Site pages

Current course

Participants

General

Module 1. Introduction to Theory of Machine

Module 2. Planar Mechanism

Module 3. Velocity and Acceleration Analysis

Topic 4

Topic 5

Topic 6

Topic 7

Topic 8

Topic 9

Topic 10

## Lesson 8.

**8.1 ACCELERATION ANALYSIS**

The rate of change of velocity with respect to time is known as acceleration. It has magnitude as well as direction. So, it is a vector quantity. Consider two points A and B on a rigid link. Point B moves relative with respect to A with an angular velocity of ω rad/s and angular acceleration of α rad/s^{2}.V_{BA} is the velocity of point B relative to A is perpendicular to the line joining A and B (as shown in figure 3.16a).V_{BA }is equal to ω × AB. Acceleration has two components radial or centripetal and tangential.

**Fig. 3.16 acceleration components**

**8.1.1 RADIAL OR CENTRIPETAL COMPONENT OF ACCELERATION**

The magnitude of radial component is given by f^{r}_{BA}= ω^{2}×BA where f^{r}_{BA}= Radial component of acceleration of point B respect to point A (as given in fig: 3.16b). Its direction is given from B to A. When the angular velocity of a rotating link or a particle moving in a circular path is not constant, but is subjected to an angular acceleration (with consequent peripheral or linear acceleration), the resultant acceleration may be found by adding the vectors representing the centripetal and peripheral (or tangential) accelerations. Since the centripetal acceleration is always radial and perpendicular to the instantaneous direction of the motion and the peripheral acceleration is perpendicular to the radius, these two accelerations are always mutually perpendicular and the resultant acceleration is easily found by applying the wall known relationship which exist between the sides of a right angled triangle, that is by extracting the square root of the sum of the squares of the respective accelerations.

**8.1.2 TANGENTIAL COMPONENT OF ACCELERATION**

It can be defined as the rate of change velocity V_{BA} in the tangential direction. Its magnitude will be equal to α×BA. It is represented by f^{t}_{BA}_{; }the tangential component of acceleration of B with respect to A. f^{t}_{BA}_{ }is perpendicular to BA and parallel to V_{BA}._{ }

The total acceleration of point B with respect to A is the vector sum of their components of radial and tangential acceleration. This can be written as

\[{{\text{f}}_{{\text{BA}}}}={\text{}}{{\text{f}}^{\text{r}}}_{{\text{BA}}}+{\text{}}{{\text{f}}^{\text{t}}}_{{\text{BA}}}={\text{}}{\omega ^{\text{2}}} \times {\text{BA}} + {\text{ }}\alpha\times {\text{BA}} = {\text{ }}\left( {{\omega ^{\text{2}}} + {\text{ }}\alpha } \right){\text{}} \times {\text{ BA}}\]

Here f_{BA }is the total acceleration of B with respect to A

**8.1.3** **How to draw acceleration diagram**

Step 1. Draw **b’x **= f^{r}_{BA} in some suitable scale parallel to AB and its direction will be from B to A.

Step 2. From point **x** draw **xa’ **= f^{t}_{BA} = α×BA which is perpendicular to AB. α is known in this case.

Step 3. Join **a’** with **b’** which shows f_{BA}, which represents the total acceleration of B with respect to A. The acceleration of B relative to A is inclined at an angle of β with AB.

\[{\text{Tan }}\beta= \frac{\alpha }{{{\omega ^{\text{2}}}}}\]

When α = 0, AB rotates at the uniform angular velocity, f^{t}_{BA }= 0 and thus f^{r}_{BA} represents the total acceleration.

When ω = 0, A has a linear motion, f^{r}_{BA }= 0 and thus the tangential acceleration is the total acceleration.

** **

**Example: 3.3** A motor-car passes round a curve of 30.5 m radius and at a given instant has a speed of 82 km/h (or 8.89 m/s). The car is accelerating at the rate of 16 km/h in 3 sec. find the resultant acceleration.

Sol:

Centripetal acceleration, ƒ^{r} = rω^{2} = v^{2}/r

ƒ^{r}_{ }= (32000/3600)^{2 }/ 30.5_{ }= 2.59 m/s^{2}

Tangential acceleration, ƒ^{t} = (16000/3600)/3 = 1.48 m/s^{2}

Resultant acceleration, ƒ = √(2.59^{2} + 1.48^{2}) = √8.8985 = 2.983 m/s^{2}

**Example 3.4 **The crank of a slider crank mechanism rotates a constant speed of 300 r.p.m. The crank is 150 mm and the connecting rod is 600 mm long. Determine: the angular velocity and angular acceleration of the connecting rod when the crank makes the angle of 45^{0} with form the inner dead centre.

**Solution.**

**Fig. 3.17**

The crank OB = 150 mm = 0.15 m

N_{BO}=300 r.p.m.

\[{\omega _{{\text{OB}}}} = \frac{{{\text{2}}\pi\times {\text{3}}00}}{{{\text{6}}0}} = {\text{ 31}}.{\text{4 rad}}/{\text{s}}\]

V_{BO}= ω _{OB} × OB= 31.4 × 0.15= 4.71 m/s

Velocity diagram is drawn by following these steps

Step 1. V_{BO,} the velocity of point B with respect to O is known and it is perpendicular to OB. Take any point **o** and draw vector **ob** which will represent V_{BO.}

Step 2. V_{AB }is perpendicular to AB, from **b** draw a vector **ba** perpendicular to AB representing V_{AB}.

Step 3. V_{A },the velocity of slider is along path OA. From **o** draw vector **oa** which will intersect vector **ba** at point** a.**

By measurement **oa**=V_{A}=4 m/s

**ba**=V_{AB}=3.34 m/s

The radial acceleration of B with respect to O,f^{r}_{BO} is given as

\[{{\text{f}}^{\text{r}}}_{{\text{BO}}}={{\text{f}}^{\text{r}}}_{\text{B}}=\frac{{{{\text{V}}^{\text{2}}}_{{\text{BO}}}}}{{{\text{OB}}}}= \frac{{{{\left( {{\text{4}}.{\text{71}}} \right)}^{\text{2}}}}}{{0.{\text{15}}}} = {\text{147}}.{\text{8 m}}/{{\text{s}}^{\text{2}}}\]

f^{t}_{BO}=0 as crank rotates with the constant speed

Radial acceleration of A with respect to B is given as

\[{{\text{f}}^{\text{r}}}_{{\text{AB}}}=\frac{{{{\text{V}}^{\text{2}}}_{{\text{AB}}}}}{{{\text{AB}}}}=\frac{{{{\left({{\text{3}}.{\text{34}}} \right)}^{\text{2}}}}}{{0.{\text{6}}0}} = {\text{18}}.{\text{5m}}/{{\text{s}}^{\text{2}}}\]

The acceleration diagram is drawn by following these steps

Step 1. Draw vector **o’b’ = **f^{r}_{BO} parallel to BO and 147.8 m/s^{2} in magnitude.

Step 2. The radial acceleration of A with respect to B, f^{r}_{AB} is known in magnitude i.e. 18.5 m/s^{2} and parallel to AB. from **b’ **draw vector** b’x**=f^{r}_{AB}. from **x **draw **xa’ **perpendicular to** b’x **representing tangential component of acceleration of A with respect to B.

Step 3. The acceleration of slider A with respect to O, f_{AO} is along the line of stroke of OA. From **o’** draw vector **o’a’ **representing f_{AO.} **o’a’ **intersects **xa’** at point **a’. **Now join **o’** to **a’**. Thus** o’a’** is f_{AO}=f_{A}

By measurement f_{A}=**o’a’**= 109 m/s^{2}

Angular velocity of the connecting rod is given by

\[{\omega _{{\text{AB}}}}=\frac{{{{\text{V}}_{{\text{AB}}}}}}{{{\text{AB}}}}= \frac{{{\text{3}}.{\text{34}}}}{{0.{\text{6}}0}} = {\text{5}}.{\text{56 rad}}/{\text{s}}\]

Angular acceleration of the connecting rod AB, α_{AB} is given by

\[{\alpha_{{\text{AB}}}}=\frac{{{{\text{f}}^{\text{t}}}_{{\text{AB}}}}}{{{\text{AB}}}}={\text{ }}\frac{{{\text{1}}0{\text{7}}.{\text{5}}}}{{0.{\text{6}}0}} = {\text{179}}.{\text{1 rad}}/{{\text{s}}^{\text{2}}}\]

(f^{t}_{AB}=**xa’=**107.5 by measurement)