤
𝑑𝜃
= −2𝑟 ( ) 𝑠𝑖𝑛𝜃 − 2𝑟𝜔(𝑐𝑜𝑠𝜃)
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝒂 = −𝟐𝒓(𝜶𝒔𝒊𝒏𝜽 + 𝝎𝟐 𝒄𝒐𝒔𝜽)
Note: The negative signs mean that v and a are in contrary to the positive x direction.
If you imagine the motion, it seems normal.
2. At a given scenario, the cylinder of radius r, presented in the figure, has an angular velocity
ω and angular acceleration α. Determine the velocity and acceleration of its center G if the
cylinder rolls without slipping.
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Solution:
The cylinder is undergoing general plane motion, as it translates and rotates
simultaneously. By inspection, point G moves in a straight line to the left, from G to G’, as the
cylinder rolls. Consequently, its new position G’ will be specified by the horizontal position
coordinate 𝑠𝐺 , which is measured from G to G’. Also, as the cylinder rolls (without slipping),
the arc length A’B on the rim which was in contact with the ground from A to B, is equivalent
to 𝑠𝐺 . Consequently, the motion requires the radial line GA to rotate 𝜃 to the position G’A’.
Since the arc A’B = 𝑟𝜃, then G travels a distance 𝑠𝐺 = 𝑟𝜃.
𝑑𝜃
,
𝑑𝑡
Taking successive time derivatives of this equation, realizing that r is constant,𝜔 =
𝑑𝜔
and 𝛼 = 𝑑𝑡 , gives the necessary relationships:
𝑠𝐺 = 𝑟𝜃
𝒗𝑮 = 𝒓𝝎
𝒂𝑮 = 𝒓𝜶
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Relative Motion Analysis: Velocity
The x, y system of coordinates is fixed and measures the absolute position of two
points A and B on the body, represented here as a bar, Fig. 2-5. The origin of the x’, y'
coordinate system is attached to the "base point" A selected, which generally has a known
motion. The axes of this coordinate system move with respect to the fixed point, but do not
rotate with the bar.
Figure 2-5
The relative velocity equation can be implemented either using Cartesian vector
analysis, or directly writing the equations of the x and y scalar components. The following
method is recommended for use.
Vector Analysis
Kinematics Diagram
Set the fixed x, y coordinates directions and draw a body kinematic diagram. Indicate
the velocities 𝑣𝐴, 𝑣𝐵 of points A and B, and angular velocity 𝜔, and the relative position
vector r𝐵/𝐴
If the magnitudes of 𝑣𝐴, 𝑣𝐵 , or 𝜔 are unknown, the sense of direction of these
vectors can be assumed.
Velocity Equation
To apply 𝑣𝐵 = 𝑣𝐴 + 𝜔 𝑥 r𝐵/𝐴 , Express the Cartesian vector form of the
vectors and replace them with the equation. Assess the cross product and then
equate the respective i and j components to get two scalar equations.
If the solution provides an unknown magnitude with a negative response, the
direction of the vector is contrary to the direction shown in the kinematic diagram.
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Scalar Analysis
Kinematics Diagram
In scalar form, if the velocity equation is to be used, then the magnitude and
direction of the relative velocity 𝑣𝐵/𝐴 must be established. Draw a kinematic diagram
which shows the relative motion. Since the body is considered to be “pinned”
momentarily at the base point A, the magnitude of 𝑣𝐵/𝐴 is 𝑣𝐵/𝐴 = 𝜔𝑟𝐵/𝐴 . The sense
of direction of 𝑣𝐵/𝐴 is always perpendicular to 𝑟𝐵/𝐴 in accordance with the rotational
motion 𝜔 of the body.
Velocity Equation
Write the equation in symbolic form, 𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴 and underneath each
of the terms represent the vectors graphically by showing their magnitudes and
directions. The scalar equations of these vectors are calculated from the x and y
components.
Note: The notation 𝑣𝐵 = 𝑣𝐴 + 𝑣𝐵/𝐴(𝑝𝑖𝑛) may be helpful in recalling that “A”
is pinned.
Examples:
1. The link listed in Fig. 2-6a is driven by two blocks at A and B moving in the fixed slots. If
the speed of A is down 2 m / s, determine the speed of B at the instant 𝜃 = 45°.
Figure 2-6a
Solution:
Vector Analysis
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Kinematic Diagram. Since points A and B are restricted to move along the fixed slots
and 𝑣𝐴 is directed downward, then velocity 𝑣𝐵 must be directed horizontally to the right, Fig.
2-6b. This motion causes the link to rotate in the counterclockwise direction; that is, the
angular velocity ω is directed outward by the right-hand rule, perpendicular to the motion
plane.
Figure 2-6b
Velocity Equation. Expressing every single vector in Fig. 2-6b as to its components i,
j, k and applying the equation 𝑣𝐵 = 𝑣𝐴 + 𝜔 𝑥 r𝐵/𝐴 to A, the base point, and B, we have
𝑣𝐵 = 𝑣𝐴 + 𝜔 𝑥 r𝐵/𝐴
𝑣𝐵 𝒊 = −2𝒋 + [𝜔𝒌 𝑥 (0.2 sin 45°𝒊 − 0.2 cos 45°𝒋)]
𝑣𝐵 𝒊 = −2𝒋 + 0.2𝜔 sin 45°𝒋 + 0.2𝜔 cos 45°𝒊
Equating the i and j components gives
𝑣𝐵 = 0.2𝜔 cos 45°
0 = −2 + 0.2𝜔 sin 45°
Thus,
𝜔 = 14.1 𝑟𝑎𝑑/𝑠
𝒗𝑩 = 𝟐 𝒎/𝒔
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Scalar Analysis
The kinematic diagram of the relative “circular motion: which produces 𝑣𝐵 is shown
𝐴
in Figure 2-6c. Here 𝑣𝐵 = 𝜔(0.2 𝑚)
𝐴
Figure 2-6c
Thus,
The solution produces the above results.
It should be emphasized that these results are valid only at the instant 𝜃 = 45°
2. In Fig 2-7a, collar C travels downwards at a pace of 2 m/s. Determine the angular speed of
CB at this moment.
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Fig. 2-7a
Solution:
Vector Analysis
Kinematic Diagram. C's downward motion causes B to pass through a curved
direction to the right. CB and AB also rotate in counterclockwise direction.
Velocity Equation. Link CB (general plane motion): See Fig. 2-7b
Fig. 2-7b
𝑣𝐵 = 𝑣𝐶 + 𝜔𝐶𝐵 × 𝑟𝐵/𝐶
𝑣𝐵 𝑖 = −2𝑗 + 𝜔𝐶𝐵 × (0.2𝑖 − 0.2𝑗)
𝑣𝐵 𝑖 = −2𝑗 + 0.2𝜔𝐶𝐵 𝑗 + 0.2𝜔𝐶𝐵 𝑖
𝑣𝐵 = 0.2𝜔𝐶𝐵
(1)
0 = −2 + 0.2𝜔𝐶𝐵
(2)
𝜔𝐶𝐵 = 10 rad/s
𝒗𝑩 = 2 m/s
Scalar Analysis
The scalar component equations of vB = vC + vB/C can be obtained directly.
The kinematic diagram in Fig. 2-7c shows the relative “circular” motion which produces
vB/C. We have
19
Fig. 2-7c
vB = vC + vB/C
Resolving these vectors in the x and y directions yields
𝑣𝐵 = 0 + 𝜔𝐶𝐵 (0.2√2𝑐𝑜𝑠45°)
0 = −2 + 𝜔𝐶𝐵 (0.2√2𝑠𝑖𝑛45°
which is the same as Eqs. 1 and 2.
20
Instantaneous Center of Zero Velocity
The velocity of any point B situated on a rigid body can be obtained rather simply by
choosing the base point A as a point defined at the moment as having zero velocity. In this
case, vA=0, and hence, the equation of speed, 𝑣𝐵 = 𝑣𝐴 + 𝜔 𝑥 r𝐵/𝐴 , becomes 𝑣𝐵 = 𝜔 𝑥 r𝐵/𝐴 .
For a body with general plane motion, point A so chosen is called zero velocity instantaneous
center (IC), and it lies on the zero-velocity instantaneous axis.
The IC for bicycle wheel in Fig 2-8, for example is at ground contact point. There the
spokes are that very obvious, while they are blurred at the top of the wheel. If one imagines
that the wheel is momentarily pinned at this point, the velocities of various points can be
found using 𝑣 = 𝜔𝑟. Here the radial distances shown in the photo, Fig. 6-8, must be
determined from the geometry of the wheel.
(© R.C. Hibbeler)
Figure 2-8
As shown on the kinematic diagram in Fig. 2-9, the body is imagined as “extended and
pinned” at the IC so that, at the instant considered, with its angular velocity 𝜔 it rotates
around this pin.
Using the equation 𝑣 = 𝜔𝑟 for each of the arbitrary points A, B, and C on the body,
the magnitude of velocity can be found, where r is the radial distance from the IC to each
point.
The action line of each vector v is perpendicular to its respective radial line r, and the
velocity has a sense of direction that appears to shift the point in a way compatible with the
radial line 's angular rotation V, Fig. 2-9.
21
Figure 2-9
Examples:
1. Block D shown in Fig. 2-10a moves with a velocity of 3 m/s. Determine the angular
velocities of connections BD and AB, at the situation show.
Figure 2-10a
Solution:
As D moves to the right, AB rotates in the clockwise direction over point A. Hence, 𝑣𝐵
is directed perpendicular to AB. The instantaneous center of zero velocity for BD is located at
the intersection of the line segments drawn perpendicular to 𝑣𝐵 and 𝑣𝐷 , Fig. 2-10b. From the
geometry.
Figure 2-10b
22
𝑟𝐵/𝐼𝐶 = 0.4 tan 45° 𝑚 = 0.4𝑚
𝑟𝐷/𝐼𝐶 =
0.4 𝑚
= 0.5657𝑚
cos 45°
Since the magnitude of 𝑣𝐷 is known, the angular velocity of link BD is
𝜔𝐵𝐷 =
𝑣𝐷
𝑟𝐷/𝐼𝐶
=
3 𝑚/𝑠
= 𝟓. 𝟑𝟎 𝒓𝒂𝒅/𝒔 (𝑐𝑐𝑤)
0.5657 𝑚
The velocity of B is therefore
𝑣𝐵 = 𝜔𝐵𝐷 (𝑟 𝐵 ) = 5.30
𝐼𝐶
𝑟𝑎𝑑
(0.4𝑚) = 2.12 𝑚/𝑠
𝑠
From Fig. 2-10c, the angular velocity of AB is
𝜔𝐴𝐵 =
𝑣𝐵
2.12𝑚/𝑠
=
= 𝟓. 𝟑𝟎 𝒓𝒂𝒅/𝒔 (𝑐𝑤)
𝑟𝐵/𝐴
0.4 𝑚
2. The cylinder listed in Fig. 2-11a rolls between two moving plates E and D, without slipping.
Determine cylinder angular velocity and the velocity of its center C..
Figure 2-11a
Solution:
As no slipping occurs, the contact points A and B on the cylinder have the same speeds
as the plates E and D , respectively. In addition, the speeds 𝑣𝐴 and 𝑣𝐵 are parallel, so that by
the proportionality of right triangles the IC is located at a point on line AB, Fig. 2-11b.
Assuming this point to be a distance x from B, we have
23
Figure 2-11b
𝑣𝐵 = 𝜔𝑋
𝑚
= 𝜔𝑋
𝑠
𝑚
0.25 = 𝜔(0.25𝑚 − 𝑋)
𝑠
0.4
𝑣𝐴 = 𝜔(0.25 𝑚 − 𝑋)
Dividing one equation into the other eliminates 𝜔 and yields
0.4(0.25 − 𝑋) = 0.25𝑋
𝑋=
0.1
= 0.1538 𝑚
0.65
Therefore, the angular velocity of the cylinder is
𝜔=
𝑣𝐵
0.4𝑚/𝑠
𝒓𝒂𝒅
=
= 𝟐. 𝟔𝟎
(𝑐𝑤)
𝑋
0.1538 𝑚
𝒔
The velocity of point C is therefore
𝑣𝐶 = 𝜔𝑟 𝐶 = 2.60
𝐼𝐶
𝑣𝐶 = 𝟎. 𝟎𝟕𝟓𝟎
𝑟𝑎𝑑
(0.1538 𝑚 − 0.125 𝑚)
𝑠
𝒎
𝒔
24
Absolute and Relative Acceleration
Velocity Analysis
Kinematic Diagram
Set the fixed x, y coordinates directions and draw the body's kinematic
diagram. Indicate on it 𝑎𝐴 , 𝑎𝐵 , 𝜔, 𝛼, 𝑎𝑛𝑑 𝑟𝐵/𝐴 .
If points A and B travel along curved paths, their accelerations in terms of
their tangential and normal components should be indicated, i.e., 𝑎𝐴 = (𝑎𝐴 )𝑡 +
(𝑎𝐴 )𝑛 and 𝑎𝐵 = (𝑎𝐵 )𝑡 + (𝑎𝐵 )𝑛 .
Acceleration Equation
To apply 𝑎𝐵 = 𝑎𝐴 + 𝛼 𝑥 𝑟𝐵/𝐴 − 𝜔2 𝑟𝐵/𝐴 , express the vectors in Cartesian
vector form and substitute them into the equation. Evaluate the cross product and
then equate the respective i and j components to obtain two scalar equations.
𝑎𝐵 = acceleration of point 𝐵
𝑎𝐴 = acceleration of the base point 𝐴
𝛼 = angular acceleration of the body
𝜔 = angular velocity of the body
𝑟𝐵/𝐴 = position vector directed from 𝐴 to 𝐵
If the solution gives a negative answer for an unknown magnitude, it means
that the vector's sense of direction is contrary to that shown on the kinematic
diagram.
Scalar Analysis
Kinematic Diagram
If the acceleration equation is applied in scalar form, then the magnitudes and
directions of the relative-acceleration components (𝑎𝐵/𝐴 )𝑡 and (𝑎𝐵/𝐴 )𝑛 must be
established. To do this, draw a kinematic diagram such as shown in Fig. 2-12. Since
the body is considered to be momentarily “pinned” at the base point A, the magnitudes
of these components are (𝑎𝐵/𝐴 )𝑡 = 𝛼𝑟𝐵/𝐴 and (𝑎𝐵/𝐴 )𝑛 = 𝜔2 𝑟𝐵/𝐴 . Their sense of
direction is established from the diagram such that (𝑎𝐵/𝐴 )𝑡 acts perpendicular to 𝑟𝐵/𝐴 ,
in accordance with the rotational motion 𝛼 of the body, and (𝑎𝐵/𝐴 )𝑛 is directed from
B toward A.
Figure 2-12
25
Acceleration Equation
Represent the vectors in 𝑎𝐵 = 𝑎𝐴 + (𝑎𝐵/𝐴 )𝑡 + (𝑎𝐵/𝐴 )𝑛 graphically by
showing their magnitudes and directions underneath each term. The scalar equations
of these vectors are calculated from the x and y components.
Consider a disk that rolls without slipping as shown in Figure 2-12a. As a
result, 𝑣𝐴 = 0 and so from the kinematic diagram in Figure 2-12b, the velocity of the
mass center G is
𝑣𝐺 = 𝑣𝐴 + 𝜔 × 𝑟𝐺/𝐴 = 0 + (−𝜔𝐤) × (𝑟𝐣)
(𝑒𝑞 2 − 3)
Figure 2-12a
Figure 2-12b
So that
𝑣𝐺 = 𝜔𝑟
Since G moves along a straight line, its acceleration in this case can be determined
from the time derivative of its velocity.
𝑑𝑣𝐺 𝑑𝜔
=
𝑟
𝑑𝑡
𝑑𝑡
𝑎𝐺 = 𝛼𝑟
(𝑒𝑞 2 − 4)
26
Examples:
1. The rod AB shown in Figure 2-13a is confined to travel along the tilted planes at A and B.
If point A has an acceleration of 3 m/s2 and a speed of 2 m / s, both of which are guided down
the plane at the moment when the rod is horizontal, evaluate the angular acceleration of the
rod at this moment.
Figure 2-13a
Solution:
Vector Analysis. We'll apply the equation of acceleration to points A and B on the
rod. For this to happen, the angular velocity of the rod must first be calculated. Show that is
is 𝜔 = 0.283 rad/s ↶ using either the velocity equation or the method of instantaneous
centers.
Kinematic Diagram. Since both points A and B travel along straight paths, they do
not have acceleration components normal to the paths. There are two unknowns in Figure 213b, namely, 𝑎𝐵 and 𝛼.
Figure 2-13b
Acceleration Equation
𝑎𝐵 = 𝑎𝐴 + 𝛼 × 𝑟𝐵/𝐴 − 𝜔2 𝑟𝐵/𝐴
𝑎𝐵 𝑐𝑜𝑠45°𝑖 + 𝑎𝐵 𝑠𝑖𝑛45°𝑗 = 3𝑐𝑜𝑠45°𝑖 − 3𝑠𝑖𝑛45°𝑗 + (𝛼𝑘) × (10𝑖) − (0.283)2 (10𝑖)
Carrying out the cross product and equating the i and j components yields
𝑎𝐵 𝑐𝑜𝑠45° = 3𝑐𝑜𝑠45° − (0.283)2 (10)
(1)
𝑎𝐵 𝑠𝑖𝑛45° = −3𝑠𝑖𝑛45° + 𝛼(10)
(2)
Solving, we have
27
𝑎𝐵 = 1.87𝑚/𝑠 2 ∡45°
𝜶 = 𝟎. 𝟑𝟒𝟒 𝒓𝒂𝒅/𝒔𝟐 ↶
2. The disk rolls without slipping and is shown in Figure 2-14a has an angular motion.
Determine point A's acceleration at this moment.
Figure 2-14a
Solution:
Vector Analysis
Kinematic Diagram. Since no slipping occurs, applying Eq. 2-4
𝑎𝐺 = 𝛼𝑟 = (4
𝑟𝑎𝑑
) (0.5𝑓𝑡) = 2𝑓𝑡/𝑠 2
𝑠2
Acceleration Equation
We will apply the acceleration equation to points G and A, Figure 2-14b
28
Figure 2-14b
𝑎𝐴 = 𝑎𝐺 + 𝛼 × 𝑟𝐴/𝐺 − 𝜔2 𝑟𝐴/𝐺
𝑎𝐴 = −2𝐢 + (4𝐤) × (−0.5𝐣) − (6)2 (−0.5𝐣)
𝑎𝐴 = 18𝒋 𝑓𝑡/𝑠2
Scalar Analysis
Using the result for 𝑎𝐺 = 2 ft/s2 determined above, and from the kinematic diagram,
showing the relative motion 𝑎𝐴/𝐺 , Figure 2-14c, we have
Figure 2-14c
𝑎𝐴 = 𝑎𝐺 + (𝑎𝐴⁄𝐺 )𝑥 + (𝑎𝐴⁄𝐺 )𝑦
Therefore,
𝑎𝐴 = √(0)2 + (18𝑓𝑡/𝑠 2 )2 = 18 ft/s
Reference
Hibbeler, R. C. (2016), Planar Kinematics of a Rigid Body, Engineering Mechanics Dynamics,
14th Edition
Meriam, J.L., Kraige, L.G., Plane Kinematics of Rigid Bodies, Engineering Mechanics Dynamics,
6th Edition
29
Assessing Learning
Activity III
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
1. If the block at C falls down at 4 𝑓𝑡/𝑠, evaluate the angular velocity of bar AB at the specified
instant.
30
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
2. The disk is originally rotating at𝜔𝑂 = 12 𝑟𝑎𝑑/𝑠. If it is subjected to a constant angular
acceleration of 𝛼 = 20 𝑟𝑎𝑑/𝑠2, calculate the velocity magnitude and the acceleration n and t
components of point B when the disk undergoes 2 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
31
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
3. Determine in this instant the angular acceleration and angular velocity of the link AB. Note:
The Guide's upward movement is in the negative y direction. At the instant 𝜃 = 50°, the
slotted guide is moving upward with an acceleration of 3 𝑚/𝑠2 and a velocity of 2 𝑚/𝑠.
32
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
4. At the given instant shown, 𝜃 = 50°, and rod AB faces a deceleration of 16𝑚/𝑠2 when the
velocity is 10 𝑚/𝑠. Evaluate the angular velocity at this instant and the angular acceleration
of the link CD.
33
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
5. Calculate the acceleration of the ladder's bottom A and the angular acceleration of the
ladder at this moment At a given instant, the top B of the ladder has an acceleration 𝑎𝐵 =
2 𝑓𝑡/𝑠2 and a velocity of 𝑣𝐵 = 4 𝑓𝑡/𝑠, both acting downward.
34
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
6. Determine the angular velocity of connection AB at the moment shown if block C moves
upward at 12 𝑖𝑛 / 𝑠.
35
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
7. Calculate the velocity of the gear rack C. The pinion gear A rolls on the fixed gear rack B
with an angular velocity 𝜔 = 4 𝑟𝑎𝑑/s.
36
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
8. At a given moment, the bottom A of the ladder has an acceleration 𝑎𝐴 = 4 𝑓𝑡/𝑠2 and
velocity 𝑣𝐴 = 6 𝑓𝑡/𝑠, they both behave to the left. Calculate the acceleration of the top of the
ladder, B, and the ladder’s angular acceleration at this same instant.
37
Name:
Course/Year/Section:
Date:
Direction: Solve the following problems.
9. The mechanism of the shaper is designed to give a slow cutting stroke and fast return to a
blade attached to the slider at C. Calculat the angular velocity of the CB link if the AB link
rotates at 4 𝑟𝑎𝑑 / 𝑠.
38