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ENGINEERING COUNCIL CERTIFICATE LEVEL
MECHANICAL AND STRUCTURAL ENGINEERING C105 TUTORIAL 10 - LINEAR AND ANGULAR MOTION
This tutorial may be skipped if you are already familiar with the topic On completion of this tutorial you should be able to
ã
Define linear motion.
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Explain the relationship between distance, velocity, acceleration and time.
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Define angular motion.
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Explain the relationship between angle, angular velocity, angular acceleration and time.
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Explain the relationship between linear and angular motion.
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D.J.DUNN freestudy.co.uk
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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.
1.3. 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.
1.4. 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.
1.5. 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.
SOLUTION
a =
∆
v/t = (26 - 2)/12 = 2 m/s2 Average velocity = (26 + 2)/2 = 14 m/s Distance travelled 14 x 12 = 168 m
©
D.J.DUNN freestudy.co.uk
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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
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1.6. GRAPHS
It is very useful to draw graphs representing movement with time. Much useful information may be found from the graph.
1.6.1. DISTANCE - TIME GRAPHS
Figure 1 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.
1.6.2. VELOCITY - TIME GRAPHS
Figure 2 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.
©
D.J.DUNN freestudy.co.uk
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