# ENGINEERING COUNCIL DYNAMICS OF MECHANICAL SYSTEMS D225 TUTORIAL 1 LINEAR AND ANGULAR DISPLACEMENT, VELOCITY AND ACCELERATION

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1 ENGINEERING COUNCIL DYNAMICS OF MECHANICAL SYSTEMS D225 TUTORIAL 1 LINEAR AND ANGULAR DISPLACEMENT, VELOCITY AND ACCELERATION This tutorial covers pre-requisite material and should be skipped if you are already familiar with the subject matter. On completion of this tutorial you should be able to Define linear motion. Explain the relationship between distance, velocity, acceleration and time. Define angular motion. Explain the relationship between angle, angular velocity, angular acceleration and time. Explain the relationship between linear and angular motion. D.J.DUNN freestudy.co.uk 1

2 1. LINEAR MOTION 1.1 MOVEMENT or DISPLACEMENT This is the distance travelled by an object and is usually denoted by x or s. Units of distance are metres. 1.2 VELOCITY This is the distance moved per second or the rate of change of distance with time. Velocity is movement in a known direction so it is a vector quantity. The symbol is v or u and it may be expressed in calculus terms as the first derivative of distance with respect to time so that v = dx/dt or u = ds/dt. The units of velocity are m/s SPEED This is the same as velocity except that the direction is not known and it is not a vector and cannot be drawn as such AVERAGE SPEED OR VELOCITY When a journey is undertaken in which the body speeds up and slows down, the average velocity is defined as TOTAL DISTANCE MOVED/TIME TAKEN ACCELERATION When a body slows down or speeds up, the velocity changes and acceleration or deceleration occurs. Acceleration is the rate of change of velocity and is denoted with a. In calculus terms it is the first derivative of velocity with time and the second derivative of distance with time such that a = dv/dt = d2x/dt2. The units are m/s2. Note that all bodies falling freely under the action of gravity experience a downwards acceleration of 9.81 m/s2. WORKED EXAMPLE No.1 A vehicle accelerates from 2 m/s to 26 m/s in 12 seconds. Determine the acceleration. Also find the average velocity and distance travelled. a = v/t = (26-2)/12 = 2 m/s 2 Average velocity = (26 + 2)/2 = 14 m/s Distance travelled 14 x 12 = 168 m SELF ASSESSMENT EXERCISE No.1 1. A body moves 5000 m in 25 seconds. What is the average velocity? (Answer 200 m/s) 2. A car accelerates from rest to a velocity of 8 m/s in 5 s. Calculate the average acceleration. (Answer 1.6 m/s 2 ) 3. A train travelling at 20 m/s decelerates to rest in 40 s. What is the acceleration? (Answer -0.5 m/s 2 ) D.J.DUNN freestudy.co.uk 2

3 1.6. GRAPHS It is very useful to draw graphs representing movement with time. Much useful information may be found from the graph DISTANCE - TIME GRAPHS Graph A shows constant distance at all times so the body must be stationary. Graph B shows that every second, the distance from start increases by the same amount so the body must be travelling at a constant velocity. Graph C shows that every passing second, the distance travelled is greater than the one before so the body must be accelerating. Figure VELOCITY - TIME GRAPHS Graph B shows that the velocity is the same at all moments in time so the body must be travelling at constant velocity. Graph C shows that for every passing second, the velocity increases by the same amount so the body must be accelerating at a constant rate. Figure SIGNIFICANCE OF THE AREA UNDER A VELOCITY-TIME GRAPH The average velocity is the average height of the v-t graph. By definition, the average height of a graph is the area under it divided by the base length. Average velocity = total distance travelled/time taken Average velocity = area under graph/base length Since the base length is also the time taken it follows that the area under the graph is the distance travelled. This is true what ever the shape of the graph. Figure 3 When working out the areas, the true scales on the graphs axis are used. D.J.DUNN freestudy.co.uk 3

4 WORKED EXAMPLE No.2 Find the average velocity and distance travelled for the journey depicted on the graph above. Also find the acceleration over the first part of the journey. Total Area under graph = A + B + C Area A = 5x7/2 = 17.5 (Triangle) Area B = 5 x 12 = 60 (Rectangle) Area C = 5x4/2 = 10 (Triangle) Total area = = 87.5 The units resulting for the area are m/s x s = m Distance travelled = 87.5 m Time taken = 23 s Average velocity = 87.5/23 = 3.8 m/s Acceleration over part A = change in velocity/time taken = 5/7 = m/s2. SELF ASSESSMENT EXERCISE No.2 1. A vehicle travelling at 1.5 m/s suddenly accelerates uniformly to 5 m/s in 30 seconds. Calculate the acceleration, the average velocity and distance travelled. (Answers m/s 2, 3.25 m/s and 97.5 m) 2. A train travelling at 60 km/h decelerates uniformly to rest at a rate of 2 m/s2. Calculate the time and distance taken to stop. (Answers 8.33 s and m) 3. A shell fired in a gun accelerates in the barrel over a length of 1.5 m to the exit velocity of 220 m/s. Calculate the time taken to travel the length of the barrel and the acceleration of the shell. (Answers s and m/s 2 ) 1.8. STANDARD FORMULAE Consider a body moving at constant velocity u. Over a time period t seconds it accelerates from u to a final velocity v. The graph looks like this. Distance travelled = s s = area under the graph = ut + (v-u)t/2 s = ut + (v - u)t/2...(1) s = ut + t/2 v - ut/2 s = vt/2 + ut/2 s = t/2 (v + u ) Figure 4 The acceleration = a = (v - u)/t from which (v - u) = at Substituting this into equation (1) gives Since v = u + the increase in velocity v = u + at and squaring we get v2 = u2 + 2a[at2/2 + ut] s = ut + at2/2 v2 = u2 + 2as D.J.DUNN freestudy.co.uk 4

5 WORKED EXAMPLE No.3 A missile is fired vertically with an initial velocity of 400 m/s. It is acted on by gravity. Calculate the height it reaches and the time taken to go up and down again. u = 400 m/s v = 0 a = -g = m/s 2 v2 = u2 + 2as 0 = (-9.81)s s = 8155 m s = (t/2)(u + v) 8155 = t/2( ) t = 8155 x 2/400 = s To go up and down takes twice as long. t = m/s WORKED EXAMPLE No.4 A lift is accelerated from rest to 3 m/s at a rate of 1.5 m/s2. It then moves at constant velocity for 8 seconds and then decelerates to rest at 1.2 m/s2. Draw the velocity - time graph and deduce the distance travelled during the journey. Also deduce the average velocity for the journey. The velocity - time graph is shown in fig.5. Figure 5 The first part of the graph shows uniform acceleration from 0 to 3 m/s. The time taken is given by t 1 = 3/1.5= 2 seconds. The distance travelled during this part of the journey is x 1 = 3 x 2/2 = 3 m The second part of the journey is a constant velocity of 3 m/s for 8 seconds so the distance travelled is x 2 = 3 x 8 = 24 m The time taken to decelerate the lift over the third part of the journey is t 3 = 3/1.2 = 2.5 seconds. The distance travelled is x 2 = 3 x 2.5/2 = 3.75 m The total distance travelled is = m. The average velocity = distance/time = 30.75/12.5 = 2.46 m/s. D.J.DUNN freestudy.co.uk 5

6 SELF ASSESSMENT EXERCISE No.3 The diagram shows a distance-time graph for a moving object. Calculate the velocity. (Answer 1.82 m/s) Figure 5 2. The diagram shows a velocity time graph for a vehicle. Calculate the following. i. The acceleration from O to A. ii. The acceleration from A to B. iii. The distance travelled. iv. The average velocity. (Answers m/s 2, -2.4 m/s 2, 35 m and 5.83 m/s) Figure 6 3. The diagram shows a velocity-time graph for a vehicle. Calculate the following. i. The acceleration from O to A. ii. The acceleration from B to C. iii. The distance travelled. iv. The average velocity. (Answers 5 m/s 2, -2 m/s 2, 85 m and 7.08 m/s) Figure 7 D.J.DUNN freestudy.co.uk 6

9 WORKED EXAMPLE No.7 A car travels around a circular track of radius 40 m at a velocity of 8 m/s. Calculate its angular velocity. v = ωr ω = v/r = 8/40 = 0.2 rad/s 2.5 EQUATIONS OF MOTIONS The equations of motion for angular motion are the same as those for linear motion but with angular quantities replacing linear quantities. replace s or x with θ replace v or u with ω replace a with α LINEAR ANGULAR s = ut + at 2 /2 θ = ω 1 t + αt 2 /2 s = (u + v)t/2 θ = (ω 1 + ω 2 )t/2 v 2 = u 2 + 2as ω 2 2 = ω αθ Also remember that the angle turned by a wheel is the area under the velocity - time graph. SELF ASSESSMENT EXERCISE No.6 1. A wheel accelerates from rest to 3 rad/s in 5 seconds. Sketch the graph and determine the angle rotated. (Answer 7.5 radian). 2. A wheel accelerates from rest to 4 rad/s in 4 seconds. It then rotates at a constant speed for 3 seconds and then decelerates uniformly to rest in 5 seconds. Sketch the velocity time graph and determine i. The angle rotated. (30 radian) ii. The initial angular acceleration. (1 rad/s 2 ) iii. The average angular velocity. (2.5 rad/s) D.J.DUNN freestudy.co.uk 9

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