Theory of Machine-I Lab DBATU manual.pdf

Theory of Machine-I Lab Manual S.Y.BTech-II
Prof. Sachin M. Shinde, KITSP Page 1
Experiment no 1
Title: Introduction of lab
Aim: To study the introduction of lab
Apparatus Required
Figure 1.1 Lab layout
1. Motorized Gyroscope Apparatus
The equipment is used for various controls in high technology vehicles. For minimizing
rolling yawing and pitching of a ship or aircraft Gyroscope is used. Balloons use gyroscope
for controlling direction, Gyroscope has applications in spacecraft, space platforms. Thus the
study of Gyroscope is of high importance for any engineering student. The motorized
Gyroscope consists of a disc rotor mounted on a horizontal shaft rotating in the ball bearings
of one frame. The rotor shaft is coupled to a motor mounted on a trunion frame having
bearings in a yoke frame which is free to rotate about the vertical axis. Thus the disc can be
rotated about three perpendicular axes. Angular scale and pointer fitted to frame help to
measure precision rate.
Features:
• Demonstration of the relationship between applied torque and precession rate along with the
direction of rotation.
• Gyroscopic couple easily varied by weights.
• Precisely balanced rotor.
• Useful to verify the relationship #T = I .w.wp
Specifications:

Prof. Sachin M. Shinde, KECSP Page 2
• Disc rotor - 30 cm. dia, 10mm thick. approx.
• Drive - FHP, AC/DC single phase motor
• Dimmerstat - 2 Amps.
• Weight set.
Experimentation:
• Observation of Gyroscopic behavior (Two lows of stability)
• Experiment justification of the education = lw.wp for calculating the Gyroscopic couple by
• observation and measurement of results of independent variations in applied couple T and
precession wp.
2. Universal Governor Apparatus
The equipment is designed to study various governors. They are normally used to control the
speed. The unit consists of the main spindle driven by a variable speed motor mounted on a
base plate. The spindle is driven by a belt and pulley system. The optional governor out of
four governor assemblies can be mounted on a spindle. The spindle speed can be controlled
by the speed control unit provided. A graduated scale is provided to measure the
displacement of the sleeve.
The center sleeve of the portal and proell governors incorporates a weight sleeve to which
weight may be added. The Hartnell governor provides means of varying spring compression
level. This enables the Hartnell governor to be operated as a stable or unstable governor.
Specifications:
• Drive unit - PMDC motor, 1/6 HP, 1500 RPM
• Speed control unit working on 230V AC supply with 0-200V DC output
• Belt and Pulley system to give spindle speed 100 to 500 rpm.
• Governor Mechanism with springs and weights for following Governors.
• Watt Governor
• Porter Governor
• Hartnell Governor
• Proell Governor
Experimentation:
• For all types of Governors.
• Determination of characteristic curve of sleeve position against the speed of the spindle.
• Derivation of actual controlling force curves from the above and comparison with
theoretically predicted force curve.
• Porter and Proell Governor - The effect of the varying mass of sleeve.

 Hartnell Governor - The effect of varying initial spring compression.
3. Static & Dynamic Balancing Apparatus
The equipment is cable to experiment balance a rotating mass system and to verify the
analytical relations. The equipment consists of a steel shaft fixed in a rectangular frame. A set
of four blocks with a clamping arrangement is provided. For static balancing, each block is
individually clamped on the shaft and its relative weight is found out using a cord and
container system in terms of the number of steel balls.
For dynamic balancing, a moment polygon is drawn using relative weights and angular and
axial positions of blocks are determined. The blocks are clamped on the shaft is rotated by a
motor to check the dynamic balance of the system.
The equipment is provided with angular and longitudinal scales and is suspended with chains
for dynamic balancing.
Specifications:
• Drive Motor, Universal Motor
• Balancing Weights - 4 Nos. with different size eccentric mass for varying unbalance.
• Cord & Container with steel balls for relative weight measurement.
Experimentation:
• Static balancing of a system using steel balls.
• Dynamic balancing of a simple rotating mass system.
• Observation of the effect of unbalance in a rotating mass system.
4. Universal vibration machine
The equipment has been designed to perform and verify the principals involved in the study
of vibrations. It consists of a basic frame and various components and sub-assembled
designed for quick changeover and easy assembly of the number of experiments in vibration.
A speed control unit is provided for forced vibration experiments. With this unit following 8
experiments can be conducted.
5. Gear tooth profile generator
This setup is used to draw a gear tooth profile.
6. Models of clutch and brakes
It includes all types of demonstration models of clutch and brakes which are helpful to
understand construction and working.
7. Models of Dynamometer.
It includes all types of demonstration models of dynamometers which are helpful to
understand construction and working.

Conclusion:
Construction and working of all machinery successfully studied and understood

Experiment no 2
Title: Graphical solution of problems on velocity, acceleration in mechanisms by relative
velocity method.
Aim: to study graphical solution of problems on velocity, acceleration in mechanisms by
relative velocity method.
Example 2.1. The crank of a slider crank mechanism rotates clockwise at a constant speed of
300 r.p.m. The crank is 150 mm and the connecting rod is 600 mm long. Determine: 1. Linear
velocity and acceleration of the midpoint of the connecting rod, and 2. angular velocity and
angular acceleration of the connecting rod, at a crank angle of 45° from inner dead centre
position.
Solution. Given : NBO = 300 r.p.m. or ωBO = 2 π × 300/60 = 31.42 rad/s; OB = 150 mm =
0.15 m ; BA = 600 mm = 0.6 m
We know that linear velocity of B with respect to O or velocity of B,
vBO = vB = ωBO × OB = 31.42 × 0.15 = 4.713 m/s...(Perpendicular to BO)
Figure 2.1
1. Linear velocity of the midpoint of the connecting rod
First of all draw the space diagram, to some suitable scale; as shown in Fig. 2.1 (a). Now the
velocity diagram, as shown in Fig. 2.1 (b), is drawn as discussed below:
1. Draw vector ob perpendicular to BO, to some suitable scale, to represent the velocity of B
with respect to O or simply velocity of B i.e. vBO or vB, such that vector ob = vBO = vB =
4.713 m/s
2. From point b, draw vector ba perpendicular to BA to represent the velocity of A with
respect to B i.e. vAB , and from point o draw vector oa parallel to the motion of A (which is
along AO) to represent the velocity of A i.e. vA. The vectors ba and oa intersect at a.
By measurement, we find that velocity of A with respect to B,
vAB =vector ba =3.4 m/ s

and Velocity of A , vA = vector oa = 4 m/ s
3. In order to find the velocity of the midpoint D of the connecting rod AB, divide the vector
ba at d in the same ratio as D divides AB, in the space diagram. In other words,
bd / ba = BD/BA
Note: Since D is the midpoint of AB, therefore d is also midpoint of vector ba.
4. Join od. Now the vector od represents the velocity of the midpoint D of the connecting
rod i.e. vD.
By measurement, we find that
vD = vector od = 4.1 m/s Ans.
Acceleration of the midpoint of the connecting rod
We know that the radial component of the acceleration of B with respect to O or the
acceleration of B,
and the radial component of the acceleration of A with respect to B
Now the acceleration diagram, as shown in Fig. 2.1 (c) is drawn as discussed below:
1. Draw vector o' b' parallel to BO, to some suitable scale, to represent the radial component
of the acceleration of B with respect to O or simply acceleration of B i.e. ar
BO or aB such that
vector ol
bl
= ar
BO or aB = 148.1 m/s2
Note: Since the crank OB rotates at a constant speed, therefore there will be no tangential
component of the acceleration of B with respect to O.
2. The acceleration of A with respect to B has the following two components:
(a) The radial component of the acceleration of A with respect to B i.e. ar
AB, and
(b) The tangential component of the acceleration of A with respect to B i.e. at
AB, these two
components are mutually perpendicular.
Therefore from point b', draw vector b' x parallel to AB to represent ar
AB =19.3 m/s2
from
point x draw vector xa' perpendicular to vector b' x whose magnitude is yet unknown.
3. Now from o', draw vector o' a' parallel to the path of motion of A (which is along AO) to
represent the acceleration of A i.e. aA . The vectors xa' and o' a' intersect at a'. Join a' b'.
4. In order to find the acceleration of the midpoint D of the connecting rod AB, divide the
vector a' b' at d' in the same ratio as D divides AB. In other words

Note: Since D is the midpoint of AB, therefore d' is also midpoint of vector b' a'.
5. Join o' d'. The vector o' d' represents the acceleration of midpoint D of the connecting rod
i.e. aD. By measurement, we find that
aD = vector o' d' = 117 m/s2
Ans.
2. Angular velocity of the connecting rod
We know that angular velocity of the connecting rod AB,
Angular acceleration of the connecting rod
From the acceleration diagram, we find that
We know that angular acceleration of the connecting rod AB,
Example 2.2. An engine mechanism is shown in Fig. 2.2. The crank CB = 100 mm and the
connecting rod BA = 300 mm with centre of gravity G, 100 mm from B. In the position
shown, the crankshaft has a speed of 75 rad/s and an angular acceleration of 1200 rad/s2
.
Find:1. velocity of G and angular velocity of AB, and 2. acceleration of G and angular
acceleration of AB.
Figure 2.2
Solution. Given : ωBC = 75 rad/s ; αBC = 1200 rad/s2
, CB = 100 mm = 0.1 m; BA = 300 mm
= 0.3 m We know that velocity of B with respect to C or velocity of B,
vBC = vB = ωBC x CB= 75 x 0.1 =7.5 m/s
Since the angular acceleration of the crankshaft, αBC = 1200 rad/s2
, therefore tangential
component of the acceleration of B with respect to C,

1. Velocity of G and angular velocity of AB
First of all, draw the space diagram, to some suitable scale, as shown in Fig. 2.3 (a). Now the
1. Draw vector cb perpendicular to CB, to some suitable scale, to represent the velocity of
B with respect to C or velocity of B (i.e. vBC or vB ), such that
2. From point b, draw vector ba perpendicular to BA to represent the velocity of A with
respect to B i.e. vAB , and from point c, draw vector ca parallel to the path of motion of A
(which isalong AC) to represent the velocity of A i.e. vA.The vectors ba and ca intersect at a.
3. Since the point G lies on AB, therefore divide vector ab at g in the same ratio as G divides
AB in the space diagram. In other words,
The vector cg represents the velocity of G.
By measurement, we find that velocity of G,
vG = vector cg = 6.8 m/s Ans.
From velocity diagram, we find that velocity of A with respect to B,
vAB = vector ba = 4 m/s
We know that angular velocity of AB,
Figure 2.3

2. Acceleration of G and angular acceleration of AB
We know that radial component of the acceleration of B with respect to C,
and radial component of the acceleration of A with respect to B,
Now the acceleration diagram, as shown in Fig. 2.3 (c), is drawn as discussed below:
1. Draw vector c' b'' parallel to CB, to some suitable scale, to represent the radial component
of the acceleration of B with respect to C,i.e, ar
BC such that
2. From point b'', draw vector b'' b' perpendicular to vector c' b'' or CB to represent the
tangential component of the acceleration of B with respect to C i.e. at
BC , such that
3. Join c' b'. The vector c' b' represents the total acceleration of B with respect to C i.e. aBC.
4. From point b', draw vector b' x parallel to BA to represent radial component of the
acceleration of A with respect to B i.e. ar
BC a such that
5. From point x, draw vector xa' perpendicular to vector b'x or BA to represent tangential
component of the acceleration of A with respect to B i.e. at
AB ,a whose magnitude is not yet
known.
6. Now draw vector c' a' parallel to the path of motion of A (which is along AC) to represent
the acceleration of A i.e. aA. The vectors xa' and c'a' intersect at a'. Join b' a'. The vector b'
a' represents the acceleration of A with respect to B i.e. aAB.
7. In order to find the acceleration of G, divide vector a' b' in g' in the same ratio as G divides
BA in Fig. 2.3 (a). Join c' g'. The vector c' g' represents the acceleration of G.
By measurement, we find that acceleration of G,
aG = vector c' g' = 414 m/s2
Ans.
From acceleration diagram, we find that tangential component of the acceleration of A with
respect to B,

∴ Angular acceleration of AB,
Conclusion
The graphical solution of velocity and acceleration by relative method successfully studied.

Experiment 3
Title: Graphical solution of problems on instantaneous center of rotation method
Aim: to study graphical solution of problems on instantaneous center of rotation method
Example 3.1. In a pin jointed four bar mechanism, as shown in Fig. 3.1, AB = 300 mm, BC
= CD = 360 mm, and AD = 600 mm. The angle BAD = 60°. The crank AB rotates uniformly
at 100 r.p.m. locates all the instantaneous centres and find the angular velocity of the link BC.
Figure 3.1
Solution. Given : NAB = 100 r.p.m or ωAB = 2 π × 100/60 = 10.47 rad/s
Since the length of crank AB = 300 mm = 0.3 m,
therefore velocity of point B on link AB,
Location of instantaneous centres
The instantaneous centres are located as discussed below:
1. Since the mechanism consists of four links (i.e. n = 4 ), therefore number of instantaneous
centres,
2. For a four bar mechanism, the book keeping table may be drawn as discussed in Art. 2.2.
3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I
12,I23, I34 and I14, as shown in Fig. 3.2.
4. Locate the remaining neither fixed nor permanent instantaneous centres by Aronhold
Kennedy’s theorem. This is done by circle diagram as shown in Fig. 2.3. Mark four points
(equal to the number of links in a mechanism) 1, 2, 3, and 4 on the circle.

Figure 3.2
Figure 3.3
5. Join points 1 to 2, 2 to 3, 3 to 4 and 4 to 1 to indicate the instantaneous centres already
located i.e. I12, I23, I34 and I14.
6 Join 1 to 3 to form two triangles 1 2 3 and 3 4 1. The side 13, common to both triangles,
is responsible for completing the two triangles. Therefore the instantaneous centre I13 lies on
the intersection of the lines joining the points I12I23 and I34 I14 as shown in Fig. 2.2. Thus
centre I13 is located. Mark number 5 (because four instantaneous centres have already been
located)on the dotted line 1 3.
7. Now join 2 to 4 to complete two triangles 2 3 4 and 1 2 4.The side 2 4, common to both
triangles, is responsible for completing the two triangles. Therefore centre I24 lies on the
intersection of the lines joining the points I23 I34 and I12 I14 as shown in Fig. 2.2. Thus centre I
24 is located. Mark number 6 on the dotted line 2 4. Thus all the six instantaneous centres are
located.
Angular velocity of the link BC
Let ωBC = Angular velocity of the link BC.
Since B is also a point on link BC, therefore velocity of point B on link BC,

vB = ωBC × I13B
By measurement, we find that I13B = 500 mm = 0.5 m
Example 3.2. Locate all the instantaneous centres of the slider crank mechanism as shown in
Fig.3.4. The lengths of crank OB and connecting rod AB are 100 mm and 400 mm
respectively. If the crank rotates clockwise with an angular velocity of 10 rad/s, find: 1.
Velocity of the slider A, and 2. Angular velocity of the connecting rod AB.
Figure 3.4
Solution. Given : ωOB = 10 rad/ s; OB = 100 mm = 0.1 m
We know that linear velocity of the crank OB,
vOB = vB = ωOB × OB = 10 × 0.1 = 1 m/s
Location of instantaneous centres
The instantaneous centres in a slider crank mechanism are located as discussed below:
1. Since there are four links (i.e. n = 4), therefore the number of instantaneous centres,
2. For a four link mechanism, the book keeping table may be drawn as discussed in Art.1.2.
3. Locate the fixed and permanent instantaneous centres by inspection. These centres are
I12,I23 and I34 as shown in Fig. 3.5. Since the slider (link 4) moves on a straight surface (link
1), therefore the instantaneous centre I14 will be at infinity.
Note: Since the slider crank mechanism has three turning pairs and one sliding pair, therefore
there will be three primary (i.e. fixed and permanent) instantaneous centres.
4. Locate the other two remaining neither fixed nor permanent instantaneous centres, by

Aronhold Kennedy’s theorem. This is done by circle diagram as shown in Fig.3.5. Mark four
points 1, 2, 3 and 4 (equal to the number of links in a mechanism) on the circle to indicate I12,
I23, I34 and I14.
Figure 3.5
Figure 3.6
5. Join 1 to 3 to form two triangles 1 2 3 and 3 4 1 in the circle diagram. The side 1 3,
common to both triangles, is responsible for completing the two triangles. Therefore the
centre I13 will lie on the intersection of I12 I23 and I14 I34, produced if necessary. Thus centre
I13 is located. Join 1 to 3 by a dotted line and mark number 5 on it.

6. Join 2 to 4 by a dotted line to form two triangles 2 3 4 and 1 2 4. The side 2 4, common to
both triangles, is responsible for completing the two triangles. Therefore the centre I24 lies on
the intersection of I23 I34 and I12 I14. Join 2 to 4 by a dotted line on the circle diagram and
mark number 6 on it. Thus all the six instantaneous centres are located.
By measurement, we find that
I13 A = 460 mm = 0.46 m ; and I13 B = 560 mm = 0.56 m
1. Velocity of the slider A
Let vA = Velocity of the slider A.
We know that
2. Angular velocity of the connecting rod AB
Let ωAB = Angular velocity of the connecting rod AB.
We know that
Conclusion
The graphical solution of velocity and acceleration by Instantaneous center method
successfully studied.

Experiment no 4
Title: Graphical solution of problems on Klein’s construction
Aim: To study graphical solution of problems on Klein’s construction
Example 4.1. The crank and connecting rod of a reciprocating engine are 200 mm and 700
mm respectively. The crank is rotating in clockwise direction at 120 rad/s. Find with the help
of Klein’s construction: 1. Velocity and acceleration of the piston, 2. Velocity and
acceleration of the midpoint of the connecting rod, and 3. Angular velocity and angular
acceleration of the connecting rod, at the instant when the crank is at 30° to I.D.C. (inner
dead centre).
Solution. Given: OC = 200 mm = 0.2 m ; PC = 700 mm = 0.7 m ; ω = 120 rad/s
Figure 4.1
The Klein’s velocity diagram OCM and Klein’s acceleration diagram CQNO as shown in
Fig. 4.1 is drawn to some suitable scale, By measurement, we find that
OM = 127 mm = 0.127 m ; CM = 173 mm = 0.173 m ; QN = 93 mm = 0.093 m ; NO = 200
mm = 0.2 m
1. Velocity and acceleration of the piston
We know that the velocity of the piston P,
vP = ω × OM = 120 × 0.127 = 15.24 m/s Ans.
and acceleration of the piston P,
aP = ω2
× NO = (120)2
× 0.2 = 2880 m/s2
Ans.
2. Velocity and acceleration of the mid-point of the connecting rod

In order to find the velocity of the mid-point D of the connecting rod, divide CM at D1 in the
same ratio as D divides CP. Since D is the mid-point of CP, therefore D1 is the mid-point of
CM, i.e.CD1 = D1M. Join OD1. By measurement,
OD1 = 140 mm = 0.14 m
∴ Velocity of D, vD = ω × OD1 = 120 × 0.14 = 16.8 m/s Ans.
In order to find the acceleration of the mid-point of the connecting rod, draw a line DD2
parallel to the line of stroke PO which intersects CN at D2. By measurement,
OD2 = 193 mm = 0.193 m
∴ Acceleration of D,
aD = ω2
× OD2 = (120)2
× 0.193 = 2779.2 m/s2
Ans.
3. Angular velocity and angular acceleration of the connecting rod
We know that the velocity of the connecting rod PC (i.e. velocity of P with respect to C),
vPC = ω × CM = 120 × 0.173 = 20.76 m/s
∴ Angular acceleration of the connecting rod PC,
We know that the tangential component of the acceleration of P with respect to C
∴ Angular acceleration of the connecting rod PC,
Example 4.2.The crank and connecting rod of a reciprocating engine are 150 mm and 600
mm respectively. The crank makes an angle of 60° with the inner dead centre and revolves at
a uniform speed of 300 r.p.m. Find, by Klein’s construction, 1. Velocity and acceleration of
the piston, 2. Velocity and acceleration of the mid-point D of the connecting rod, and 3.
Angular velocity and angular acceleration of the connecting rod.[Ans. 4.6 m/s, 61.7 m/s2 ;
4.6 m/s, 93.8 m/s2 ; 4.17 rad/s, 214 rad/s2]
Example 4.3. In a slider crank mechanism, the length of the crank and connecting rod are
100 mm and 400 mm respectively. The crank rotates uniformly at 600 r.p.m. clockwise.
When the crank has turned through 45° from the inner dead centre, find, by analytical

method: 1. Velocity and acceleration of the slider, 2. Angular velocity and angular
acceleration of the connecting rod. Check your result by Klein’s construction. [Ans. 5.2 m/s;
279 m/s2; 11 rad/s; 698 rad/s2]
Conclusion
The graphical solution of velocity and acceleration by Klein’s construction method

Experiment no 5
Title: Graphical solution of problems on Corioli’s component of acceleration
Aim: To study graphical solution of problems on Corioli’s component of acceleration
Theory: Coriolis Component of Acceleration
When a point on one link is sliding along another rotating link, such as in quick return motion
mechanism, then the coriolis component of the acceleration must be calculated.
Consider a link OA and a slider B as shown in Fig. 5.1 (a). The slider B moves along the link
OA. The point C is the coincident point on the link OA.
Let ω = Angular velocity of the link OA at time t seconds.
v = Velocity of the slider B along the link OA at time t seconds.
ω.r = Velocity of the slider B with respect to O (perpendicular to the link OA) at time t
seconds, and
(ω + δω), (v + δv) and (ω + δω) (r + δr) = Corresponding values at time (t + δt) seconds.
Figure 5.1
Let us now find out the acceleration of the slider B with respect to O and with respect to its
coincident point C lying on the link OA. Fig. 5.1 (b) shows the velocity diagram when their
velocities v and (v + δv) are considered. In this diagram, the vector bb1 represents the change
in velocity in time δt sec ; the vector bx represents the component of change of velocity bb1
along OA (i.e. along radial direction) and vector xb1 represents the component of change of
velocity bb1 in a direction perpendicular to OA (i.e. in tangential direction). Therefore
Since δθ is very small, therefore substituting cos δθ = 1, we have

Since δθ is very small, therefore substituting sin δθ =δθ, we have
Neglecting δv.δθ being very small, therefore
Fig. 5.1 (c) shows the velocity diagram when the velocities ω.r and (ω + δω) (r + δr) are
considered. In this diagram, vector bb1 represents the change in velocity ; vector yb1
represents the component of change of velocity bb1 along OA (i.e. along radial direction) and
vector by represents the component of change of velocity bb1 in a direction perpendicular to
OA (i.e. in a tangential direction). Therefore
Since δθ is very small, therefore substituting sin δθ = δθ in the above expression, we have
Since δθ is small, therefore substituting cos δθ = 1, we have
by = ω.r + ω.δr + δω.r + δω.δr – ω.r= ω.δr + r.δω ...(Neglecting δω.δr)...(Perpendicular to OA
and towards left)
Therefore, total component of change of velocity along radial direction
∴ Radial component of the acceleration of the slider B with respect to O on the link OA,
acting radially outwards from O to A,
Also, the total component of change of velocity along tangential direction
∴ Tangential component of acceleration of the slider B with respect to O on the link OA,

acting perpendicular to OA and towards left,
Now radial component of acceleration of the coincident point C with respect to O, acting in
a direction from C to O,
and tangential component of acceleration of the coincident point C with respect to O, acting
in a direction perpendicular to CO and towards left,
Radial component of the slider B with respect to the coincident point C on the link OA,
acting radially outwards,
and tangential component of the slider B with respect to the coincident point C on the link
OA acting in a direction perpendicular to OA and towards left,
This tangential component of acceleration of the slider B with respect to the coincident point
C on the link is known as Corioli’s component of acceleration and is always perpendicular to
the link.
∴ Corioli’s component of the acceleration of B with respect of C,
Where ω = Angular velocity of the link OA, and
v = Velocity of slider B with respect to coincident point C.
In the above discussion, the anticlockwise direction for ω and the radially outward direction
for v are taken as positive. It may be noted that the direction of Corioli’s component of
acceleration changes sign, if either ω or v is reversed in direction. But the direction of
Corioli’s component of acceleration will not be changed in sign if both ω and v are reversed
in direction. It is concluded that the direction of Corioli’s component of acceleration is
obtained by rotating v, at 90°, about its origin in the same direction as that of ω.

Figure 5.2 Direction of Corioli’s component of acceleration.
The direction of Corioli’s component of acceleration (2 ω.v) for all four possible cases, is
shown in Fig. 5.2. The directions of ω and v are given.
Example 5.1 A mechanism of a crank and slotted lever quick return motion is shown in Fig.
5.3. If the crank rotates counter clockwise at 120 r.p.m., determine for the configuration
shown, the velocity and acceleration of the ram D. Also determine the angular acceleration of
the slotted lever. Crank, AB = 150 mm ; Slotted arm, OC = 700 mm and link CD = 200 mm.
Solution. Given : NBA = 120 r.p.m or ωBA = 2 π × 120/60= 12.57 rad/s ; AB = 150 mm =
0.15 m; OC = 700 mm = 0.7 m; CD = 200 mm = 0.2 m
Figure 5.3
We know that velocity of B with respect to A,

Velocity of the ram D
First of all draw the space diagram, to some suitable scale, as shown in Fig. 5.4 (a). Now the
1. Since O and A are fixed points, therefore these points are marked as one point in velocity
diagram. Now draw vector ab in a direction perpendicular to AB , to some suitable scale, to
represent the velocity of slider B with respect to A i.e.vBA, such that
vector ab = vBA = 1.9 m/s
Figure 5.4

2. From point o, draw vector ob' perpendicular to OB' to represent the velocity of coincident
point B' (on the link OC) with respect to O i.e. vB′O and from point b draw vector bb' parallel
to the path of motion of B' (which is along the link OC) to represent the velocity of
coincident point B' with respect to the slider B i.e. vB'B. The vectors ob' and bb' intersect at b'.
Note: Since we have to find the Corioli’s component of acceleration of the slider B with
respect to the coincident point B', therefore we require the velocity of B with respect to B'
i.e. vBB'. The vector b'b will represent vBB' as shown in Fig. 5.4 (b).
3. Since the point C lies on OB' produced, therefore, divide vector ob' at c in the same ratio as
C divides OB' in the space diagram. In other words,
The vector oc represents the velocity of C with respect to O i.e. vCO.
4. Now from point c, draw vector cd perpendicular to CD to represent the velocity of D with
respect to C i.e. vDC ,and from point o draw vector od parallel to the path of motion of D
(which is along the horizontal) to represent the velocity of D i.e. vD.The vectors cd and od
intersect at d.
By measurement, we find that velocity of the ram D,
vD = vector od = 2.15 m/s Ans.
From velocity diagram, we also find that
Velocity of B with respect to B',
vBB' = vector b'b = 1.05 m/s
Velocity of D with respect to C,
vDC = vector cd = 0.45 m/s
Velocity of B' with respect to O
vB′O = vector ob' = 1.55 m/s
Velocity of C with respect to O,
vCO = vector oc = 2.15 m/s
∴ Angular velocity of the link OC or OB',
Acceleration of the ram D
We know that radial component of the acceleration of B with respect to A,
Coriolis component of the acceleration of slider B with respect to the coincident point B',

Radial component of the acceleration of D with respect to C,
Radial component of the acceleration of the coincident point B' with respect to O,
Now the acceleration diagram, as shown in Fig. 5.4 (d), is drawn as discussed below:
1. Since O and A are fixed points, therefore these points are marked as one point in the
acceleration diagram. Draw vector a'b' parallel to AB, to some suitable scale, to represent the
radial component of the acceleration of B with respect to A i.e. BA ar
BA or aB, such that
2. The acceleration of the slider B with respect to the coincident point B' has the following
two components :
(i) Coriolis component of the acceleration of B with respect to B' i.e.
and
(ii) Radial component of the acceleration of B with respect to B' i.e. .
These two components are mutually perpendicular. Therefore from point b' draw vector b'x
perpendicular to B'O i.e. in a direction as shown in Fig. 5.4 (c) to represent direction of ac
BB
is obtained by rotating vBB′ (represented by vector b'b in velocity diagram) through 90° in the
same sense as that of link OC which rotates in the counter clockwise direction. Now from
point x, draw vector xb'' perpendicular to vector b'x (or parallel to B'O) to represent ar
BB
whose magnitude is yet unknown.
3. The acceleration of the coincident point B' with respect to O has also the following two
components:
(i) Radial component of the acceleration of coincident point B' with respect to O i.e.
(ii) Tangential component of the acceleration of coincident point B' with respect to O,i.e,

These two components are mutually perpendicular. Therefore from point o', draw vector o'y
parallel to B'O to represent and from point y draw vector yb''
perpendicular to vector o'y to represent The vectors xb'' and yb'' intersect at b''. Join
o'b''. The vector o'b'' represents the acceleration of B' with respect to O, i.e. aB′O.
4. Since the point C lies on OB' produced, therefore divide vector o'b'' at c' in the same ratio
as C divides OB' in the space diagram. In other words,
5. The acceleration of the ram D with respect to C has also the following two components:
(i) Radial component of the acceleration of D with respect to C i.e and
(ii) Tangential component of the acceleration of D with respect to C, i.e.
The two components are mutually perpendicular. Therefore draw vector c'z parallel to CD
to represent from z draw zd' perpendicular to vector zc' to represent
a whose magnitude is yet unknown.
6. From point o', draw vector o'd' in the direction of motion of the ram D which is along the
horizontal. The vectors zd' and o'd' intersect at d'. The vector o'd' represents the acceleration
of ram D i.e. aD.
By measurement, we find that acceleration of the ram D,
aD = vector o'd' = 8.4 m/s2
Ans.
Angular acceleration of the slotted lever
By measurement from acceleration diagram, we find that tangential component of the
coincident point B' with respect to O,
We know that angular acceleration of the slotted lever,
Conclusion
The graphical solution of Corioli’s component of acceleration successfully studied.

Experiment no 6
Title : Experimental determination of velocity and acceleration of Hooke’s joint.
Aim : To study Hook (Cardan) Joint or Universal Joint
1. Introduction:
The simplest means of transferring motion between non-axial shafts is by means of one or
two universal joints, also known as Cardan joints in Europe and Hook’s joints in Britain. The
shafts are not parallel to one another and they may be free to move relative to one another.
For this reason, this very simple spherical mechanism appears in an enormous variety of
applications. The most common application is Cardan joint used in the trucks as shown in
Figure 6.1. A universal joint is a simple spherical four-bar mechanism that transfers rotary
motion between two shafts whose axes pass through the concurrency points. The joint itself
consists of two revolute joints whose axes are orthogonal to one another. They are often
configured in a cross-shape member as shown in Figure 6.2.
2. Objective of Universal Joint
 Coupling is used to connect two intersecting shafts.  Consists of 2 yokes and a cross-link
Figure 6.1: Cardan Joint, used in Trucks
Figure 6.2: Schematic of Cardan Joint
The transmission behavior of this joint is described by Equation (5.1).

Where α2 is the momentary rotation angle of the driven shaft 2
Figure 6.3: Schematic of Cardan Joint
The angular velocity ratio can be described by:
where β is the angular misalignment of the shafts and Ө is the angle of the driving shaft.
It is noted that:
ω4/ω2 will vary between a minimum and a maximum during each revolution.
ω2 = ω4 only at 4 instants as shown in Figure 6.4.
This is a very big disadvantage, in case of automotive vehicles, this means that rear
wheels will rotate at variable speed.
Figure 6.4: Relation between ω4/ω2 and Ө
ω2 is constant (because engine and flywheel) speed of car cannot be variable each revolution
because its inertia, and so this means tires will slip and severe wear will happen.

The effect of the angle β can be shown by plotting ω4/ω2 for different values of β (Figure 5.5).
6
Figure 6.5: Relation between ω4/ω2 and β
The acceleration expression of the follower of universal joint for constant ω2 is given by:
For acceleration or retardation
-acceleration when 2Ө lies between 0 -180o
-deceleration when 2Ө lies between 180 -360o
Maximum acceleration occurs when dα4 / dӨ = 0. Thus
It is possible to connect 2 shafts by 2 Hook's couplings and an intermediate shaft that the
uneven velocity ratio of the first coupling will be cancelled out by the second Condition
(Figure 5.6).
Yoke 32 lies in plane containing shaft 2, 3
Yoke 34 lies in plane containing shaft 3, 4
Figure 6.6: Two shafts by two Hook's couplings and an intermediate shaft

2. Objective:
Testing the effect of changing of the angle between the shaft and universal joint on the input
and output shaft speed and calculate the speed ration between output and input shaft speed
3. Experimental Procedure
1) Fixing the input shaft angle β2 = 0o
.
2) Setting the output shaft angle β4 = 15o
.
3) Filter setting: cutoff freq. 4. Scope setting: ch1 500mv, ch2: 500mv, M 50ms. run scope,
turn on motor. After stable fluctuate signal appear on screen, stop scope then turn off motor.
4) Check the input and output shaft speed by using Cursor: paired. We assume that at θ = 0,
N4 is min. setting paired cursor at this point as t=0 ms. Moving cursor along ch2 signal and
record t and N4 by ∆: __ms, @:__v. Fill table 5-1. N2 is input shaft speed. It is constant.
5) Repeat step #2, setting the output shaft angle β4 = 30o
. Fill table 5-2.
6) Plot the graph of velocity ratio during one cycle for both cases.
Table 6. 1: Test results of Cardan joint β4 = 15o.
θ0
0 45 90 135 180 225 270 315 360
t (ms)
N4(v)
N2(v)
N4/N2
Table 6. 2: Test results of Cardan joint β4 = 30o.
θ0
0 45 90 135 180 225 270 315 360
t (ms)
N4(v)
N2(v)
N4/N2
Conclusion
Experimental determination of velocity and acceleration of Hooke’s joint.

Experiment No.-7
Title : Experiment on Corioli’s component of acceleration.
Aim: To find out experimentally the Corioli’s and component of acceleration and compare
with theoretical values.
Requirement: Corioli’s Component of Acceleration apparatus.
Theory:
Introduction: The apparatus has been designed to enable students to measure the various
parameters comprising the Corioli’s of Acceleration.
To maintain this acceleration long enough for measurements to be taken the conventional
slider mechanism is replaced by two streams of water flowing radially outwards from an
inverted ‘T’ shaped tube, which is rotated about its vertical axis so that the water in passing
along the tube is subjected to a Corioli’s Components of Acceleration. The total acceleration
of a point with respect to another point in a rigid link is the vector sum of its centripetal and
tangential components. This holds true when the distance between two points is fixed and the
relative acceleration of the two points on a moving rigid link has been considered. If the
distances between two points vary, that is the second point, which was stationary, now slides;
the total acceleration will contain one additional component, known as Corioli’s component
of acceleration.
A mechanism (shown in fig.) consisting translating pair i.e. blocks B, which is free to slide in
straight path fixed in direction. If the translating pair itself revolves, its acceleration will
include the Corioli’s component of acceleration due to change in relative distance between
two points.
Let link OA oscillate about the fixed center O with constant angular velocity ω, from OA to
OA’ in time dt, angle between OA and OA’ being dθ. The link consists of a slider B that
interval of time. Now the slider can be considered to have moved from B to E follows: From
B to C due to outward velocity v of the slider. D to E due to acceleration perpendicular to the
rod. The third movement of the slider is due to Corioli’s acceleration, which can be analyzed
as under: - Arc DE = arc EF – arc FD
= Arc EF – arc BC
= FO x dθ - BO x dθ
= dθ(FO – BO)
= BF x dθ
= CD x dθ
Now linear displacement

CD = v. dt
And angular displacement
dθ = ω. Dt
Arc DE = (v. dt) (ω. dt = v . ω . (dt)2
However, DE= ½ fcc
. (dt)2
(If fcc
the acceleration of the particle is constant).
∴ ½ fcc
. (dt)2
= (dt)2
or fcc
= 2 v . θ
This is the required Corioli’s component of acceleration and is always perpendicular to the
link.
Hydraulic Analogy:
Consider a short column of the fluid of length δr at distance r from the axis of rotation of the
tube. Then if the velocity of the fluid relative to the tube is v and the angular velocity of the
tube is ω the Corioli’s component of acceleration of the column is 2vω in a direction
perpendicular to, and in the plane of rotation of the tube. The torque δT applied by the tube to
produce this acceleration is then
Where δw is the weight of fluid of the short column. If (w) is the specific weight of the fluid
and (a ) is the cross-section area of the tube outlet, then: δw = wa x δr
and the complete torque applied to a column of length L is given by
T = 2vω – (w x a x L2
)/2g
T = (Cc x w x a x L2
)/g
or Coriolli’s of acceleration
Cc = 2 g T /Wa L2
(Considering both tubes)
Description:
The Apparatus consists of two stainless steel tunes, projecting radially from a central Perspex
tube are rotated by a DC swinging field motor, mounted vertically in a pillow blocks and
bearings. A spring balance attached to a fixed swinging field motor with a fixed armed length
measures the torque supplied by the motor. A digital rpm indicator is provided to measure the
speed of the motor. Water from the pump flows to Perspex tube through the control valve.
The water flow rate is measured with the help of rotameter. The water leaving the radial tube

circulate continuously by the water pump. The splash tank and all the accessories are
mounted on a fabricated M.S. frame.
Procedure:
1. Supply the main power to the control panel.
2. Switch on the motor and then increase the speed of the motor with the help of Variac up to
certain speed.
3. Now start the water pump and maintain a constant water level in the vertical Perspex tube
with the help of bye pass valve.
4. Note down the readings of spring balance for actual torque.
5. Note down the discharge and RPM from rotameter and digital RPM indicator respectively.
6. Now switch off the water pump and adjust the speed of the motor to its previous value.
7. Note down the reading on the spring balance for frictional torque.
8. Repeat the procedure for the different speeds and flow rates.
Formulae:
1. Corioli’s Component of acceleration (actual) = 2gt/ 2 x pw x a L2
m/s2
2. Torque, T = F x R kgm
3. Coriolli’s Component of acceleration (theo) = 2 ωv m/s2
4. Angular velocity, ω = 2 π N /60 rad/sec
Velocity, v = Q/2a m/s
Observation & Calculations: Data:
g = Acceleration due to gravity = 9.81 m/s2
pw = Density of water = 1000 kg/m3
d = Internal diameter of pipe = 6.0 mm = 6.0 x 10-3
m
a = Cross sectional area of pipe = 2.827 x 10-5 m2
L = Length of pipe = 300 mm = 0.300 m
R = Swinging field arm = 137 mm = 0.122 m
1 LPH = 2.77 x 10-7
m3
/sec
Observation Table:
S. No. Speed N LPH Force Kg
1
2
3

Calculations
S. No. Torque (T)
Kgm
Angular
Velocity (ω)
Rad/sec
Velocity of
Water (v) m/s
Cor. Comp. of
acceleration
(theo) m/sec2
Cor. Comp. of
acceleration
(act.) m/sec2
1
2
3
Precautions & Maintenance Instructions:
1. Don’t exceed the discharge more than 2500 LPH.
2. To control the overflow in the central tube, increase the speed of the motor simultaneously
as the discharge increases.
3. Bye pass valve should fully open before the experiment start.
4. Variac should be at zero position before starting the experiment.
Troubleshooting:
1. If pump gets jam, open the back cover of pump and rotate the shaft manually.
2. If pump gets heat up, switch off the main power for 15 minutes and avoid closing the flow
control valve and bye pass valve simultaneously during operation.
Result: Compared experimentally the Corioli’s component of acceleration with theoretical
values.

Experiment no 8
Title : Computer programming for velocity and acceleration of slider crank mechanism
Aim : To study Computer programming for velocity and acceleration of slider crank
mechanism
PROGRAM TO FIND THE VELOCITY AND ACCELERATION IN A SLIDER CRANK
MECHANISM
READ (*, 5) A, B, E, VA, ACC, THA 5
FORMAT (6F5.1)
PI= 4* ATAN (1.)
TH= 0
IH= 180/THA
DTH= PI/IH
DO 10 I=1 ,2 *IH
TH= (I-1)*DTH
BET= ASIN (E-A*SIN (TH) /B)
VS= -A*VA*SIN (TH-BET)/ (COS (BET) *1000)
VB= -A*VA*COS (TH)/B* COS (BET)
AC1=A*ACC*SIN (BET-TH) - B*VB**2
AC2= A*VA**2*COS (BET-TH)
ACS= (ACI-AC2)/ (COS (BET)* 1000)
AC3= A*ACC*COS (TH)-A*VA**2*SIN (TH)
AC4= B*VB**2*SIN (BET)
ACB= - (AC3-AC4)/ (B*COS (BET))
IF (I.EQ.1) WRITE (*, 9)
FORMAT (3X, ‘TH’, 5X, ‘BET’, 4X, ‘VS’, 4X, ‘VB’, 4X, ‘ACS’, 4X, ‘ACB’)
WRITE (*, 8) TH *180/PI, BET* 180/PI, VS, VB, ACS, ACB 8
FORMAT (6F8.2)
STOP
END

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