Dynamics lecture5

Planar Kinematics of a Rigid Body

Chapter Objectives

• To classify the various types of rigid-body planar

motion.

• To investigate rigid-body translation and show how

analyze motion about a fixed axis.

• To study planar motion using an absolute motion

analysis.

Chapter Objectives

• To provide a relative motion analysis of velocity and

acceleration using a translating frame of reference.

• To show how to find the instantaneous center of zero

velocity and determine the velocity of a point on a

body using this method.

• To provide a relative-motion analysis of velocity and

acceleration using a rotating frame of reference.

• Rigid-Body Motion

• Translation

• Rotation about a Fixed Axis

• Absolute General Plane Motion Analysis

• Relative-Motion Analysis: Velocity

• Instantaneous Center of Zero Velocity

• Relative-Motion Analysis: Acceleration

• Relative-Motion Analysis Using Rotating Axis

Chapter Outline

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 rigid body planar motion:

1)Translation – This type of motion occurs if every line segment on the body remains parallel to it original direction during the motion.

Rectilinear translation occurs when the paths of motion for any two particles of the body are along equidistant straight lines.

Curvilinear translation occurs when the paths of motion are along curves lines which are equidistant.

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, move along circular paths.

Rigid Body Motion

4) General plane 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.

Rigid Body Motion

• Consider a rigid body which is subjected to either rectilinear or curvilinear translation in the x-y plane.

Translation

Position. The locations of points A and B in the body are defined from the fixed x, y reference frame by using position vectors rA and rB.

The translating x’, y’ coordinate system is fixed in the body and has its origin located at A, hereafter referred to as the base point.

The position of B with respect to A is denoted by the relative-position vector rB/A. By vector addition,

ABAB /rrr

Translation

Velocity. A relationship between the instantaneous velocities A and B is obtained by taking the time derivative of the position equation, which yields vB = vA + drB/A/dt

The term drB/A/dt = 0, since the magnitude of rB/A is constant by definition of a rigid body. Therefore,

Translation

Acceleration. Taking time derivative of the velocity equation yields a similar relationship between the instantaneous accelerations of A and B:

It indicate that all points in a rigid body subjected to either rectilinear or curvilinear translation move with the same velocity and acceleration.

Translation

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. A point is without dimension, and so it has no angular motion. Only lines or bodies undergo angular motion. Consider the body shown and the angular motion of a radial line r located with the shaded plane and directed from point O on the axis of rotation to point P.

Angular Position. At the instant shown in the figure, the angular position of r is defined by the angle θ, measured between a fixed reference line and r.

Angular Displacement. The change in the 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. Since dθ occurs during an instant of time dt, then,

dtd ( +)

Angular Acceleration. The angular acceleration α measure the time rate of change of the angular velocity. The magnitude of this vector may be written as

( +) 2

The line of action of α is the same as that for ω. However, it sense of direction depends on whether ω is increasing or decreasing.

In particular, if ω is decreasing, then α is called angular deceleration and it therefore has a sense of direction which is opposite to ω.

By eliminating the two equations, we obtain a differential relation between the angular acceleration, angular velocity and angular displacement,

dd ( +)

Constant Angular Acceleration. If the angular acceleration of the body is constant, α = αc,

Constant Angular Acceleration

Motion of Point P. As the rigid body rotates, point P travels along a circular path of radius r and center at center at point O. This path is contained within the shaded plane

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 found from its polar coordinate components and . Since r is constant, the radial components and so

rvr rv 0rvr

The direction of v is tangent to the circular path.

Both the magnitude can direction of v can also be accounted for by using the cross product of ω and rp. Here rp is directed from any point on the axis of rotation to point P

Acceleration. The acceleration of P can be expressed in terms of its normal and tangential components

The tangential component of acceleration represents the time rate of change in the velocity’s magnitude.

If speed of P increases => at same direction as v

If speed of P decreases => at opposite direction as v

If speed of P constant => at 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 center of the circular path.

The acceleration of point P may be expressed in terms of the vector cross product.

The first term on the R.H.S has a magnitude of

rra Pt sin

By the right-hand rule, α x rP is in the direction of at. The second term has a magnitude

rra Pn22 sin

Applying right-hand rule twice, first to determine the result vP = ω x rP then ω x rP , it can be seen that this result is in the same direction as an which is also in the same direction as –r, which lies in the same plane of motion.

Therefore,

Since at and an are perpendicular to one another, if needed the magnitude of acceleration can be determined from the Pythagorean theorem,

22tn aaa

PROCEDURE FOR ANALYSIS

Angular Motion.

• Establish the positive sense of direction along the axis of rotation and show it alongside each kinematics equation as it is applied.

• If a relationship is known between any two of the four variables α, ω, θ and t, then a third variable can be obtained by using one of the following kinematics equations which relates all three variables

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

• Once the solution is obtained, the sense of α, ω and θ is determined from algebraic signs of their numerical quantities.

Motion of P.

• In most cases the velocity of P and its two components of acceleration can be determined from the scalar equations

rat rv tan2

• If the geometry of the problem is difficult to visualize, the following vector equations should be used:

Example

A cord is wrapped around a wheel which is initially at rest. 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 the line OP in radians.

Solution

Part (a). The wheel is subjected to rotation about a fixed axis passing through point O. Thus, point P on the wheel has a motion about a circular path, and the acceleration of this point has both tangential and normal components. The tangential component is (aP)t = (4t) m/s2, since the cord is wrapped around the wheel and moves tangent to it.

Hence the angular acceleration of the wheel is

)2.0()4(

Using this result, the wheel’s angular velocity ω can now be determine from α = dω/dt

Integrating, with the initial condition that ω = 0, t = 0,

Using this result, the angular position θ of OP can be found from ω = dθ/dt, since this equation relates θ, ω, and t.

Integrating, with the initial condition θ = 0 at t = 0,

Example

The motor is used to turn a wheel and attached blower contained with the housing. If the pulley A connected to the motor begins rotating from rest with an angular acceleration of αA = 2 rad/s2, determine the magnitudes of the velocity and acceleration of point P on the wheel, after the wheel B has turned one revolution.

Angular Motion. Converting revolution to radians,

radB 283.6

Since the belt does not slip, an equivalent length of belt s must be unraveled from both the pulley and wheel at all times. Thus,

)4.0(283.6)15.0(;

Since αA is constant, the angular velocity of pulley A is therefore

/188.8

)076.16)(2(20

The belt has the same speed and tangential component of acceleration as it passes over the pulley and wheel. Thus,

2/750.0

/070.3

sradrra

sradrrv

BBBAAt

Motion of P. As shown in the diagram, we have,

/78.3)77.3()3.0(

/77.3)(

/3.0)(

Absolute Motion Analysis

• A body subjected to general plane motion undergoes a simultaneous translation and rotation.

• One way to define these motions is to use a rectilinear position coordinate s to locate the point along its path and an angular position coordinate θ to specify the orientation of the line.

• By direct application of the time-differential equations v = ds/dt, a = dv/dt, ω = dθ/dt, α = dω/dt, the motion of the point and the angular motion of the line can be related.

Position Coordinate Equation.

• Locate point P using a position coordinate s, which is measured 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 θ of a line lying in the body.

• From the dimensions of the body, relate s to θ, s = f(θ), using geometry and/or trigonometry.

Position Coordinate Equation.

• Take the first derivative of s = f(θ) w.r.t time to get a relationship between v and ω.

• Take the second derivative to get a relationship between a and α.

• In each case the chain rule of calculus must be used when taking the derivatives of the position coordinate equation.

Example

The end of rod R maintains contact with the cam by means of a spring. If the cam rotates about an axis through point O with an angular acceleration α and angular velocity ω, determine the velocity and acceleration of the rod when the cam is in the arbitrary position θ.

Position Coordinate Equation. Coordinates θ and x are chosen in order to relate the rotational motion of the line segment OA on the cam to the rectilinear motion of the rod. These coordinates are measured from the fixed point O and may be related to each other using trigonometry.

Since OC = CB = r cos θ,

cos2rx

Time Derivatives. Using chain rule of calculus, we have,

)cossin(2

)(cos2sin2

)(sin2

Example

At a given instant, the cylinder of radius r, has an angular velocity ω and angular acceleration α. Determine the velocity and acceleration of its center G if the cylinder rolls without slipping.

Position Coordinate Equation. By inspection, point G moves horizontally to the left from G to G’ as the cylinder rolls. Its new location G’ will be specified by the horizontal position coordinate sG, which is measured from the original position (G) of the cylinder’s center.

As the cylinder rolls, points on its surface contact the ground such that the arc length A’B of contact must be equal to the distant sG.

Consequently, the motion requires the radial line GA to rotate θ to position G’A’. Since the arc A’B = rθ, then G travels a distance

Time Derivatives. Taking successive time derivatives of this equation, realizing that r is constant, ω = dθ/dt, and α = dω/dt, gives the necessary relationships:

rarvrs GGG

ExampleThe large window 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 θ = 30°

Position Coordinate Equation. The angular motion of the window can be obtained using the coordinate θ, whereas the extension or motion along the hydraulic cylinder is defined using a coordinate s, which measures the length from the fixed point A to moving point B. These coordinates can be related using the law of cosines, namely,

cos)1)(2(2)1()2(2

When θ = 30°,

ms 239.1

Time Derivatives. Taking the time derivatives of Eq. 1,

/620.0

)(sin2)(

)sin(402

Since vs = 0.5 m/s, then at θ = 30°

Taking the time derivatives of Eq. 2 yields,

/415.0

30sin2)620.0(30cos20)5.0(

)(sin2)(cos2

dtdvsv

Since as = dvs/dt = 0, then

Relative–Motion Analysis: Velocity

• The general plane motion of a rigid body can be described as a combination of translation and rotation.

• To view these “component” motions separately, we use a relative-motion analysis involving two sets of coordinate axes.

• The x, y coordinate system is fixed and measures the absolute position 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 do not rotate with the body; rather they will only be allowed to translate with respect to the fixed frame.

Position.

• The position vector rA specifies the location of the “base point” A, and the relative-position vector rB/A locates point B with respect to point A.

• By vector addition, the position of B is

ABAB /rrr

Displacement.

• During an instant of time dt, point A and B undergo displacements drA and drB.

• If we consider the general plane motion by its component parts then the entire body first translates by an amount drA so that A, the base point, moves to its final position and point B to B’.

• The body is then rotated about A by an amount dθ so that B’ undergoes a relative displacement drB/A and thus moves to its final position B.

• Due to the rotation about A, drB/A = rB/A dθ, and the displacement of B is

ABAB ddd /rrr

due to rotation about A

due to translation about A

due to translation and rotation

Velocity.

• To determine the relationship between the velocities of points A and B, it is necessary to take the time derivative of the position equation, or simply divide the displacement equation by dt. This yields,

dtd ABAB /rrr

• The terms drB/dt = vB and drA/dt = vA are measured from the fixed x, y axes and represent the absolute velocities of points A and B, respectively.

• The body appears to move as if it were rotating with an angular velocity ω about the z’ axis passing through A

• vB/A has a magnitude of vB/A = ωrB/A and a direction which is perpendicular to rB/A.

ABAB /vvv

• The velocity of B is determined by considering the entire body to translate with a velocity of vA, and rotate about A with an angular velocity ω.

• Vector addition of these two effects, applied to B, yields vB.

• The relative velocity vB/A represents the effect of circular motion, about A. It can be expressed by the cross product

ABAB // rv

ABAB /rvv • Hence,

VECTOR ANALYSIS

Kinematics Diagram.

• Establish the directions of the fixed x, y coordinates and draw a kinematics diagram of the body. Indicate on it the velocities vA, vB of points A and B, the angular velocity ω, and the relative-position vector rB/A

• If the magnitudes of vA, vB or ω are unknown, the sense of the direction of these vectors may be assumed.

Velocity Equation.

• To apply vB = vA + ω x rB/A, express the vectors in Cartesian vector form and substitute them into the equation. Evaluate the cross product and then equate the respectively i and j components to obtain two equations.

• If the solution yields a negative answer for an unknown magnitude, it indicates the sense of direction of the vector is opposite to that shown on the kinematics diagram.

SCALAR ANALYSIS

Kinematics Diagram.

• If the velocity equation is to be applied in scalar form, then the magnitude and direction of the relative velocity vB/A must be established.

• Draw a kinematics diagram, which shows the relative motion. Since the body is considered to be “pinned” momentarily at the base point A, the magnitude is vB/A = ωrB/A.

• The sense of the direction of vB/A is established from the diagram, such that vB/A acts perpendicular to rB/A in accordance with the rotational motion ω of the body.

Velocity Equation.

• Write in symbolic form, and underneath each of the terms represent the vectors graphically by showing their magnitudes and directions. The scalar equations are determined from the x and y components of these vectors.

ABAB /vvv

Example

The link is guided by two block 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 = 45°.

Solution (Vector Analysis)

Kinematic Diagram.

• Since points A and B are restricted to move along the fixed slots and vA is directed downward, the velocity vB must be directed horizontally to the right.

• This motion causes the link to rotate CCW; by right-hand-rule the angular velocity ω is directed outward, perpendicular to the plane of plane.

• Knowing the magnitude and direction of vA and the lines of action of vB and ω, it is possible to apply the velocity equation to points A and B in order to solve for the two unknown magnitudes vB and ω.

ABAB /vvv

Velocity Equation.

45cos2.045sin2.02

)]45cos2.045sin2.0([2

Equating the i and j components gives

45sin2.02045cos2.0

Example

The cylinder rolls without slipping on the surface of a conveyor belt which is moving at 2 m/s. Determine the velocity of point A. The cylinder has a clockwise angular velocity ω = 15 rad/s at the instant.

Kinematics Diagram.

• Since no slipping occurs, point B on the cylinder has the same velocity as the conveyor.

• The angular velocity of the cylinder is known, so we can apply the velocity equation to B, the base point, and A to determine vA.

Velocity Equation.

jikiji

50.750.72)()(

)]5.05.0()15(2)()(/

So that

/50.7)(

/50.950.72)(

3.3850.950.7

/1.12)50.7()50.9(

Same results can be obtained using Scalar Analysis.

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

Kinematics Diagram.

• The downward motion of C causes B to move to the right.

• CB and AB rotate counterclockwise, to solve, we will write the appropriate kinematics equation for each link.

Velocity Equation.

Link CB (general plane motion):

CBCBCB

2.02.02

)2.02.0(2/

Link AB (rotation about a fixed axis):

sradAB

)2.0(2

Same results can be obtained using Scalar Analysis.

Example

The bar AB of the linkage has a clockwise angular velocity of 30 rad/s when θ = 60°. Determine the angular velocities of member BC and the wheel at this instant.

Kinematics Diagram.

• the velocities of point B and C are defined by the rotation of link AB and the wheel about their fixed axes

Velocity Equation.

Link AB (rotation about fixed axis):

/}0.320.5{

)60sin2.060cos2.0()30(

Link BC (general plane motion):

BCBCBC

0.32.00

)0.32.0(20.5

)2.0()(0.320.5/

Wheel (general plane motion):

1.020.5

)1.0()(20.5

Instantaneous Center of Zero Velocity

• The velocity of any point B located on a rigid body can be obtained in a very direct way if one choose the base point A to be a point that has zero velocity at the instant considered.

• Since vA = 0, therefore vB = ω x rB/A.

• Point A 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 planre of 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/A and so point B moves momentarily anout the IC in a circular path.

• The magnitude of vB is vB = ωrB/IC.

• Due to the circular motion, the direction of vB must always be perpendicular to rB/IC

• Consider the wheel as shown, if it rolls without slipping, then the point of contact with the ground has zero velocity.

• Hence this point represents the IC for the wheel.

• If it is imagined that the wheel is momentarily pinned at this point, the velocities of points B, C, O and so on, can be found using v = ωr.

• The radial distance rB/IC, rC/IC and rO/IC must be determined from the geometry of the wheel.

Location of the IC. To locate the IC, we use the fact that the velocity of a point on the body is always perpendicular to the relative-position vector extending from the IC to the point. Several possibilities exist:

• 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 to the right of A since vA must cause a clockwise angular velocity ω about the IC.

• Given the line of action of two nonparallel velocities vA and vB. Construct at points A and B line segments that are perpendicular to vA and vB. Extending these perpendicular to their point of intersection as shown locates the IC at the instant considered.

• Given the magnitude and direction of two parallel velocities vA and vB. Here the location of the IC is determined by proportional triangles.

In both cases rA/IC = vA/ω and rB/IC = vB/ω. If d is a known distance between point A and B, then rA/IC + rB/IC = d for first diagram, and rB/IC - rA/IC = d for second diagram. As a special case, note that if the body is translating, vA = vB, then the IC would be located at infinity, in which case rA/IC = rB/IC → ∞. This being the case, ω = (vA/rA/IC) = (vA/rA/IC) → 0, as expected.

• the point chosen 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.

• 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 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 on a body

The velocity of a point on a body which is subjected to general plane ,motion can be determined with reference to its instantaneous center of zero velocity provided the location of IC is first established using one of the three methods described in previous segment.

• As shown in the diagram, the body is imagined as “extended and pinned” at the IC such that, at the instant considered, it rotates about this pin with its angular velocity ω.

• 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 = ωr, where r is the radial distance from the IC to each point.

• 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 ω of the radial line.

Example

Show how to determine the location of the instantaneous center of zero velocity for (a) member BC shown in (a); and (b) the link CB shown in (b).

Solution

Part (a).

• Point B has a velocity vB, which is caused by the clockwise rotation of link AB.

•Point B moves in a circular path such that vB is perpendicular to AB, and so it acts at an angle θ from the horizontal as shown.

• The motion of point B causes the piston to move forward horizontally with a velocity vC

• When the line are drawn perpendicular to vB and vC, the intersect at the IC.

Part (b).

• Points B and C follow circular paths of motion since rods AB and DC are each subjected to rotation about a fixed axis.

• Since the velocity is always tangent to the path, at the instant considered, vC on rod DC and vB on rod AB are both directed vertically downward, along the axis of link CB.

• Radial lines drawn perpendicular to these two velocities form parallel lines which intersect at “infinity”.

0)/( /

ICCCCB

ICBICC

• As a result, rod CB momentarily translates.

• At an instant later, however, CB will move to a tilted position, causing the instantaneous center to move to some finite location.

ExampleBlock D moves with a speed of 3 m/s. Determine the angular velocities of links BD and AB, at the instant shown.

Solution

• As D moves to the right, it causes arm AB to rotate clockwise about 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

From the geometry,

566.045cos

4.045tan4.0

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

sradrv

DBD /30.5

566.03

The velocity of B is therefore

smrv ICBBDB /12.2)4.0(30.5)( /

From the figure, the angular velocity of AB is

sradrv

BAB /30.5

4.012.2

ExampleThe cylinder rolls without slipping between the two moving plates E and D. Determine the angular velocity of the cylinder and the velocity of its center C at the instant shown.

Solution

• Since no slipping occurs, the contact points A and B on the cylinder have the same velocities as the plate E and D, respectively.

• The velocities vA and vB are parallel, so that by the proportionally of the right triangles the IC is located at a point on line AB.

Assuming this point to be a distance x form B, we have

)25.0(25.0);25.0(

Dividing one equation into the other eliminates ω and yields

154.065.01.0

25.0)25.0(4.0

Hence the angular velocity of the cylinder is

sradxvB /60.2

154.04.0

The velocity of point C is therefore

rv ICCC

/0750.0

)125.0154.0(6.2/

Relative-Motion Analysis: Acceleration

• An equation that relates the accelerations of two points on a rigid body subjected to general plane motion,

dtd ABAB /vvv

• The terms dvB/dt = aB and dvA/dt = aA are measured from a set of fixed x, y axes and represent the absolute accelerations of points B and A.

• The last term represents the acceleration of B w.r.t A as measured by an observer fixed to translating x’, y’ axes which have their origin at the base point A.

• To this observer, point B appears to move along a circular arc that has a radius of curvature rB/A.

• aB/A can be expressed in terms of its tangential and normal components of motion

nABtABAB )()( // aaaa

• Since points A and B move along curved paths, the accelerations of these points will have both tangential and normal components.

• The relative-acceleration components represent the effect of circular motion observed from translating axes having their origin at the base point A, and can be expressed as (aB/A)t = α x rB/A and (aB/A)n = -ω2rB/A

ABABAB /2

/ rraa

Velocity Analysis.

• Determine the angular velocity ω of the body by using a velocity analysis as discuss in previous section. Also determine the velocities vA and vB of points A and B if these points moves along curved paths.

VECTOR ANALYSIS

Kinematics Diagram.

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

• If points A and B move along curved paths, then their accelerations should be indicated in terms of their tangential and normal components.

Acceleration Equation.

• To apply aB = aA + α x rB/A – ω2rB/A express the vectors in Cartesian vector form and substitutide them into the equation. Evaluate the cross product and then equate the respective i and j components to obtain two scalar equations.

• If the solution yields a negative answer for an unknown magnitude, it indicates that the sense of direction of the vector is opposite to that shown on the kinematics diagram.

SCALAR ANALYSIS

Kinematics Diagram.

• If the equation is applied, 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.

• Since the body is considered to be momentarily “pinned” at the base point A, the magnitudes are (aB/A)t = αrB/A and (aB/A)n = ω2rB/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 towards A.

• Represent the vectors in graphically by showing their magnitudes and directions underneath each term. The scalar equations are determined from the x and y components of these vectors.

Example

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.

• We can obtain ω = 0.283 rad/s using either the velocity equation or the method of instantaneous centers.

Kinematic Diagram. Since points A and B both move along straight-line paths, they have no components of acceleration normal to the paths. There are two unknowns, aB and α.

)10()238.0()10()(

45cos345sin45cos2

ABABAB

Carrying out the cross product and equating the i and j components yields

)10(45sin345sin

)10()283.0(45cos345cos 2

Solving, we have

/344.0

ExampleAt a given instant, the cylinder of radius r has an angular velocity ω and angular acceleration α. Determine the velocity and acceleration of its center G if it rolls without slipping.

• As the cylinder rolls, point G moves along a straight line, and point A, located on the rim of the cylinder, moves along a curved path called a cycloid.

• We will apply the velocity and acceleration equations to these two points.

Velocity Analysis. Since no slipping occurs, at the instant A contacts the ground, vA = 0. Thus, from the kinematic diagram,

Note the same result can also be obtained directly by noting that point A represents the instantaneous center of zero velocity

Kinematic Diagram. • The acceleration of point G is horizontal since it moves along a straight-line path.• Just before point A touches the ground, its velocity is directly downward along the y axis, just after contact, its velocity is directed upward.• Therefore, point A begins to accelerate upward when it leaves the ground at A.• The magnitudes of aA and aG are unknown.

)()()( 2

rraa AG

AGAGAG

Evaluating the cross product and equating the i and j components yields

Example

The ball rolls without slipping and has the angular motion shown in figure. Determine the acceleration of point B and point A at this instant.

Kinematic Diagram. Using the results of the previous example, the center of the ball has an acceleration of aO = αr = (4)(0.15) = 0.6 m/s2. We apply the acceleration equation to points O and B and points O and A.

For point B,

/}6.06{

)15.0()6()15.0()4(6.0

OBOBOB

For point A,

/}4.52.1{

)15.0()6()15.0()4(6.0

OAOAOA

Example

The spool 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.

• The spool “appear” to be rolling downward without slipping at point A. The acceleration of point G is aG = αr = (4)(0.15) = 0.6 m/s2.

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 as shown.

)225()3()225()4(600)()( 2

jjkjji

GBGBGB

Equating the i and j terms, the component equations are

/625.2/26252025600)(

/9.0/900)225(4)(

smsmma

The magnitude and direction of aB are therefore

1.719.0

625.2tan

/775.2)625.2()9.0(

Example

The collar is moving 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 ωAB = ωCB = 10 rad/s. Determine the angular accelerations of CB and AB at this instant.

• The kinematic diagrams of both links AB and CD are as shown. To solve, we will apply the appropriate kinematic equation to each link.

Link AB (rotation about to a fixed axis):

)2.0()10()2.0()( 2

BABBABB

Note that aB has two components since it moves along a curved path.

Link BC (general plane motion):

jiijjji

jikjji

20202.02.01202.0

)2.02.0()10(

)2.02.0()(1202.02

CBCBAB

CBCBCBCBCB

202.0120

202.02.0

/95/95

sradsrad

Solving,

Example

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

Kinematic Diagram. The kinematic diagrams of both AB and BC are as shown. Here aC is vertical since C moves along a straight-line path.

}729.0176.0{

}6.13cos75.06.13sin75.0{

}177.0177.0{

}45cos25.045sin25.0{

Acceleration Equation. Expressing each of the position vectors in Cartesian vector form

Crankshaft AB (rotational about a fixed axis):

/}14.1421.21{

)177.0177.0()10()177.0177.0()20(

BABBABB

Connecting Rod BC (general plane motion):

45.18176.0

729.017.200

)729.0176.0()43.2()729.0176.0(

)(14.1421.212

BCBCBCBCBC

Solving yields

• Translating coordinate system

describes relative motion analysis for velocity and acceleration

determines the motion of the points on the same rigid body

determines the motion of points located on several pin-connected rigid bodies

• Rigid bodies are constructed such that sliding occur at their connections

Relative-Motion Analysis Using Rotating Axes

• Coordinate system

Use for kinematics analysis

use for analyzing motion of two points on a mechanism which are not located in the same rigid body

use for specifying kinematics of particle motion when the particle is moving along a rotating path

• In the following analysis, 2 equations are developed to relate the velocity and acceleration of 2 points, one of which is the origin of a moving frame of reference subjected to both a translation and rotation in the plane

• The 2 points can represent either 2 points moving independently of one another or 2 points located on the same (or different rigid bodies)

Position

• Consider 2 points A and B, whose location are specified by rA and rB, measured from the fixed X, Y, Z coordinate system

• Base point A represent the origin of the x, y, z coordinate system assumed to be both translating and rotating with respect to X, Y and Z system

• Position of B with respect to A is specified by the relative position vector rB/A

• Components of this vector can either be expressed in unit vectors along the X, Y axes i.e. I, J or by unit vectors along the x, y axes i.e. i and j

• For developed rB/A will be measured relative to the moving x, y frame of reference

• If B has coordinates (xB, yB)jir BBAB yx /

• Using vector addition,

ABAB /rrr

• At the instant considered, point A has a velocity vA and an acceleration aA, while angular velocity and angular acceleration of the x, y and z axes are Ω and respectively

• All these vectors are measured from the X, Y and Z axes of reference although they may be expressed in terms of either I, J and K or i, j or k components

• Since planar motion is specified, by the right hand rule, and are always directed perpendicular to the reference plane of motion whereas vA and aA lie on this plane

Velocity

• For velocity of point B,

dtd AB

• The last term of this equation is evaluated as

• The two terms in the first set of parentheses represent the components of velocity of point B as measured by an observer attached to the moving x, y and z coordinate system, being denoted by vector (vB/A)xyz

• In the second set of parentheses, the instantaneous time rate of change of unit vectors i and j is measured by an observer located in a fixed X, Y and Z system

• These changes di and dj are due to only an instantaneous rotation dθ of the x, y and z axes, causing i to become i’ = i + di and j to become j’ = j + dj

• Magnitudes of both di and dj = 1 (dθ) since i = i’ = j = j’ =1

• The direction of di is defined by +j since di is tangent to the path described by the arrowhead of i in the limit as Δt →dt

• Likewise, dj acts in the –i direction, hence

jji )()(

• Viewing the axes in 3D, noting that Ω = Ωk,

• Using the derivative property of the vector cross product

ABxyzABBBxyzABAB yx

//// )()()( rvjiv

• Hence

xyzABABAB )( // vrvv

Acceleration

• Acceleration of B, observed from the X, Y and Z coordinate system, may be expressed in terms of its motion measured with respect to the rotating or moving system of coordinates by taking the time derivative

dtd xyzABAB

ABAB)( //

• Here is the angular acceleration of the x, y, z coordinate system

• For planar motion, Ω is always perpendicular to the plane of motion and therefore measures only the change in the magnitude of Ω

• For the derivative of drB/A/dt,

)()( ///

ABxyzABAB

• Finding the time derivative of (vB/A)xyz = (vB/A)xi + (vB/A)yj

yABxAB

yABxABxyzAB

)()()(

• The first two terms in the first set of brackets represent the components of acceleration of point B as measured by an observer attached to the moving coordinate system, as denoted by (aB/A)xyz

xyzABxyzABxyzAB

• The terms in the second bracket can be simplified by

xyzABxyzAB

ABABAB

• Rearranging terms,

• The term 2Ω x (vB/A)xyz is called the Coriolis acceleration, representing the difference in the acceleration of B as measured from the non-rotating and rotating x, y, z axes

• As indicated by the vector cross-product, the Coriolis acceleration will always be perpendicular to both Ω and (vB/A)xyz

ExampleAt the instant θ = 60°, the rod has an angular velocity of 3 rad/s and an angular acceleration of 2 rad/s2. At the same instant, the collar C is travelling outward along the rod such that when x = 2 m the velocity is 2 m/s and the acceleration is 3 m/s2, both measure relative to the rod. Determine the Coriolis acceleration and the velocity and acceleration of the collar at the instant.

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

xyzOCxyzOCOCOCOC

xyzOCOCOC

)()(2)(

avrraa

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

smsrad

xyzOCO

/3)(/3

Motion of moving reference

Motion of C with respect to moving reference

Therefore Coriolis acceleration is defined as

2/ /12)2()3(2)(2 smxyzOCCor jikva

This vector is shown in figure. If desired, it may be resolved in I, J components acting along the X and Y respectively.

The velocity and acceleration of the collar are determined by substituting the data in the previous 2 equations and evaluating the cross products, which yields,

/4.1220.1

3)2()3(2)2.0()3()3()2.0()2(0

)()(2)(

2)2.0()3(0

xyzOCxyzOCOCOCOC

xyzOCOCOC

iikikkik

avrraa

Example

The rod AB, shown in Fig. 16–34, rotates clockwise such that it has an angular velocity and angular acceleration when . Determine the angular motion of rod DE at this instant. The collar at C is pin connected to AB and slides over rod DE.

sradwAB /32/4 sradAB 45

SolutionCoordinate Axes. The origin of both the fixed and moving framesof reference is located at D, Fig. 16–34. 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.Kinematic Equations.

xyzDCxyzDCDCDCOC

xyzDCDCOC

)()(2)(

avrraa

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

xyzDCxyzDCDE

xyzDCxyzDCD

Motion of moving reference

Motion of C with respect to moving reference

Motion of C: Since the collar moves along a circular path, its velocity and acceleration can be determined.

/2.12.1

ACABACAB

Substituting the data into Eqs. 1 and 2, we have

xyzDCDE

xyzDCDCDC

4.002.12.1

4.0334.002.52

sradsrad

xyzDCxyzDCDCDCDC

ikkikji

avrraa

Example

Two planes A and B are flying at the same elevation and have the motions shown in Fig. 16–35. Determine the velocity and acceleration of A as measured by the pilot of B.

SolutionCoordinate Axes. Since the relative motion of A with respect to the pilot in B is being sought, the x, y, z axes are attached to plane B, Fig. 16–35. At the instant considered, the origin B coincides with the origin of the fixed X,Y, Z frame.

Kinematic Equations.

xyzBAxyzBABABABA

xyzBABABA

)()(2)(

avrraa

SolutionMotion of Moving Reference:

/25.0400

/5.1400

/100900

/900400

hkmjiaaa

Rigid-Body Planar Motion

• A rigid body undergoes three types of planar motion: translation, rotation about a fixed axis and general plane motion.

Translation

• When a body has rectilinear translation, all the particles of the body travel along straight-line paths.

CHAPTER REVIEW

If the paths have the same radius of curvature, then curvilinear translation occurs. Provided we know the motion of one particles, then the motion of all others is also known.

CHAPTER REVIEW

Rotation about a Fixed Axis

• For this type of motion, all of the particles moves along circular paths

• Here, all segments in the body undergo the same angular displacement, angular velocity and angular acceleration.

• The differential relationships between these kinematic quantities are

dddtddtd //

CHAPTER REVIEW

• If the angular acceleration is constant, α = αc, then these equations can be integrated and become

CHAPTER REVIEW

• Once the angular motion of the body is known, then the velocity of any particle a distance r from the axis of rotation is

• The acceleration of the particle has two components. The tangential component accounts for the change in the magnitude of the velocity

rv orrv

ra tt orra

CHAPTER REVIEW

• The normal component accounts for the change in the velocity direction

General Plane Motion

• When a body undergoes general plane motion, it simultaneously translates and rotates.

• There are several types of methods for analyzing this motion:

ra 22 nn orra

CHAPTER REVIEW

Absolute Motion Analysis

• If the motion of a point on a body or the angular motion of a line is known, then it may possible to relate this motion to that of another point or line using an absolute motion analysis

• To do so, linear position coordinates s or angular position coordinates θ are established (measured from a fixed point or line).

CHAPTER REVIEW

• These position coordinates are then related using the geometry of the body .

• The time derivative of this equation gives the relationship between the velocities and/or the angular velocities

• A second time derivative relates the accelerations and/or the angular accelerations.

CHAPTER REVIEW

Relative Velocity Analysis

• General plane motion can also be analyzed using a relative-motion analysis between two points A and B.

• This method considers the motion in parts; first a translation of the selected base point A, then a relative “rotation” of the body about point A, measured from a translating axis.

CHAPTER REVIEW

• The velocities of the teo points A and B are then related using

• This equation can be applied in Cartesian vector form, written as

ABAB /vvv

ABAB /rvv

CHAPTER REVIEW

• In similar manner, for acceleration,

• Since the relative motion is viewed as circular motion bout the base point, point B will have a velocity vB/A, that is tangent to the circle.

ABABAB

nABtABAB

// )()(

CHAPTER REVIEW

• It also has two components of acceleration, (aB/A)t, and (aB/A)n.

• It is important to also realize that aA and aB may have two components if these points move along curved paths.

CHAPTER REVIEW

• If the base point A is selected as having zero velocity, then the relative velocity equation becomes

• In this case, motion appears as if the body is rotating about an instantaneous axis.

ABB /rv

CHAPTER REVIEW

• The instantaneous center of rotation (IC) can be established provided the directions of the velocities of any two points on the body are known.

• Since the radial line r will always be perpendicular to each velocity, then the IC is at the point of intersection of these two radial lines.

• Its measured location is determined from the geometry of the body.

CHAPTER REVIEW

• Once it is established, then the velocity of any point P on the body can be determined from v = ωr, where r extends from IC to point P.

Relative Motion Using Rotating Axes

• Problems that involve connected members that slide relative to one another, or points not located on the same body, can be analysed using a relative motion analysis referenced from a rotating frame.

CHAPTER REVIEW

• The equations of relative motion are

• In particular, the term 2 Ω x (vB/A)xyz is called the Coriolis Acceleration.

xyzABABAB )( // vrvv

xyzABxyzAB

ABABAB

CHAPTER REVIEW

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IsoldeShellModelCourseforNonPractitioners: Lecture5 ... · IsoldeShellModelCourseforNonPractitioners: Lecture5: LanczosStrengthFunctions E. Caurier, G. Martínez-Pinedo, F. Nowacki,