Statement (1): In a quick return motion mechanism, Coriolis acceleration exists.

Statement (2): Two links in this mechanism oscillate with one sliding relative to the otherOption 3 : Statement I) is true but Statement II) is false

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.Option 3 : 90°

__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.

When a crank rotates with uniform speed it has

Option 1 : only radial acceleration

**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 = v ^{2}/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 = v** ^{2}**/r.

∴ radial/centripetal acceleration = ω^{2}r (∵ v = ωr).

**∴ Crank rotating with uniform speed has only radial acceleration.**

The component of the acceleration, perpendicular to the velocity of the particle, at the given instant is called:

Option 1 :

Radial component

__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

Option 4 : rω^{2}

__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 × ω^{2}**

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

Option 3 : So that necessary centripetal force may be obtained from the horizontal component & weight of the train

__Explanation:__

The outer part of a railway track near the bend or a curve is generally raised, i.e., the outer track of the bend is slightly higher than the inner. This is known as banking of the rails or tracks, the following three kinds of force act on it:

- The weight of the train acting vertically downwards,
- The normal reaction provided by the track, acting upwards, to support the weight of the train, and
- The frictional force acting between the contact area of wheels and the track surface helps the train move forward.

When a fast-moving train takes a curved path, it tends to move away tangentially off the track. In order to prevent this, **the curved tracks are banked on the outside to produce the necessary additional force termed as centripetal force, which is required to keep the train moving in a curved path**. If there is no banking of the track, this centripetal force that is obtained from the friction between the rim of the wheels and rails which is generally small may cause the train to jump off the rails.

**So that railway tracks are banked on curves, so that necessary centripetal force may be obtained from the horizontal component & weight of the train **

Option 3 : It deflects the wind to the right direction : in the southern hemisphere.

The correct answer is __ It deflects the wind to the right direction: in the southern hemisphere__.

- The
**Coriolis effect**describes the**pattern of deflection taken by objects**not firmly connected to the ground as they travel long distances around Earth. - The Coriolis effect is responsible for
**many large-scale weather patterns**. - The key to the Coriolis effect
**lies in Earth’s rotation**. Specifically,**Earth rotates faster at the Equator**than it**does at the poles**.**Equatorial regions**race nearly**1,600 kilometers (1,000 miles)/hour**.**Near the poles**, Earth rotates at a sluggish**0.00008 kilometers (0.00005 miles)/hour**.

- The Coriolis effect becomes
**more extreme as you move further away from the equator toward the poles**. - Due to this Coriolis force, the winds in the
**Northern Hemisphere get deflected towards the right,**and in**the Southern Hemisphere get deflected towards the left**. - Wind and ocean currents are strongly affected by the Coriolis effect.

Option 4 : Along a line rotated 90° from the sliding velocity vector in a direction same as that of the angular velocity of the slotted lever

__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.

A rigid link PQ of length 1.0 m is pinned at P. It rotates about P in a vertical plane with a uniform angular acceleration of 1.0 rad/s^{2} . At an instant when the angular velocity of the link is 1.0 rad/s, the magnitude of total acceleration (in m/s^{2} ) of point Q relative to point P is

Option 1 : 1.41

**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

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

The direction of Coriolis’s component of acceleration is obtained by :

Option 2 : Rotating V (Velocity of slider w.r.t. coincident point) at 90 degrees about its origin in the same direction as that of of (angular velocity of the slotted lever)

__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, a**_{c }**= 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.**

Option 2 : 2ωv perpendicular to the link

__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

Option 1 : Quick Return motion mechanism

__Explanation:__

**Coriolis component** comes into the picture whenever **a point moves along a path that has rotational motion**.

In the **Quick return motion mechanism**, the sliders are attached to the link and these **sliders slide when the link is rotating**. Hence, there **exists a Coriolis component**.

**Note:** Though the **slider-crank mechanism has a slider**, the **slider slides on a link without any rotation.**

Coriolis component of acceleration depends on

1. angular velocity of the link

2. acceleration of the slider

3. angular acceleration of the link

Which of the above is/are correct?

Option 1 : 1 only

Coriolis component of acceleration exists when there is a sliding motion of a slider which is sliding on a link which itself is rotating.

The magnitude of the Coriolis component of acceleration is given by

a^{c} = 2Vω

Where **V is the velocity of the slider sliding on the rotating link **

and **ω is the angular velocity of the slider.**

The direction can be found by following steps

- The sense of ω is taken,
- and the velocity vector is rotated in that sense by 90°.

Option 3 : 60 m/s^{2} and direction is such as to rotate slider velocity in the same sense as the angular velocity

__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/s^{2}

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.

Option 3 : 3000

**Concept:**

Coriolis component of acceleration = 2 × v × ω

v = velocity of the slider along the rotating link ‘AB’

ω = Angular velocity of the link ‘AB’

__Calculation:__

Given, v = 30 m/s, ω = 50 rad/s

a_{c }= 2 × v × ω = 2 × 30 × 50 = 3000 m/s^{2}

A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plane. At the instant shown, edge QR is parallel to the x-axis, and the body moves such that velocities of points P and R are V_{P} and V_{R}, in the x and y directions, respectively. The magnitude of the angular velocity of the body is

Option 1 : 2V_{R}

__Concept: __

If V_{A} and V_{B} are velocity at point A and B respectively then the I-centre of this system will be finding by the meeting point of perpendicular drawn on the both velocity vector.

If w is the angular velocity of system, then

\(\frac{{{{\vec V}_A}}}{{A{I_C}}} = \frac{{{{\vec V}_B}}}{{B{I_C}}} = \omega \)

__Calculation:__

∵ AI = AC sin 60° = AC × √3/2 = √3/2

CI = AC cos 60° = AC × 1/2 = 1/2

\(\therefore {V_p} = \left( {AI} \right)w \Rightarrow w = \frac{{{V_P}}}{{AI}} = \frac{2}{{\sqrt 3 }}{V_p}\)

V_{R} = (CI)w ⇒ w = V_{R}/CI = 2V_{R}

Option 3 : Quick return motion mechanism

**Coriolis component of acceleration** exists when there is a** sliding motion of a slider** **which is sliding on a link which itself is rotating.**

In the case of the shaper, the **quick return mechanism** is used which has slider sliding on the rotating link. So the Coriolis component of acceleration exists. The Coriolis component of acceleration depends on the crank position.

As at the extreme ends, the angular velocity of the crank is zero and so Coriolis component of acceleration is also zero and at the center velocity of sliding is zero again Coriolis acceleration is zero. So it depends on the crank position.

Option 2 : linear scale of configuration diagram multiplied by square of angular velocity of crank

__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.

- Velocity of connecting rod AB, V

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 }= ω^{2}× OG. - Tangential acceleration of connecting rod, ft
_{BA}= ω^{2}× FG. - Normal acceleration of connecting rod, fn
_{BA}= ω^{2}× AF. - Total acceleration of connecting rod, f
_{BA}= ω^{2}× OG.

- Acceleration of piston (Slider B), f
- 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 = ω**^{2}× 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.

Option 4 : Iα

__Concept__:

- Angular acceleration (α): It is defined as the time rate of change of angular velocity of a particle is called its angular acceleration.
- If Δω is the change in angular velocity time Δt, then average acceleration is

\(\vec α = \frac{{{\rm{\Delta }}\omega }}{{{\rm{\Delta }}t}}\)

Moment of Inertia (I):

- Moment of inertia plays the same role in rotational motion as mass plays in linear motion. It is the property of a body due to which it opposes any change in its state of rest or of uniform rotation.
- The moment of inertia of a particle is

I = mr2

where r = perpendicular distance of the particle from the rotational axis.

Torque (τ):

- It is the twisting force that tends to cause rotation.
- The point where the object rotates is known as the axis of rotation.
- Mathematically it is written as,

The relationship between the angular acceleration (α), torque (τ) and moment of inertia (I) is given by

⇒ τ = α × I

\( \Rightarrow \alpha = \frac{T }{I}\)

In the figure, link 2 rotates with constant angular velocity ω_{2}. A slider link 3 moves outwards with a constant relative velocity V_{Q/P}, where Q is a point on slider 3 and P is a point on link 2. The magnitude and direction of Coriolis component of acceleration is given by

Option 1 : 2ω_{2} V_{Q/P} ; direction of V_{Q/P }rotated by 90° in the direction ω_{2}

**Concept:**

**Coriolis acceleration: **Coriolis component of acceleration exists when there is a sliding motion of a slider which is sliding on a link which itself is rotating. The magnitude of the Coriolis component is given by 2Vω in the direction obtained by rotating the velocity vector by 90° in the direction of rotation of the link.

a_{c} = 2Vω

a_{c} = Coriolis Acceleration

V = Component of velocity which causes sliding or velocity of the slider i.e. V_{Q/P}

ω = Angular velocity of the link on which slider is present i.e. ω_{2}

**Calculation:**

In the given question,

Link 2 is rotating in the clockwise direction, So the slider velocity will be rotated by 90° in the clockwise direction.