Velocity Analysis MCQ Quiz in বাংলা - Objective Question with Answer for Velocity Analysis - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Single slider crank mechanism:

Velocity of piston is \({v_p} = \omega r\left( {\sin θ + \frac{{\sin 2θ }}{{2n}}} \right)\)

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)

∴ Vmax = 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:

Referring to the velocity diagram,

• The point P, S is the fixed point so while drawing the velocity diagram, these points will be considered as reference points. The direction of the velocity of point R on the link SR will be perpendicular to SR and will pass through point s, on the velocity diagram, as shown.
• The velocity vector of the point Q on link PQ will pass through p and will be perpendicular to the link PQ.
• The point R is also on the link QR so, the line of action of the velocity vector of point R will be perpendicular to point QR and will pass through q as shown on the velocity diagram.
• r is the intersection of the line of Action (LOA) perpendicular to RS, through point s and Line of Action (LOA) perpendicular to RQ, through q

Calculation:

Given:

PQ = 72 cm, QR = 130 cm, RS = 120 cm, PS = 122 cm, ω­1 = 9 rad/s (Anticlockwise)

From Δ QSR,

\( \Rightarrow \tan \left( {\rm{\theta }} \right) = \frac{{120}}{{50}} = \frac{12}{5}{\rm{\;\;\;\;\;}} \ldots \left( 1 \right)\)

From velocity triangle,

∵ ∠RQS = ∠rqs = θ

As the angle between RQ and SQ will be the same as the angle between the perpendiculars drawn on RQ and SQ.

pq = Velocity of the point Q with respect to P = PQ × ω3 = 72 × ω3

sr = Velocity of the point R with respect to S = RS × ω1 = 120 × 9 = 1080 cm/s

Also, from the velocity triangle,

\( \Rightarrow \tan \left( {\rm{\theta }} \right) = \frac{{{\rm{RS}} \times {{\rm{\omega }}_1}}}{{{\rm{PQ}} \times {{\rm{\omega }}_3}}} = \frac{{120 \times 9}}{{72 \times {{\rm{\omega }}_3}}}{\rm{\;\;\;\;\;}} \ldots \left( 2 \right)\)

By equating (1) and (2),

\( \Rightarrow \frac{{120 \times 9}}{{72 \times {\omega _3}}} = \frac{12}{5}\)

⇒ ω3 = 6.25 rad/s

Since, the sense of rotation of input crank will be clockwise so a negative sign will be mentioned in answer box.

Concept:

The instantaneous center of rotation of the link can be found by drawing perpendiculars on the line of action of velocity vectors.

Calculation:

Here,

\(\frac{{{V_A}}}{{AI}} = \frac{{{V_B}}}{{BI}} = \omega\)

\(⇒ \frac{{{V_A}}}{{{V_B}}} = \frac{{AI}}{{BI}} = \tan 30^\circ = \frac{1}{{\sqrt 3 }}\)

⇒ V_B = V_A × √3 = 3√3 = 5.2 m/s

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-

\({v_{BA}} = \omega \times AB\)

Concept:

Tangential acceleration (at):

• It acts along the tangent to the circular path in the plane of the circular path.
• Mathematically Tangential acceleration is written as

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where α = angular acceleration and r = radius

Calculation:

Given:

α = Angular acceleration = 50 rad/s2

r = radius of crank = 50 cm = 0.5 m

Tangential Acceleration at = rα

∴ at = rα = 50 × 0.5 = 25 m/s²

Explanation:

The relative velocity of Points on the Same link:

The velocity of any point on a rigid link with respect to another point on the same link is always perpendicular to the line joining these points on the 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-

\({v_{BA}} = \omega \times AB\)

Additional Information

Relative velocity:

The velocity of one body with respect to the other body is termed as relative velocity.

The relative velocity of two objects moving in the same direction is given by:

V = |v1 - v2|

The relative velocity of two objects moving in the different direction is given by​

V = |v1 + v2|

Concept:

Velocity of slider,

V_s = ℓ_DE × ω_DE

Calculation:

As shown in configuration, link AB and CD form parallel linkages

∴ V_B = V_C

i.e. \(\frac{{{\omega _{BA}}}}{{{\omega _{CD}}}} = \frac{{{\ell _{CD}}}}{{{\ell _{AB}}}}\)

\(\therefore \frac{{{\omega _{AB}}}}{{{\omega _{CD}}}} = \frac{{30}}{{10}}\)

\(\therefore {\omega _{CD}} = \frac{4}{3}\;rad/s\)

As, link CD and DE are single link, ω_CD = ω_DE = 4/3 rad/s

Velocity of slider,

V_s = ℓ_DE × ω_DE

\(= 20\left( {cm} \right) \times \frac{4}{3}\;\left( {rad/s} \right)\)

∴ V_S = 26.67 cm/sec

Concept used:

i) Instantaneous Centre → It is a point in common between two members where the velocities are equal in direction & magnitude

ii) Relative velocity method → Velocity of any point on a link is always perpendicular to the line joining these points on the configuration diagram

Here, from the velocity diagram

\({V_A} = \omega .IA \Rightarrow w = \frac{{{V_A}}}{{{IA}}}\)

\({V_M} = \omega .IM \Rightarrow w = \frac{{{V_M}}}{{{IM}}}\)

From i) & ii)

\(\frac{{{V_A}}}{{IA}} = \frac{{{V_M}}}{{IM}}\)

\(\Rightarrow {V_M} =V_A \times \frac{{{}IM}}{{IA}}\)

\(\frac{{IM}}{{IA}} = \sin 30^\circ \)

VA = 2 m/s

V_M = 2 × sin 30°

⇒ V_M = 1 m/s

Concept:

Single slider crank mechanism:

The velocity of the piston is \({v_p} = \omega r\left( {\sin θ + \frac{{\sin 2θ }}{{2n}}} \right)\)

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

Calculation:

Given:

The velocity of piston v_p = \(2~\frac{{m}}{{s}}\) , Angular velocity of the crank, \(\omega=\;20~\frac{{rad}}{{sec}}\) , θ = 90° 

\({V_p} = \omega r\left( {\sin θ + \frac{{\sin 2θ }}{{2n}}} \right)\)

V_p = ωr

\(r = \frac{{{V_p}}}{\omega } = \frac{2}{{20}}\)

⇒ 0.1 m or 10 cm

Concept:

Velocity of any point on a link at distance r, V = ω × r 

ω = angular velocity of the link

Calculation:

Given:

Length of the link AB, (r) = 30 cm, Speed of the link, N = 100 rpm

Velocity of B = r × ω = \(r ~\times~\frac{{2\pi N}}{{60}} \) 

\(V =30 ~\times~\frac{{2~\times~\pi~\times100}}{{60}}\) = 314.2 cm/sec