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Kinematics of Rigid Bodies

Unit IV

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Introduction• Involves study of the equations of motion, which relate the forces on

the body to the body’s motion.

• This study is important for the design of gears, cams, and mechanismsused for many mechanical and electrical operations.

• Depending on the motion of the object, the rigid body planar motion

is divided into 3 types :

TranslationThis type of motion occurs when a line in the body remains parallel

to its original orientation throughout the motion.

When the paths of motion for

any two points on the body are

parallel 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

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Rotation about a fixed axis:

– When a rigid body rotates about a fixedaxis, all the particles of the body, except

those which lie on the axis of rotation,

move along circular paths.

General plane motion:

When a body is subjected to general plane

motion, it undergoes a combination oftranslation and rotation. The translation occurs

within a reference plane, and the rotation

occurs about an axis perpendicular to the

reference plane.

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Translation

• Consider a rigid body subjected to

translation (rectilinear or curvilinear).

Position:

The locations of points A and B on the

body are defined with respect to fixed

x, y reference frame using position

vectors r A and r B.

The translating coordinate system (x`,y`) is fixed in the body and

has its origin at A, (base point). The position of B with respect to A

is denoted by the relative-position vector r B/A. By vector addition

rB = rA + rB/A

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• Velocity

A relation between the instantaneous velocities of A and B is obtained

by taking the time derivative of this equation, which yields

vB = vA + d (rB/A)/dtvA and vB denote absolute velocities since these vectors are measured

with respect to the x, y axes. In a rigid body, the magnitude of rB/A is

constant and as the body is translating the direction of rB/A is also

constant. Therefore,

vA = vB.• Acceleration

The time derivative of the velocity equation yields a similar relationshipbetween the instantaneous accelerations of A and B:

aA = aB

The above two equations indicate that all points in a rigid body subjectedto either rectilinear or curvilinear translation move with the same

velocity and acceleration.

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Rotation

• When a body rotates about a fixed axis, anypoint P located in the body travels along a

circular path. To study this motion it is firstnecessary to discuss the angular motion of thebody about the axis.

• Angular Motion

Since a point is without dimension, it cannot

have angular motion. Only lines or bodiesundergo angular motion. For the rigid bodyshown in Fig, the angular motion of point P isalong the circle formed by radial line r locatedwithin the shaded plane.

• Angular Position

At the given condition, the angular position of ris defined by the angle measured from a fixedreference line to r.

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• Angular Displacement

– The change in the angular position, which can be measured asa differential d is called the angular displacement.

– This vector has a magnitude of d measured in degrees,radians, or revolutions, where 1 rev = 2 rad. Since motion isabout a fixed axis, the direction of is always along this axis.

– The direction is determined by the right-hand rule; that is, thefingers of the right hand are curled with the sense of rotation,so that in this case the thumb, or points upward.

• Angular Velocity

– The time rate of change in the angular position is called theangular velocity, denoted by (omega). Since d occurs during

an instant of time dt, then, = d/dt. – This vector has a magnitude which is often measured in rad/s.

It is expressed here in scalar form since its direction is alsoalong the axis of rotation.

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• Angular Acceleration

– The angular acceleration, a, (alpha) measures the time rate of

change of the angular velocity. The magnitude of this vector is

a = d/dt = d

2

/dt

2

. – The line of action is the same of , but the direction depends

on increasing or decreasing value of , i.e. If is decreasing, it

is angular deceleration.

–

From the equations of and a, the following relationship canbe obtained, a d = d.

• Constant Angular Acceleration

If the angular acceleration of the body is constant, the above

equations, when integrated, yield a set of formulae which relatethe body’s angular velocity, angular position, and time. These

equations are similar to that used for rectilinear motion.

= o + a t

= o

+ o

t + 1/2 ac

t2

2 = o2 + 2 ac ( - 0)

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Motion of Point P

As the rigid body rotates, point P travels

along a circular path of radius r with centre

at point O.

Position and Displacement

The position of P is defined by the position

vector r, which extends from O to P. If the

body rotates d then P will displace by ds =

r d.Velocity

The velocity of P has a magnitude which can be found by dividing so

that v = r, the direction of v is tangent to the circular path. In vector

form, it can be expressed as

v = x rPAcceleration

The acceleration of P can be expressed in terms of its normal and

tangential components. Since at = dv/dt and an = v2/ where = r, v= r

and a = d/dt we have at = a r and an = 2 r.

In vector form, a = at + an, i.e. a = arP + (rP)

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• Motion of Point P

As the rigid body in Fig. 16 –4c rotates, point P travels along a

circular path of radius r with center at point O. This path is

contained within the shaded plane shown in top view

The position of P is defined by the position vector r, which extends from O

to P. If the body rotates d, then P will displace by ds (= r d).

Position and Displacement

VelocityThe velocity of P has a magnitude which can be computed by dividing ds

by dt, so that v = r, by cross product, v = × rp.

For the direction of the velocity vector, the fingers are of the right hand

are curled from toward rp ( “cross” rp). The thumb indicates the

correct direction of v, which is tangent to the path in the direction ofmotion. The magnitude of v is given by v = rp sin f. and since, r = rp sin

f, then v = r .

As a special case, the position vector r can be chosen for rp. Here r lies in

the plane of motion and again the velocity of point P is v = × r .

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Acceleration

• The acceleration of P can be expressed in terms ofits normal and tangential components. Since at =

dv/dt and an = v

2

/ where, = r, v = r, a =d/dt, we have

at = a r and an = 2 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 isdecreasing, atacts in the opposite direction of v;and finally, if the speed is constant, is zero.

• The normal component of acceleration represents

the time rate of change in the velocity’s direction.The direction of an is always toward O, the centreof the circular path.

• In vector form, a =

=

× + ×

i.e. a= a × rp + × ( × rp)

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• From the definition of the cross product, the first term on the right

has a magnitude at = a rp sinf = a r and by the right-hand rule, a × rpis in the direction of at. Likewise, the second term has a magnitude

an=2rpsinf = 2r and applying the right-hand rule twice, first todetermine the result vp = × rp then to × vp, it can be seen that

this result is in the same direction as an , i.e. – r (as seen in fig).

therefore, a = a r - 2 r.

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Procedure for Analysis• The velocity and acceleration of a point located on a rigid body that is

rotating about a fixed axis can be determined using the followingprocedure.

Angular Motion.• Establish the positive sense of rotation about the axis of rotation and

show it alongside each kinematic equation as it is applied.

• If a relation is known between any two of the four variables a, , andt , then a third variable can be obtained by using one of the following

kinematic equations which relates all three variables.

• If the body’s angular acceleration is constant , then the followingequations can be used:

• Once the solution is obtained, the sense of , and a is determinedfrom the algebraic signs of their numerical quantities.

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Motion of Point P.

• In most cases the velocity of P and its two components of acceleration

can be determined from the scalar equations

• If the geometry of the problem is difficult to visualize, the following

vector equations should be used

• Here rp is directed from any point on the axis of rotation to point P,

whereas r lies in the plane of motion of P. Either of these vectors,

along with and a should be expressed in terms of its i, j, k

components, and, if necessary, the cross products determined using a

determinant expansion.

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Numerical

A cord is wrapped around a wheel in Fig. 16 –5, which is

initially at rest when = 0. 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, and (b) the angular

position of line OP .

Solution:since the cord is wrapped around the wheel and moves tangent to it,

ie tangential component is at = 4t m2/s. Hence the

angular acceleration of

the wheel is

angular velocity of the

wheel is angular position of the

wheel is

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Absolute Motion Analysis

• A body subjected to general plane motion undergoes a

simultaneous translation and rotation.• The motion can be completely specified by knowing both the

angular rotation of a line fixed in the body and the motion of apoint on the body.

• One way to relate these motions is to use a rectilinear position

coordinate s to locate the point along its path and an angular position coordinate to specify the orientation of the line.

• By direct application of the time-differential equations, v =ds/dt, a = dv/dt, = d/dt and a = d/dt the motion of the

point and the angular motion of the line can then be related.

• In some cases, this same procedure may be used to relate themotion of one body, undergoing either rotation about a fixedaxis or translation, to that of a connected body undergoinggeneral plane motion.

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Procedure• The velocity and acceleration of a point P undergoing rectilinear

motion can be related to the angular velocity and angular acceleration

of a line contained within a body using the following procedure.Position Coordinate Equation.

• Locate point P on the body using a position coordinate s, which ismeasured from a fixed origin and is directed along the straight-line path of motion of point P.

•

Measure from a fixed reference line the angular position of a line lyingin the body.

• From the dimensions of the body, relate s to using geometry and/ortrigonometry.

Time Derivatives

•

Take the first derivative of s = f() with respect to time to get a relationbetween v and .

• Take the second time derivative to get a relation between a and a

• In each case the chain rule of calculus must be used when taking

• the time derivatives of the position coordinate equation.

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At a given instant, the cylinder of radius r, has an angular velocity and

angular acceleration Determine the velocity and acceleration of its centre

G if the cylinder rolls without slipping.

The cylinder undergoes general plane

motion since it simultaneously translates

and rotates. 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 positioncoordinate 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 groundfrom A to B, is equivalent to sG.

Consequently, the motion requires the

radial line GA to rotate to the position

G’A’ . Since the arc A’B = r , then G travels

a distance

sG = r

Therefore,

vG = r

aG = r a

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• The large window in Fig. 16 –9 is opened using a hydraulic cylinder

AB. If the cylinder extends at a constant rate of 0.5 m/s, determine

the angular velocity and angular acceleration of the window at

the instant = 30o

.

Position Coordinate Equation. The angular

motion of the window can be obtained using the

coordinate , whereas the extension or motion

along the hydraulic cylinder is defined using a

coordinate s, which measures its length from the

fixed point A to the moving point B. From the

law of cosines,

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Relative Velocity

• The general plane motion of a rigid body can be

described as a combination of translation androtation.

• The relative position vector, r B = r A + r B/A

• Displacement: dr B = dr A + dr B/A

• Therefore, velocity

would be

where rB is due to translation and

rotation, r A is due to rotation about A, r

B/A due to translation of A

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• The earlier equation takes the form,

vB = vA + vB/A

• vB/A being the rotational velocity can be writtenas v = r w, i.e. vB/A = rB/A i.e.

vB = vA + rB/A

P d f A l i

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Procedure for Analysis• Vector Analysis

Kinematic Diagram• Establish the directions of the fixed x, y coordinates and draw a

kinematic diagram of the body. Indicate on it the velocities vA,vB of points A and B, the angular velocity, and the relativeposition vector rB/A.

•If the magnitudes of vA, vB or are unknown, the sense ofdirection of these vectors can be assumed.

Velocity Equation

• To apply the equation, vB

= vA

+ × rB/A

express the vectors inCartesian vector form and substitute them into the equation.Evaluate the cross product and then equate the respective iand j components to obtain two scalar equations.

• If the solution yields a negative answer for an unknownmagnitude, it indicates the sense of direction of the vector isopposite to that shown on the kinematic diagram.

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• Scalar Analysis

Kinematic Diagram

• If the velocity equation is to be applied in scalar form, then

the magnitude and direction of the relative velocity mustbe established. Draw a kinematic diagram such as shown inFig. 16 –10g, which shows the relative motion. Since thebody is considered to be “pinned” momentarily at the base

point A, the magnitude of is The sense of direction of isalways perpendicular to in accordance with the rotationalmotion of the body.*

Velocity Equation

• Write Eq. 16 –15 in symbolic form, and underneath each ofthe terms represent the vectors graphically by showingtheir magnitudes and directions. The scalar equations aredetermined from the x and y components of these vectors.

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Solution• Since points A and B re restricted to move along the

fixed slots and is directed downward, the velocity must

be directed horizontally to the right

• This motion causes the link to rotate counterclockwise;

that is, by the right-hand rule the angular velocity is

directed outward, perpendicular to the plane ofmotion.

The link shown in Fig. 16 –13a is guided by two blocks at A

and B, which move in the fixed slots. If the velocity of A is

downward, determine the velocity of B at the instant =

45°.

Expressing each of the vectors in terms of their i, j, k

components and applying velocity Eq. to A, the base point,

and B, we have

i.e.

Since both results are positive, the directions of and are

indeed correct as shown in Fig. 16 –13b. It should be

emphasized that these results are valid only at the instant

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The collar C in Fig. 16 –15a is moving downward with a

velocity of 2 m/s. Determine the angular velocity of CB at this

instant

Kinematic Diagram. The downward motion of C causes B tomove to the right along a curved path. Also, CB and AB rotate

counterclockwise.

Vector form:

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Scalar form:

Resolving these vectors in the x and y directions yields

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The bar AB of the linkage shown in Fig. 1, has a clockwise

angular velocity of 30 rad/s when = 60o Determine the angular

velocities of member BC and the wheel at this instant

Kinematic Diagram: By inspection, the velocities of points B

and C are defined by the rotation of link AB and the wheel

about their fixed axes. The position vectors and the angular

velocity of each member are shown on the kinematic diagram

in Fig. 2. To solve, we will write the appropriate kinematic

equation for each member.

Fig 1

Fig 2

Velocity Equation for Link AB (rotation about a fixed axis):

Velocity eq for Link BC (general plane motion)Wheel (rotation about a fixed axis):

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Instantaneous Centre of Zero Velocity

• The velocity of any point B located on a rigid body can be obtained in

a very direct way by choosing 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 +

rB/A becomes vB = rB/A .

• For a body having general plane motion, point A so chosen is called

the instantaneous centre of zero velocity (IC), and it lies on theinstantaneous axis of zero velocity.

• This axis is always perpendicular to the plane of motion, and the

intersection of the axis with this plane defines the location of the IC.

• Since point A coincides with the IC, then vB = rB/A 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 v B of is simply where is vB = rB/A the angular velocity

of the body. Due to the circular motion, the direction of vB must

always be p erpendicular to rB/A.

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• For example, the IC for the bicycle wheel in Fig. is at the contact

point with the ground.

• There the spokes are somewhat visible, whereas at the top of

the wheel they become blurred.

• If one imagines that the wheel is momentarily pinned at thispoint, the velocities of various points can be found using v = r

• Here the radial distances shown in the photo, must be

determined from the geometry of the wheel.

Location of the IC

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Location of the IC

• To locate the IC we can use the fact that the velocity of a point on the

body is always perpendicular to the relative position vector directed

from the IC to the point.• Several possibilities are

The velocity v A of a point A on the body and theangular velocity of the body are known: In thiscase, the IC is located along the line drawn

perpendicular to vA at A, such that the distance fromA to the IC is rA/IC = vA/. Note that the IC lies up andto the right of A since must cause a clockwise angularvelocity about the IC.

The lines of action of two nonparallel velocities v A and v B and are known: Construct at points A and B line

segments that are perpendicular to vA and vB. Extending

these perpendiculars to their point of intersection as

shown locates the IC at the instant considered.

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The magnitude and direction of two parallel velocities and are known:

Here the location of the IC is determined by proportional triangles.

In both cases, rA/IC = vA/ and rB/IC = vB/

d = rA/IC + rB/IC d = rA/IC - rB/IC

h b d l d d d h l f

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As the board slides downward to the left it is

subjected to general plane motion. Since the

directions of the velocities of its ends A and B are

known, the IC is located as shown. At this instant

the board will momentarily rotate about thispoint. Draw the board in several other positions

and establish the IC for each case.

• The point chosen as the instantaneous centre of zero velocity for the body

can only be used at the instant considered since the body changes its

position from one instant to the next.

• The locus of points which define the location of the IC during the body’s

motion is called a centrode and so each point on the centrode acts as the IC

for the body only for an instant.

• Although the IC may be conveniently used to determine the velocity of any

point in a body, it generally does not have zero acceleration and therefore it

should not be used for finding the accelerations of points in a body.

Points to remember

Procedure for Analysis

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• Establish a kinematic diagram in which the body

is imagined as “extended and pinned” at the IC

so that, at the instant considered, it rotatesabout this pin with its angular velocity.

• The magnitude of velocity for each of the

arbitrary points A, B, and C on the body can be

determined by using the equation v = r where

r is the radial distance from the IC to each point.

The velocity of a point on a body which is subjected to general plane motion

can be determined with reference to its instantaneous centre of zero

velocity provided the location of the IC is first established using one of thethree methods described above.

• The line of action of each velocity vector v is perpendicular to its

associated radial line r, and the velocity has a sense of direction which

tends to move the point in a manner consistent with the angular

rotation V of the radial line

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Block D shown in Fig. moves with a speed of 3m/s. Determine the angular velocities of

links BD and AB, at the instant shown.

Solution:

As D moves to the right, it causes AB to

rotate clockwise about point A. Hence, is

directed perpendicular to AB. The

instantaneous centre of zero velocity for

BD is located at the intersection of the

line segments drawn perpendicular to vA

and vD . From the geometry,

Since the magnitude of vD is known, the angular velocity of link BD is

Therefore,

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From Fig. the angular velocity of AB is

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The crankshaft AB turns with a clockwise angular velocity of 10 rad/s. Determine the

velocity of the piston at the instant shown.

The rotational sense of BC must be the same as the rotation caused by about the IC,

which is counterclockwise. Therefore,

Therefore, the velocity of the piston is

Relative Motion Analysis Acceleration

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Relative-Motion Analysis: Acceleration

• An equation that relates the accelerations of two points on a bar(rigid body) subjected to general plane motion may be determined by

differentiating the velocity equation with respect to time. This yields

• The terms dvB/dt = aB and dvA/dt = aA are measured with respect to aset of fixed x, y axes and represent the absolute accelerations of

points B and A.• The last term represents the acceleration of B with respect to A as

measured by an observer fixed to translating x’, y’ axes which havetheir origin at the base point A.

• Earlier, we have studied that to this observer point B appears to move

along a circular arc that has a radius of curvature rB/A.• Therefore, aB/A, as angular acceleration has two components,

tangential and normal accelerations, thereby relative acceleration fora body in general plane motion can be written as

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= +

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• Since the relative-acceleration components represent the effect

of circular motion observed from translating axes having their

origin at the base point (aB/A )T = a r B/A , (aB/A )N = - 2 r B/A.

• The equation reduces to

•

Note: the acceleration of a point is tangent to the path onlywhen the path is rectilinear or when it is an inflection point on

a curve.


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• Tp determine the angular velocity of the body by using a velocityanalysis as discussed and also, determine the velocities of points A

and B if these points move along curved paths.Kinematic Diagram.

• Establish the directions of the fixed x, y coordinates and draw thekinematic diagram of the body. Indicate on it aA, aB, , a and rB/A.

• If points A and B move along curved paths, then their accelerations

should be indicated in terms of their tangential and normalcomponents, aA,B = (aA,B)t +(aA,B)n

Acceleration Equation.

• To apply aB = aA+ arB/A - 2rB/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 toobtain two scalar equations.

• If the solution yields a negative answer for an unknown magnitude, itindicates that the sense of direction of the vector is opposite to thatshown on the kinematic diagram.

Th d AB h i Fi i fi d

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• The rod AB shown in Fig. is confined to move

along the inclined planes at A and B. If point A

has an acceleration of 3m/s2 and a velocity of

2m/s, both directed down the plane at the

instant the rod is horizontal, determine theangular acceleration of the rod at this instant

Solution:

Kinematic Diagram. Since points A and B both

move along straight-line paths, they have no

components of acceleration normal to the

paths.

To depict , first determine the magnitude

using velocity equation.

= v/rB/A, = 2/(10 sin 45o) = 0.238 rad/s

Acceleration Equation

Solving, we have

Th l h i Fi l f th d h

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The spool shown in Fig. unravels from the cord, such

that at the instant shown it has an angular velocity of 3

rad/s and an angular acceleration of 4 rad/s2.

Determine the acceleration of point B.

SOLUTION:

The spool “appears” to be rolling downward without slipping at

point A. the acceleration of point G, i.e.,

Kinematic Diagram: Point B moves along a curved path

having an unknown radius of curvature. Its acceleration will

be represented by its unknown x and y components


Equating the i and j terms

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The collar C in Fig. moves downward with an

acceleration of 1 m/s2. At the instant shown, it has

a speed of 2 m/s which gives links CB and AB an

angular velocity AB= CB= 10 rad/s. Determine the

angular accelerations of CB and AB at this instant.

Solution:

Kinematic Diagram. The kinematic diagrams of both

links AB and CB are shown in Fig. To solve, apply the

appropriate kinematic equation to each link.


Link BC (general plane motion): Using the result for

Link AB (rotation about a fixed axis):

Relative-Motion Analysis using Rotating Axes

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Relative Motion Analysis using Rotating Axes

• Earlier, the relative-motion analysis for velocity and acceleration

was described using a translating coordinate system.

•

This type of analysis is useful for determining the motion of pointson the same rigid body, or the motion of points located on several

pin-connected bodies.

• In some problems, rigid bodies (mechanisms) are constructed such

that sliding will occur at their connections. The kinematic analysis

for such cases is best performed if the motion is analysed using a

coordinate system which both translates and rotates.

• Furthermore, this frame of reference is useful for analysing the

motions of two points on a mechanism which are not located in

the same body and for specifying the kinematics of particle motionwhen the particle moves along a rotating path.

• In the following analysis two equations will be developed which

relate the velocity and acceleration of two points, one of which is

the origin of a moving frame of reference subjected to both atranslation and a rotation in the lane.

• Consider the two points A and B shown in Fig

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Consider the two points A and B shown in Fig.

Their location is specified by the position

vectors (rA, rB) which are measured with

respect to the fixed X,Y, Z coordinate system.

• As shown in the figure, the “base point” Arepresents the origin of the x, y, z coordinate

system, which is assumed to be both

translating and rotating with respect to the

X,Y, Z system.

• The position of B with respect to A is specified by the relative-position

vector rB/A.

• The components of this vector may be expressed either in terms of unit

vectors along the X, Y axes, i.e., I and J, or by unit vectors along the x, y

axes, i.e., i and j.• For the development which follows rB/A, will be measured with respect

to the moving x, y frame of reference.

• Thus, if B has coordinates (xB, yB), then rB/A = xB i+ yB j

• For obtaining the position of B we use the equation

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• For obtaining the position of B, we use the equation

rB = rA+ rB/A

•

At the instant considered, point A has a velocity vA and an accelerationaA, while the angular velocity and angular acceleration of the x, y axes

are (omega) and ’ = d /dt respectively.

Velocity

• To obtain the velocity of point B, the time derivative of the above

equation gives

VB = VA+ drB/A/dt

•

Where, the last term can be further written as

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• In the above equation, The two terms in the first set of parentheses

represent the components of velocity of point B as measured by anobserver attached to the moving x, y, z coordinate system.

• In the second set of parentheses the instantaneous time rate of

change of the unit vectors i and j is measured by an observer

located in the fixed X,Y, Z coordinate system.• These changes, d i and d j, are due only to the rotation d of the x,

y, z axes, causing i to become i’ = i + di and j to become j’ = j + dj.

• The magnitudes of both d i and d j equal 1

d , since i’ = i = j’ = j = 1.• The direction of d i is defined by +j, since d i

is tangent to the path described by the

arrowhead of i in the limit as t dt

Likewise, d j acts in the -i direction,

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•

Viewing the axes in three dimensions, andnoting that = k, we can express the

above derivatives in terms of the cross

product as

• Solving the above equations,

• Therefore, the final velocity equation can

be expressed as

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Comparing the above equation with that of a translating frame of reference, it can be seen

that the only difference between these two equations is represented by the term (vB/A)xyz.

Acceleration

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Acceleration

• The acceleration of B, observed from the X, Y, Z coordinate system, may

be expressed in terms of its motion measured with respect to the

rotating system of coordinates by taking the time derivative of velocity

equation

• Here ’ (=d/dt) is the angular acceleration of the x, y, z coordinate

system. Since is always perpendicular to the plane of motion, then ’

measures only the change in magnitude of .

• The derivative term drB/A/dt is expresses as earlier form,

• Finding the time derivative of

’

• The two terms in the first set of brackets represent the components of

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The two terms in the first set of brackets represent the components of

acceleration of point B as measured by an observer attached to the

rotating coordinate system. These terms will be denoted by (aB/A)xyz.

• The terms in the second set of brackets can be simplified using

• Hence, the final form of acceleration is given by

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NOTE:

•

The equation, 2 (vB/A)xyz is called as Coriolis acceleration, named after theFrench engineer G. C. Coriolis, who was the first to determine it.

• This term represents the difference in the acceleration of B as measured fromnonrotating and rotating x, y, z axes.

• As indicated by the vector cross product, the Coriolis acceleration will alwaysbe perpendicular to both and (vB/A)xyz.

• It is an important component of the acceleration which must be consideredwhenever rotating reference frames are used.

• For example, when studying the accelerations and forces which act on rockets,long-range projectiles, or other bodies having motions whose measurementsare significantly affected by the rotation of the earth.

The following interpretation of the terms in acceleration Eq may be useful when applying

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The following interpretation of the terms in acceleration Eq. may be useful when applying

this equation to the solution of problems


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Coordinate Axes.

• Choose an appropriate location for the origin and proper orientation of the axes forboth fixed X , Y , Z and moving x , y , z reference frames.

• Most often solutions are easily obtained if at the instant considered:

1. the origins are coincident

2. the corresponding axes are collinear

3. the corresponding axes are parallel

• The moving frame should be selected fixed to the body or device along which therelative motion occurs.

Kinematic Equations.

• After defining the origin A of the moving reference and specifying the moving point B,Eqs. 16 –24 and 16 –27 should be written in symbolic form

• The Cartesian components of all these vectors may be expressed along either the X, Y, Zaxes or the x, y, z axes. The choice is arbitrary provided a consistent set of unit vectorsis used.

• Motion of the moving reference is expressed by and and motion of B with respect tothe moving reference is expressed


At the instant 60o the rod in Fig has an angular velocity of 3 rad/s and an angular

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At the instant = 60o, the rod in Fig. has an angular velocity of 3 rad/s and an angular

acceleration of 2 rad/s2. At this same instant, collar C travels outward along the rod such

that when x= 0.2m, the velocity is 2m/s and the acceleration is 3 m/s2 both measured

relative to the rod. Determine the Coriolis acceleration and the velocity and acceleration

of the collar at this instant.

SOLUTION

Coordinate Axes: The origin of both coordinate systems is

located at point O. Since motion of the collar is reported

relative to the rod, the moving x, y, z frame of reference is

attached to the rod.

Kinematic Equations

It will be simpler to express the data in terms of i, j, k component vectors rather than I, J, K

components.

The Coriolis acceleration is

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The Coriolis acceleration is

The velocity and acceleration of the collar are determined by substituting the

data into respective equations Eqs

Rod AB, shown in Fig. rotates clockwise such that it has an angular velocity AB = 3

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Rod AB, shown in Fig. rotates clockwise such that it has an angular velocity AB 3

rads/s and angular acceleration aAB = 4 rads/s2, when = 45o. Determine the

angular motion of rod DE at this instant. The collar at C is pin connected to AB and

slides over rod DE

Solution:

Coordinate Axes. The origin of both the fixed and

moving frames of reference is located at D.

Furthermore, the x, y, z reference is attached to

and rotates with rod DE so that the relative motion

of the collar is easy to follow.

All vectors will be expressed in terms of i, j, k components

M ti f C Si th ll l i l th f di AC it l it d

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Motion of C: Since the collar moves along a circular path of radius AC, its velocity and

acceleration can be determined using Eqs of rotating bodies.

Substituting the above values in velocity and acceleration eqns, we have

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ic_240_unit4

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