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• Transmission angle
• Instantaneous center method for Velocity
analysis
Summary of previous lecture
Today’s lecture
• Relative Velocity Method for Velocity analysis
• Acceleration in Mechanisms
Relative Velocity of Two Bodies
Moving in Straight Lines
(Velocity A w.r.t B)AB A Bv at v v 
(Velocity w.r.t )BA B Av at B A v v 
(i)
(ii) BA B Av v v 
AB A Bv v v 
Motion of a Link
Velocity of any point on a link with respect to
another point on the same link is always
perpendicular to the line joining these points on
the configuration (or space) diagram.
Velocity of a Point on a Link by Relative
Velocity Method
Magnitude
And
Direction
Known
Only
direction
Known
???
Steps:
1. Take some convenient point o, known as the pole.
2. Through o, draw oa parallel and equal to vA, to some suitable scale.
3. Through a, draw a line perpendicular to AB. This line will represent the
velocity of B with respect to A, i.e. vBA.
4. Through o, draw a line parallel to vB intersecting the line of vBA at b.
5. Measure ob, which gives the required velocity of point B (vB), to the scale.
Velocities in Slider Crank Mechanism
Magnitude
And
Direction
Known
???
Velocities in Slider Crank Mechanism
Velocity diagram
Steps:
1. From any point o, draw vector ob parallel to the direction of vB (or
perpendicular to OB) such that ob = vB = ωxr, to some suitable scale.
2. Since AB is a rigid link, therefore the velocity of A relative to B is
perpendicular to AB. Now draw vector ba perpendicular to AB to
represent the velocity of A with respect to B i.e. vAB.
3. From point o, draw vector oa parallel to the path of motion of the slider A
(which is along AO only). The vectors ba and oa intersect at a. Now oa
represents the velocity of the slider A i.e. vA, to the scale.
Acceleration Diagram for a Link
Acceleration of a particle whose velocity
changes both in magnitude and direction
at any instant has the following two
components
(i) Centripetal or radial component
which is perpendicular to the velocity
of the particle at the given instant.
(ii) Tangential component which is
parallel to the velocity of the particle
at the given instant.
Acceleration of a Point on a Link
Magnitude
And
Direction
Known
Only direction
Known
???
Steps:
1. From any point o', draw vector o'a' parallel to the direction of absolute acceleration at point A i.e.
aA , to some suitable scale, as shown in Fig. (b).
2. We know that the acceleration of B with respect to A i.e. aBA has the following two components:
1. Radial component of the acceleration of B with respect to A i.e. ar
BA and
2. Tangential component of the acceleration B with respect to A i.e. at
BA. These two components
are mutually perpendicular.
3. Draw vector a'x parallel to the link AB (because radial component of the acceleration of B with
respect to A will pass through AB), such that vector a’x= a’
BA =v2
BA / AB.
Acceleration of a Point on a Link
4. From point x, draw vector xb’ perpendicular to AB or vector a'x (because tangential component
of B with respect to A i.e. ar
BA, is perpendicular to radial component a r
AB ) and through o'
draw a line parallel to the path of B to represent the absolute acceleration of B i.e. aB. The
vectors xb' and o' b' intersect at b'. Now the values of aB and at
BA at may be measured, to the
scale.
5. By joining the points a' and b' we may determine the total acceleration of B with respect to A
i.e. aBA. The vector a' b' is known as acceleration image of the link AB.
6. For any other point C on the link, draw triangle a' b' c' similar to triangle ABC. Now vector
b'c' represents the acceleration of C with respect to B i.e. aCB, and vector a'c' represents the
acceleration of C with respect to A i.e. aCA. As discussed above, aCB and aCA will each have
two components as follows
1. aCB has two components; ar
CB and at
CB as shown by triangle b'zc' in Fig. (b), in which b'z
is parallel to BC and zc' is perpendicular to b'z or BC.
2. aCA has two components ; ar
CA and at
CA as shown by triangle a'yc' in Fig. (b), in which a'y
is parallel to AC and yc' is perpendicular to a'y or AC.
7. The angular acceleration of the link AB is obtained by dividing the tangential components of
the acceleration of B with respect to A (at
BA ) to the length of the link. Mathematically, angular
acceleration of the link AB
Acceleration in Slider Crank Mechanisms
Magnitude
And
Direction
Known
???
Steps:
Acceleration in Slider Crank Mechanisms
Steps:
Next lecture
• Acceleration analysis for Mechanisms
Contact details:
Rohit Kumar Singla
Email ID: rohit.kumar@thapar.edu

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Relative Velocity Method for Velocity and Acceleration analysis

  • 1.
  • 2. • Transmission angle • Instantaneous center method for Velocity analysis Summary of previous lecture
  • 3. Today’s lecture • Relative Velocity Method for Velocity analysis • Acceleration in Mechanisms
  • 4. Relative Velocity of Two Bodies Moving in Straight Lines (Velocity A w.r.t B)AB A Bv at v v  (Velocity w.r.t )BA B Av at B A v v  (i) (ii) BA B Av v v  AB A Bv v v 
  • 5. Motion of a Link Velocity of any point on a link with respect to another point on the same link is always perpendicular to the line joining these points on the configuration (or space) diagram.
  • 6. Velocity of a Point on a Link by Relative Velocity Method Magnitude And Direction Known Only direction Known ??? Steps: 1. Take some convenient point o, known as the pole. 2. Through o, draw oa parallel and equal to vA, to some suitable scale. 3. Through a, draw a line perpendicular to AB. This line will represent the velocity of B with respect to A, i.e. vBA. 4. Through o, draw a line parallel to vB intersecting the line of vBA at b. 5. Measure ob, which gives the required velocity of point B (vB), to the scale.
  • 7. Velocities in Slider Crank Mechanism Magnitude And Direction Known ???
  • 8. Velocities in Slider Crank Mechanism Velocity diagram Steps: 1. From any point o, draw vector ob parallel to the direction of vB (or perpendicular to OB) such that ob = vB = ωxr, to some suitable scale. 2. Since AB is a rigid link, therefore the velocity of A relative to B is perpendicular to AB. Now draw vector ba perpendicular to AB to represent the velocity of A with respect to B i.e. vAB. 3. From point o, draw vector oa parallel to the path of motion of the slider A (which is along AO only). The vectors ba and oa intersect at a. Now oa represents the velocity of the slider A i.e. vA, to the scale.
  • 9. Acceleration Diagram for a Link Acceleration of a particle whose velocity changes both in magnitude and direction at any instant has the following two components (i) Centripetal or radial component which is perpendicular to the velocity of the particle at the given instant. (ii) Tangential component which is parallel to the velocity of the particle at the given instant.
  • 10. Acceleration of a Point on a Link Magnitude And Direction Known Only direction Known ??? Steps: 1. From any point o', draw vector o'a' parallel to the direction of absolute acceleration at point A i.e. aA , to some suitable scale, as shown in Fig. (b). 2. We know that the acceleration of B with respect to A i.e. aBA has the following two components: 1. Radial component of the acceleration of B with respect to A i.e. ar BA and 2. Tangential component of the acceleration B with respect to A i.e. at BA. These two components are mutually perpendicular. 3. Draw vector a'x parallel to the link AB (because radial component of the acceleration of B with respect to A will pass through AB), such that vector a’x= a’ BA =v2 BA / AB.
  • 11. Acceleration of a Point on a Link 4. From point x, draw vector xb’ perpendicular to AB or vector a'x (because tangential component of B with respect to A i.e. ar BA, is perpendicular to radial component a r AB ) and through o' draw a line parallel to the path of B to represent the absolute acceleration of B i.e. aB. The vectors xb' and o' b' intersect at b'. Now the values of aB and at BA at may be measured, to the scale. 5. By joining the points a' and b' we may determine the total acceleration of B with respect to A i.e. aBA. The vector a' b' is known as acceleration image of the link AB. 6. For any other point C on the link, draw triangle a' b' c' similar to triangle ABC. Now vector b'c' represents the acceleration of C with respect to B i.e. aCB, and vector a'c' represents the acceleration of C with respect to A i.e. aCA. As discussed above, aCB and aCA will each have two components as follows 1. aCB has two components; ar CB and at CB as shown by triangle b'zc' in Fig. (b), in which b'z is parallel to BC and zc' is perpendicular to b'z or BC. 2. aCA has two components ; ar CA and at CA as shown by triangle a'yc' in Fig. (b), in which a'y is parallel to AC and yc' is perpendicular to a'y or AC. 7. The angular acceleration of the link AB is obtained by dividing the tangential components of the acceleration of B with respect to A (at BA ) to the length of the link. Mathematically, angular acceleration of the link AB
  • 12. Acceleration in Slider Crank Mechanisms Magnitude And Direction Known ??? Steps:
  • 13. Acceleration in Slider Crank Mechanisms Steps:
  • 14. Next lecture • Acceleration analysis for Mechanisms
  • 15. Contact details: Rohit Kumar Singla Email ID: rohit.kumar@thapar.edu