Analysis of Velocity and Accelerations Components

8/8/2019 Analysis of Velocity and Accelerations Components

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Analysis Of Velocity And Accelerations Components.

a) Velocity

In the diagram the point P moves in the Plane XOY. The length OP = r and the. Then:-

(1)

And

(2)

Differentiating with respect to time :-

(3)

(4)

The radial component of Velocity v ( i.e. in the direction of OP) is given by:-

(5)

And using equations (3) and (4)Which is the rate of increase of OP

The tangential component of velocity (i.e.Perpendicular to OP in the direction of

increasing)

(6)

(7)
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Where

b) Acceleration

Differentiating (3) and (4)

(8)

(9)

The Radial components of acceleration

(10)

And from equation (5)

(11)

The Tangential component of acceleration

(12)

By substitution:-

(13)

(Note )

Of these four terms in equations (11) and (13)

is the rate of change of radial velocity

is the centripetal acceleration due to the rotation of OP

is due to the change in angular velocity.

is called the compound supplementary acceleration orCoriolis Component

Notice that the direction of this is the same as when v is radially outwards.

The Velocity And Acceleration Of A Piston By Analysis

In the following analysis is the uniform angular velocity of the crank. The positive

direction of velocity and acceleration is away from the crankshaft.
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(14)

And

(15)

(16)From the above three equations:-

(17)

(18)

(19)

(20)

Instantaneous Centre Method For Velocities.

This has been covered in the section on Mechanisms.

The Vector Method For Velocity And Acceleration.The Law of addition of velocities states that:-

(21)

(22)
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"Absolute velocities (or accelerations) are given from O to the corresponding point on the

diagram.

For Velocities

The relative velocity between two points A and B on the same link of amechanism must be perpendicular to the line joining the points and is equal to

(Equation (4))since r is constant and is zero.

The relative velocity for two points sliding over one another is along the common

tangents of their paths and represents the component , is zero since

for Acceleration

The relationships for acceleration are similar to those given for velocity:-

Acceleration of B = Acceleration of A + Acceleration of B relative to A

Equations (11) and (13) are the general expressions for the radial and tangentialcomponents of relative acceleration.

For two points on the same link leaving centripetal component

(which can be calculated when the velocities are determined) and thetangential component

For a uniformly rotating Crank leaving the centripetal as the only term.

The Coriolis component arises when a point on one link is sliding along another

link wwhich is itself rotating.

If A,B,C, are three points on the same link of a mechanism and a,b,c, are the

corresponding points on the velocity (or acceleration) diagram, it can be shown that the

triangles ABC and abc are similar. abc is called the Velocity (or acceleration) image ofthe link.


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The above is a diagramatic sketch of a piston; connecting rod; crank assembly where:-

PC is the connecting rod with C the Crank Pin.

OC is the crank. OP is the line of stroke.

P is the gudgeon pin

The Construction is as follows:-

Extend PC to meet the line through O perpendicular to the line of stroke. Let thepoint of intersection be N.

Draw a circle centre C and radius CN.

Draw a circle with CP as diameter.

Let the common cord cut the line CP at L and the line of stroke PO at M.

Then the quadrilateral OCLM represents, to a certain scale, the acceleration diagram for

OCP. It can be shown that this scale is .

The Cewntripetal acceleration of the crank pin is

The piston acceleration is

CL is the Centripetal component and LM the Tangential component of the

acceleration of P relative to C, so that CM is the acceleration image of CP.

For any point Q on CP draw a line parallel to OP cutting CM in q. The

acceleration of in magnitude and direction.

Worked Examples

The workings associated with the following Examples have been "Hidden". Th view

them pease click om the red buttons.

Example 1:


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An aeroplane A flying at 180 m.p.h. in a direction North of West sights

another B due North of A. After 30 seconds flying B is seen to be in a North-Easterly direction from A and after a further 45 seconds B is directly astern of A.

If B is flying at a constant speed in a direction due South, find:- (a) The speed of

B (b)For how long B is within 2 miles of A (U.L.)

In the following diagram the Paths of A and B are shown. Their three particular

positions are respectively.

(23)

By scaling or by calculation and occupies 45 seconds of

flying time.

(24)

Relative displacement of B to A = displacement of B - displacement of A

(25)

In the first 30 seconds

Assuming that A is at rest at , the relative path of B is produced and a

circle centre and of radius 2 miles cuts this path at 1 and m. Between 1 and m B

lies within 2 miles of A.

(26)

(27)

This represents 30 seconds of relative displacement
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(28)

Example 2:

The quick return mechanism for a shaping machine is shown in the diagram. The

upper end of the slotted lever is pin jointed to the ram and tool box at at D so thatthis point moves in a horizontal straight line whilst the lower end slides over a

block at C mounted on trunnion bearings.

It is driven by the crank BA which turns at uniform angular velocity about thefixed centre B, through the slide block at A. If the ratio of the lengths BA/BC is

denoted by k and is the angle BA has turned from the upwards vertical, showfrom the geometry that the displacement of the tool box from its mid point is :-

(29)And determine the corresponding velocity and acceleration.
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From the diagram it can be seen that:-

(30)

(31)

(32)

Combining equations (31) and (33)

(33)

Differentiating to find the velocity:-

(34)

(35)

The second differentiation gives the acceleration:-

(36)

(37)

At the given values:-

(38)

(39)

(40)

(41)
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(42)

Example 3:

In the mechanism shown in the diagram, the crank AC is 5 in. long and rotatesclockwise about a centre A with a speed of 100 r.p.m.. The slotted lever BC

rotates about a fixed centre B, 10 in. vertically below A. Its centre of gravity G is

9 in. from B, its weight 25 lb. and its radius of gyration about G is 8 in. For theposition shown in which the angle BAC is

(43)

determine graphically the angular velocity of BC, the linear velocity of G, the

velocity of the sliding block in slot, the angular velocity of the pin at C relative tothe block. Also calculate the kinetic energy stored in the lever BC.

Let the point at the end of the crank be and the "coincident" point on BC be

Note that these two points are slidding over each other in the direction of the

common Tangent BC

Velocity of

(44)

The Construction of the vector diagram (b)


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Draw perpendicular to AC to represent

From draw a line parallel to CB to represent the velocity of

sliding of relative to .

From O draw a line perpendicular to BC to represent the velocity

of which is turning about the fixed centre at B.

The intersection of these last two lines gives the point

The angular velocity of BC

(45)

Velocity of G

(46)

(47)

Angular Velocity of pin relative to the block

(48)

kinetic Energy;storedTotal K.of BC = Linear + Angular Energy

(49)

(Note For any of you not used to the Imperial system of measurements, the 32.2 is

the acceleration due to gravity in ft./sec sq. and the 12 converts this to in/sec,sq.)

Example 4:

In the linkwork shown in the following Diagram, the crank AB rotates about A at

a uniform speed of 120 r.p.m. The lever DC oscillates about the fixed point D

being connected to AB by the coupler BC. The block F moves in horizontalguides and is driven by the link EF.


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When the angle determine (a) the velocity of F (b) the angular velocity of DCand (c) the rubbing speed at the pin C which is 2 in. diameter. (U.L.)

AB = 6 in. BC = 18 in. CD = 18 in. DE = 6 in. and Ef = 15 in.

The configuration of the links when is shown in diagram (a). The velocity

diagram (b) is drawn as follows:-

and is drawn perpendicular to

AB

where bc isperpendicular to BC

B must move perpendicular to DC and hence its absolute velocity is oc which isfound by intersection with bc.

D is a fixed point and coincides with o. Hence co is the velocity image of CD. e

divides co in the same proportion as E divides CD.

i.e.

The velocity of E (ef) is perpendicular to EF, but F must move horizontally, hence

the point f is determined.
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Velocity diagram

Answers
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anti

-clockwise.

This is anti-clockwise movement since C, relative to B, is moving "upwards".

The rubbing velocity at C = Radius of pin X difference of angular velocity = 0

Example 5:

In the mechanism shown in the diagram, the crank rotates at a constant

speed of 60 r.p.m., in a clockwise direction, imparting a vertical reciprocating

motion to the rack R, by means of the toothed quadrant Q. are fixed

centres and the slotted bar BC and the quadrant Q rock on .

Determine (aq) the linear speed of the rack when the angle

(b)The ratio of the times of lowering and raising the rack and (c) The length of the

stroke of the rack. (U.L.)

Example 6:

The diagram shows a mechanism in which the crank AB turns uniformly at 180

r.p.m., the blocks at D and E working in frictionless guides. AB = 1.5 ft.; BD = 5

ft. ; BC = 3 ft. ; CE = 3 ft.

Draw the velocity vector diagram and state the velocities of the blocks D and E in

their guides.


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Find the turning Moment at A if a force of 100 lb. acts on D in the direction of the

arrow X and a force of 150 lb. acts on E in the direction of arrow Y (U.L.)

The mechanism is drawn to scale in the following diagram

The Velocity diagram, drawn to scale

(54)

The relative velocity bd is perpendicular to BD and od is parallel to

the movement of D

and the relative velocity ce is perpendicular to CE.

oe is in the direction of motion of E.
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Velocity of D = od = 31.4 ft./sec. to the right.

Velocity of E = oe = 5.5 ft./sec. upwards.

Power input = Force X velocity.The Force at E is opposing motion

(55)

Thus the Power Input = The power output ( Neglecting losses)

(56)

(57)

Example 7:

For the engine shown in the Diagram, the crank radius CB is 2.25 in. and thelength of the connecting rod Ab is 9.25 in. between centres.

The centre of gravity of the rod is at G which is 3 in. from B. The engine speed is1200 r.p.m. For the position shown, in which CB is turned from CA, findgraphically the velocity of G and the angular acceleration of AB. Indicate the

direction of each of these values.

The velocity of B, . Note that v_Bis perpendicular to CB

Draw cb to represent v_B to scale

The velocity of A relative to B must be perpendicular to BA and is

represented by ba. A must move in the direction CA and hence its absolute velocity is

represented by ca
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ba is the velocity image of BA and g divides it in the ratio BG/BA.The line blm is used for this construction.

The Angular velocity of AB is given by:-

(58)

The acceleration is anti-clockwise since the velocity of A relative to B is parallelto ba.

The Acceleration of

cb is drawn on the acceleration diagram (b) to represent the

acceleration of B

The acceleration of A relative to B has the following components.

Centripetal represented

by ba' in the direction of AB. Tangentiallyrepresented by a'a

perpendicular to AB. The Absolute acceleration A is ca parallel to CA. abis the image of AB and g divides it in the ratio BG/BA as before.

The acceleration of

The acceleration of

clockwise since the

tangential acceleration of A relative to B is parallel to a'a

Example 8:


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In a four bar chain ABCD, A and D are fixed centres 2.5 in. apart on a horizontal

line. The driving crank AB = 1 in., the driven crank DC = 1.5 in. and the coupler

BC = 1.5 in. with its centre of gravity G at 0.5 in. from C. When AB is turned

through anti-clockwise from AD, B and C are on the same side as AD. If, for

this position, the angular velocity of AB is 20 rad. / sec. anti- clockwise, find the

angular velocity of BC and DC and the linear velocity of G.If also for this position the angular acceleration of AB is anti-

clockwise, find the angulkar acceleration of BC and DC and the linear

acceleration of G.

On the velocity Diagram (a) let I be the instantaneous centre of BC.

(59)

(60)

(61)

(62)

On the acceleration Diagram (b)XXXXXXXXXX

(63)

(64)

(65)


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(66)

(67)

(68)

(69)

(70)

The point c is found from the intersection of the two lines of unknown magnitude.

The Angular acceleration of BC is given by:-

(71)Similarly the Angular acceleration of CD

(72)

The linear acceleration of G = og where

Example 9:

The diagram shows a slider weighing 4 lb. moving along horizontal guides whilst

another slider B, weighing 6 lb. moves in vertical guides. The two sliders areconnected by a light connecting-rod 8 in. long/ The coefficient of friction at the

sliding faces is 0.08 whilst the friction at the turning pairs may be neglected.
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Find the horizontal force P required to drive the mechanism, at an instant when

, the velocity of A being then 3 ft/sec to the right and theacceleration of A is 5 ft/sec.sq. to the right. (U.L.)

I is the instantaneous centre of AB and hence the angular velocity of AB is as follows:-

(73)The following is the acceleration diagram.

(74)

(75)

(76)
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(77)

(78)

Total reaction at inclined at a friction angle

to the normal.

Resolving perpendicular to

(79)

(80)

Similarly at A the inertia force to the left and is the total

reaction.

Resolving perpendicularly to for A.

Substituting in the value of T from above:-

Example 10:

If in a mine shaft at the equator, the mine cage is allowed to drop freely down theshaft, all frictional effects being ignored, a side thrust develops between the cage

and the shaft wall. Derive an expression for this side thrust in terms of the cage

weight W ad the period t during which the cage has dropped. Assume tat the earth

rotates once in 24 hours.Hence if the mine cage is replaced by a small heavy weight, dropped from ground

level from the exact centre of the circular cross section of the shaft, determine the

position at which the weight first touches the side or bottom of the shaft. The shaftis 20 ft. diameter and 10,000 ft. deep. Note the radius of the Earth is irrelevant.

(U.L.)

The cage is subjected to the Coriolis component of acceleration which is

determined by its linear velocity and the angular rotation of the Earth. After a

time t the velocity of the cage = gt and the angular velocity of the Earth
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(81)

Side thrust = mass X acceleration

(82)Note that any point which moves in a radial direction towards the centre of the

Earth will have a tangential component of acceleration . But a weight

dropped from the centre of the shaft can not be acted upon by any tangential force

and hence its initial tangential acceleration is zero. It will therefore deviate froman imaginary point (moving with an equal radial velocity v) at a rate of tangential

acceleration of subject to the condition that that the rate of deviation is zero

at the instant of release.Since acceleration is the second differential of displacement, the tangential

displacement after a time t from release.

(83)Substituting from equation (82)

(84)

But since the radius of the shaft is 10 ft. and assuming that the weight hits the

walls before the bottom, it is possible to solve the above equation for t.

(85)

The vertical distance fallen

(86)

Example 11:

In the crank and slotted lever mechanism shown in the diagram, the crank OP is

driven at a uniform speed of

(87)

radians per second. If OL is the perpendicular from O onto XQ, the centre line ofthe slotted lever, prove that the angular acceleration of the slotted lever is given

by:-


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(88)

Hence or otherwise find the acceleration of the point Q in magnitude and

direction when the crank angle

(U.L.)

From the diagram:-

(89)Differentiate w.r.t.t.

(90)

(91)

(92)

(93)

Differentiating equation (92)


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(94)

Substituting for from equation (94) and making use of the geometry of the

mechanism, this reduces to:-

(95)

(96)From equation (90)

(97)

(98)

(99)

(100)

Hence

(101)

(102)

(103)

Using equation (94)

(104)

From equation (97)

(105)

The Acceleration components of Q are:-

(106)

(107)


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(108)

(109)

Example 12:

The diagram shows a link mechnanism which the link OA rotates uniformly in an

anti-clockwise direction at 10 rads. /sec. and AC is pivoted at B. The length of the

various links are OA = 3 in.; OB = 6 in.,; BC= 6 in.; CD = 12 in.

Determine, for the position shown, the instantaneous acceleration of D. (U.L.)

Let be the point at the end of the crank OA and let be the coincident pointon the link ABC.

The velocity diagram
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(110)

(111)

(112)

(113)

(114)


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(115)

(116)

(117)

The velocity of sliding v is given by:-

(118)

Acceleration diagram

(119)

(120)

The tangential acceleration of perpendicular to AB

and of unknown magnitude.

The acceleration of relative to has two components,


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in the same direction as is moving outwards relative to

along the direction BA. This is drawn as perpendicula to BA and

the intersection of components along BA and

perpendicula to . Then is the acceleration

image of ABC

(121)

(122)

= The tangential acceleration of D to C (Perpendicular to CD)

(123)

Analysis of Velocity and Accelerations Components

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