Acceleration Analysis MCQ Quiz - Objective Question with Answer for Acceleration Analysis - Download Free PDF

Explanation:

At the instant when AP is horizontal and OP is vertical, V_P will come in the line of AP, then the perpendicular component of V_P will be zero.

⇒ V_P = (OP) × ω_OP = (OP) × 7 m/s; (direction is along AP)

Hence, ω_ABC = 0;

No perpendicular component of velocity

Explanation:

Analyzing Acceleration in a Rotating Link:

• When analyzing the acceleration of a point in a rotating link, it is essential to consider the different components of acceleration that contribute to the overall motion.
• The relative acceleration equation is a fundamental tool used in this analysis, breaking down the acceleration into various components that act on the point in question.

Components of Acceleration: The relative acceleration equation typically considers the following components:

• Tangential Acceleration: This component is due to the change in the magnitude of the velocity of the point. It acts tangentially to the path of the point and is given by the product of the angular acceleration and the radius of rotation.
• Centripetal (or Radial) Acceleration: This component is due to the change in direction of the velocity of the point. It acts radially inward towards the center of rotation and is given by the square of the angular velocity times the radius of rotation.
• Coriolis Acceleration: This component arises when a point moves in a rotating reference frame. It acts perpendicular to the velocity of the point and the axis of rotation. It is given by twice the product of the angular velocity and the relative velocity of the point in the rotating frame.

Gravitational Acceleration

• When analyzing the acceleration of a point in a rotating link using the relative acceleration equation, gravitational acceleration is NOT directly considered. The relative acceleration equation focuses on the components of acceleration that arise due to the rotational motion and the relative motion of the point within the rotating link. Gravitational acceleration, while acting on the point, is a constant acceleration that affects all points equally and is not specific to the rotational motion being analyzed.

Concept:

• Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction.
  • At all instances, the object is moving tangent to the circle. Since the direction of the velocity vector is the same as the direction of the object's motion, the velocity vector is directed tangent to the circle as well.

• Centripetal acceleration is the acceleration of a body traversing a circular path.
  • Because velocity is a vector quantity (that is, it has both a magnitude, speed, and direction), when a body travels on a circular path, its direction constantly changes, and thus its velocity changes, producing acceleration.
• It is given by the formula, \(a_c = {v^2 \over r}\)
  • where v is the velocity of the body & r is the radius of the circular path.
• Apart from centripetal acceleration, there is another acceleration called as tangential acceleration, a_t = rα 
  • where α  is the angular acceleration of the body & r is the radius of the circular path.

Calculation:

Given:

Initially, a₁ = a and v₁ = v,

⇒ \(a_1 = {v_1^2 \over r} = {v^2 \over r}\)

Finally, a₂ = to be calculated and v₂ = 2v,

\(a_2 = {v_2^2 \over r} = {4v^2 \over r}\)

Then, taking the ratio, we have

\({a_2 \over a_1} = {{4v^2 \over r}\over {v^2\over r}} = {4\over 1}\)

a₂ : a₁ = 4 : 1

Concept:
The relationship between linear acceleration (a) and angular acceleration (alpha) is given by the equation:

a = α  * r

where:

a is the linear acceleration,
α   is the angular acceleration, and
r is the radius.
We can rearrange the equation to solve for alpha:

α   = a / r
Calculation:
Given:

In this case, a = 3 m/s², and r = 750 mm = 0.75 m (since 1 m = 1000 mm).

α   = 3 m/s² / 0.75 m = 4 rad/s²

So, the correct answer is option 3.

Explanation:

Let a link AR rotate about a fixed point A on it. P is a point on a slider on the link.

At any given instant, Let:
ω = Angular velocity of the link,

α = Angular acceleration of the link,

v = Linear velocity of the slider on the link,

f = Linear acceleration of the slider on the link,

r = Radial distance of point P on the slider

In a short interval of time δt, let δθ be the angular displacement of the link and δr be the displacement of the slider in the outward direction.

After the short interval of time δt, let
ω′ = ω + α ∙ δt = angular velocity of the link.
v′ = v + f∙ δt = Linear velocity of the slider on the link.
r′ =r + δr = Radial distance of the slider.
Acceleration of P perpendicular to AR:
Initial velocity of P perpendicular to AR = ωr
Final velocity of P perpendicular to AR = v′sin δθ + ω′r′ cos δθ
Change of velocity perpendicular to AR = (v′sinδθ + ω′r′cosδθ) ‒ ωr

Acceleration of P perpendicular to AR = \(\frac{(v~+~δ t)\cos δ θ~-~(ω~+~α δ t)(r~+~δ r)\cos δ θ~-~ω r}{δ t}\)

In the limit, as δt → 0, cos δθ → 1 and sin δθ → δθ

Acceleration of P perpendicular to AR = \(v\frac{d\theta}{dt}~+~ω\frac{dr}{dt}~+~rα\)

⇒ Acceleration of P perpendicular to AR = vω + vω + αr

⇒ Acceleration of P perpendicular to AR = 2vω + αr

And Acceleration of P parallel to AR = f - ω2r

Explanation:

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 acceleration comes into account.
• It is the tangential component of the acceleration of the slider with respect to the coincident point on the link.
• Mathematically, Coriolis component of acceleration, ac = 2vω
• The direction of the Coriolis component of acceleration is given by rotating the velocity of the slider by 90° in the direction of the angular velocity of the rotating link.

Concept:

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 acceleration comes into account. It is the tangential component of the acceleration of the slider with respect to the coincident point on the link.

Mathematically,

Coriolis component of acceleration, ac = 2vω

The direction of the Coriolis component of acceleration is given by rotating the velocity of the slider by 90° in the direction of the angular velocity of the rotating link.

Concept:

Tangential acceleration (at):

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

Where, α = angular acceleration and r = radius

at = rα

Centripital acceleration (ac):

ac = ω2r

where,  ω = Angular Velocity

Calculation:

Given:

α = Angular acceleration = 1 rad/s2

r = radius of crank = 1 m

ω = Angular Velocity = 1 rad/s

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

∴ at = rα = 1 × 1 = 1 m/s2 

∴  ac = ω2r = (1)2 × 1 = 1 m/s2 

\(\therefore\ \text{Total acceleration a} = \sqrt {a_t^2 + a_c^2} = \sqrt {{1^2} + {1^2}} = \sqrt 2 =1.41 \;{m}/{{{s^2}}}\)

Concept:

Let a crank OA, of radius r, rotate in a circular path in the clockwise direction. It has an instantaneous angular velocity ω and an angular acceleration α in the same direction.

Tangential velocity of A is v_a and v_a = ωr.

In a short interval of time 'dt', let OA assumes a new position OA' by rotating through small angle 'dθ'.

Angular velocity of OA' is ω' and ω' = ω + αdt.

Tangential velocity of A' is v_a' and va' = (ω + αdt)r.

The tangential velocity of A' may be considered to have two components, one perpendicular to OA and the other parallel to OA.

Due to these velocity differences in different directions, there will be two components of acceleration also.

Acceleration of A ⊥ to OA = Tangential acceleration of A relative to O.

It is denoted by f^t_ao.

Tangential acceleration = αr.

Accleration of A / to OA = Centripetal or radial acceleration of A relative to O which is directed towards centre.

It is denoted by f^c_ao.

Centripetal or radial acceleration = v²/r.

Coriolis acceleration:

It comes into picture when a point is in motion relative to a moving body for example, the motion of a slider on a rotating link.

Calculation:

Given:

The crank rotates with angular speed i.e the motion is relative to a fixed body, ∴ no Coriolis acceleration will be present.

Tangential acceleration = αr

∵ α = 0, tangential acceleration is zero. [α = dω/dt and ω = constant].

Centripetal or radial acceleration = v²/r.

∴ radial/centripetal acceleration = ω²r (∵ v = ωr).

∴ Crank rotating with uniform speed has only radial acceleration.

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 \(\alpha_{AB}=f^{t}_{BA}/AB=ω^{2}\frac{FG}{AB}\;\;\;\;(ccw)\)
• 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.

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.

Concept:

Linear velocity V is given by:

\(V=r×ω\)

where ω is angular velocity.

Tangential acceleration in the no-slip condition is

\(\frac{{dv}}{{dt}} = r\frac{{dω }}{{dt}} = r×α \)

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 

Concept:

The Coriolis Effect occurs when an object traveling in a straight path is viewed from a moving frame of reference. The moving frame of reference causes the object to appear as if it is traveling along a curved path.

Direction of Coriolis acceleration: The direction is determined by rotating the velocity vector by 90° in the sense of angular velocity.

Magnitude of Coriolis acceleration = 2 × v × ω

Calculation:

Given, ω = 20 rad/s, v = 1.5 m/s

a = 2 × 1.5 × 20

a = 60 m/s²

The direction of the Coriolis acceleration is obtained by rotating the velocity vector in the sense of angular velocity of the link by 90°, the direction is shown below

It seems that the Coriolis component is trying to rotate the slider in the direction of the angular velocity of the link.