Velocity and Acceleration Analysis MCQ Quiz in मल्याळम - Objective Question with Answer for Velocity and Acceleration Analysis - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Use the concept of instantaneous center of rotation.

Instantaneous center of rotation is the point about which all particles of a body perform pure rotation motion.

Velocity is then given by,

V = rω,

Where, r = radius and ω = angular velocity

Construction:

Calculation:

V_A = ω_AB.× (AI)

V_p = ω_AB × PI

By LAMI’S Theorem,  ​​

∴ PI = 2.82 m

Now,

∴ V_p = 2.82 × 2.876

Explanation:

Klein's Construction:

• It is used to draw the velocity and acceleration diagrams for a single slider crank mechanism.
• The velocity and acceleration of piston of a reciprocating engine mechanism can be determined by the figure given below.

Velocity diagram:

• Draw the configuration diagram OAB for the slider-crank mechanism.
• Draw OI perpendicular to OB and produce BA to meet OI at C. Then the triangle formed, is known as Klein's velocity diagram.
  • Velocity of connecting rod AB, V_BA = ω × AC.
  • Velocity of piston B,VP = V_BO = ω × OC.
  • Velocity of crankpin A, V_AO = ω × OA.

Acceleration diagram:

• With A as centre and AC as radius, draw a circle.
• Locate D as the midpoint of AB.
• With D as centre, and DA as radius, draw a circle to intersect the previously drawn circle at points H and E.
• Join HE intersecting AB at F.
• Produce HE to meet OB at G.
• Then OAFG is Klein's acceleration diagram.
  • Acceleration of piston (Slider B), f_BO = ω² × OG.
  • Tangential acceleration of connecting rod, ft_BA = ω² × FG.
  • Normal acceleration of connecting rod, fn_BA = ω² × AF.
  • Total acceleration of connecting rod, f_BA = ω² × OG.
• Angular acceleration of connecting rod, AB 
• To find the acceleration of any point x in AB, draw a line X-X parallel to OB to intersect at X. Join OX.
• Then, Acceleration of point x, fx = ω² × OX.

Though, we can calculate both velocity and acceleration through Klein's construction but mainly it is used in calculating the linear acceleration of the piston.

Concept:

If the relative motion between two links is pure sliding, then the relative instantaneous centre is at the infinity on a line perpendicular to the direction of sliding.

Location of Instantaneous Centres:

a) When the two links are connected by a pin joint (or pivot joint), the instantaneous centre lies on the centre of the pin.

b) When the two links have a pure rolling contact the instantaneous centre lies on their point of contact.

c) When the two links have a sliding contact, the instantaneous centre lies on the common normal at the point of contact.

• When the link 2 (slider) moves on fixed link 1 having straight surface the instantaneous centre lies at infinity.
• When the link 2 (slider) moves on fixed link 1 having a curved surface, the instantaneous centre lies on the centre of curvature of the curvilinear path.

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 is to be calculated.

So, statement (1) is correct.

Whitworth quick return motion mechanism is an inversion of single slider crank chain. In this mechanism, one link rotates and other oscillates and slider (third link) will slides.

So, statement (2) is incorrect.

Explanation:

The 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.

When A and B are two points on a given link, the velocities VA and VB cannot be chosen arbitrarily for both points A and B because point A and B remain apart at a fixed distance throughout the motion transmission.

Thus the relative motion between points A and B is not possible along with line AB.

It is possible only in a direction perpendicular to AB which is given by-

Concept:

Single slider crank mechanism:

Velocity of piston is 

Where r = crank radius, ω = crank speed, n = Obliquity ratio

 Velocity of slider will be maximum at θ = 90°, (∵ sin function is maximum at 90° i.e. sin 90° = 1)

∴ V_max = rω (1 + 0) = rω

When θ = 90°, i.e. crank is perpendicular to the line of stroke, the velocity of the slider is maximum.

Concept:

Location of Instantaneous Centres:

a) When the two links are connected by a pin joint (or pivot joint), the instantaneous centre lies on the centre of the pin.

b) When the two links have a pure rolling contact the instantaneous centre lies on their point of contact.

c) When the two links have a sliding contact, the instantaneous centre lies on the common normal at the point of contact.

• When the link 2 (slider) moves on fixed link 1 having straight surface the instantaneous centre lies at infinity
• When the link 2 (slider) moves on fixed link 1 having curved surface, the instantaneous centre lies on the centre of curvature of the curvilinear path

Concept:

Linear velocity V is given by:

where ω is angular velocity.

Tangential acceleration in the no-slip condition is

a_t = r × α

Centripetal acceleration is given by:

a_c = r × ω²

The instantaneous velocity of the point of contact is zero.

So at the point of contact, Instantaneous tangential acceleration is also zero.

∴ Only centripetal acceleration is there at the point of contact.

∴ Net acceleration of the point of contact is

ac = r × ω2

Explanation:

Coriolis Acceleration

• If the distance between the two points does not remain fixed and the second point slides, the total acceleration will contain an additional component of acceleration, known as Coriolis Acceleration.
• Coriolis component of acceleration is equal to 2vω, where, v is the sliding velocity and ω is the angular speed. 
• The direction of Coriolis acceleration is such as to rotate the slider velocity vector in the same sense as the angular velocity of the link.
• This acceleration is experienced by a point on one link that is sliding along another rotating link.

CONCEPT:

• Uniform circular motion: The movement of a body following a circular path is called a circular motion.
  • The motion of a body moving with constant speed along a circular path is called Uniform Circular Motion.
  • Here, the speed is constant but the velocity changes.
  • For a particle to move along the circular path it should have acceleration acting towards the center, which makes it move in a circular path.
  • As acceleration is perpendicular to the velocity of a particle at every instant, it is only changing the direction of velocity and not magnitude and that’s why the motion is uniform circular motion.
  • This acceleration is centripetal acceleration (or radial acceleration), and the force acting towards the center is called centripetal force. The radial acceleration is perpendicular to the velocity of particle.

Diagram: 

• Centripetal acceleration (ac): It is the acceleration towards the center when an object is moving in a circle.
  • Though the speed may be constant, the change in direction results in a non-zero acceleration.

Formula:

ac = v2/r

where, v = velocity(m/s), r = radius