LINEAR AND ANGULAR DISPLACEMENT, VELOCITY AND
ACCELERATION
MOVEMENT or DISPLACEMENT
This is the distance travelled by an object and is usually denoted by x or s. Units of
distance are metres
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 = 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.
Average velocity/speed of a moving object can be calculated as
v = s / t (1a)
where
v = velocity or speed (m/s, ft/s)
s = linear distance traveled (m, ft)
t = time (s)
- distance is the length of the path a body follows in moving from one point to another – displacement is the straight line distance between the initial and final positions of the body
- we use velocity and speed interchangeable – but be aware that speed is a measure of how fast or slow a distance is covered, the rate at which distance is covered – velocity is a vector, specifying how fast or slow a distance is covered and the direction
Velocity can be expressed as (velocity is variable)
v = ds / dt (1f)
where
ds = change in distance (m, ft)
dt = change in time (s)
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 = d2s/dt^2. The units are m/s^2.
Note that all bodies falling freely under the action of gravity experience a downwards
acceleration of 9.81 m/s^2.
Acceleration can be expressed as
a = dv / dt (1g)
where
dv = change in velocity (m/s, ft/s)
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 = Δv /t = (26 – 2)/12 = 2 m/s^2
Average velocity = (26 + 2)/2 = 14 m/s
Distance travelled 14 x 12 = 168 m
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.
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.
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. When working out the
areas, the true scales on the graphs axis are used.
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.
SOLUTION
Total Area under graph = A + B + C
Area A = 5×7/2 = 17.5 (Triangle)
Area B = 5 x 12 = 60 (Rectangle)
Area C = 5×4/2 = 10 (Triangle)
Total area = 17.5 + 60 + 10 = 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 = 0.714 m/s^2.
If acceleration is constant then velocity can be expressed as:
v = v0 + a t (1b)
where
v0 = initial linear velocity (m/s, ft/s)
a = acceleration (m/s2, ft/s2)
Linear distance can be expressed as (if acceleration is constant):
s = v0 t + 1/2 a t2 (1c)
Combining 1b and 1c to express the final velocity
v = (v0^2 + 2 a s)^1/2 (1d)
ANGULAR MOTION
ANGLE θ
Angle has no units since it a ratio of arc length to radius. We use the names revolution,
degree and radian. Engineers use radian. Consider the arc shown.
The length of the arc is r θ and the radius is r .
The angle is the ratio of the arc length to the radius.
θ = arc length/ radius hence it has no units but it is called radians.
If the arc length is one radius, the angle is one radian so a radian is
defined as the angle which produces an arc length of one radius
ANGULAR VELOCITY ω
Angular velocity is the rate of change of angle per second. Although rev/s is commonly
used to measure angular velocity, we should use radians/s (symbol ω). Note that since a
circle (or revolution) is 2π radian we convert rev/s into rad/s by ω= 2πN.
Also note that since one revolution is 2π radian and 360o we convert degrees into radian
as follows. θ radian = degrees x 2π/360 = degrees x π/180
DEFINITION
angular velocity = ω = angle rotated/time taken = θ/t
EXAMPLE No.3
A wheel rotates 200o in 4 seconds. Calculate the following.
i. The angle turned in radians?
ii. The angular velocity in rad/s
SOLUTION
θ = (200/180)π = 3.49 rad.
ω = 3.49/4 = 0.873 rad/s
Angular velocity can be expressed as (angular velocity = constant):
ω = θ / t (2)
where
ω = angular velocity (rad/s)
θ = angular distance (rad)
t = time (s)
Angular velocity and rpm:
ω = 2 π n / 60 (2a)
where
n = revolutions per minute (rpm)
π = 3.14…
ANGULAR ACCELERATION α
Angular acceleration (symbol α) occurs when a wheel speeds up or slows down. It is
defined as the rate of change of angular velocity. If the wheel changes its velocity by Δω
in t seconds, the acceleration is
α = Δω / t rad/s2
EXAMPLE No.4
A disc is spinning at 2 rad/s and it is uniformly accelerated to 6 rad/s in 3 seconds.
Calculate the angular acceleration.
SOLUTION
α = Δω/t = (ω2 – ω1)/t = (6 -2)/3 = 1.33 rad/s2
Angular velocity can also be expressed as (angular acceleration = constant):
ω = ωo + α t (2c)
where
ωo = angular velocity at time zero (rad/s)
α = angular acceleration or deceleration (rad/s2)
Angular Displacement
Angular distance can be expressed as (angular acceleration is constant):
θ = ωo t + 1/2 α t2 (2d)
Combining 2a and 2c:
ω = (ωo^2 + 2 α θ)^1/2
Angular Acceleration
Angular acceleration can be expressed as:
α = dω / dt = d2θ / dt2 (2e)
where
dθ = change of angular distance (rad)
dt = change in time (s)
LINK BETWEEN ANGULAR AND LINEAR MOTION
Consider a point moving on a circular path as shown.
The length of the arc = s metres.
Angle of the arc is θ radians
The link is s = r θ
Suppose the point P travels the length of the arc in tme t seconds. The wheel rotates θ
radians and the point travels a distance of r θ.
The velocity along the circular path is v = r θ/t =r ω
Next suppose that the point accelerates from angular velocity ω1 to ω2.
The velocity along the curve also changes from v1 to v2.
Angular acceleration = α = (ω2 – ω1)/t
Substituting ω = v /r
α = (v 2/r – v 1/r )/t = a/r hence a =r α
EXAMPLE No.5
A car travels around a circular track of radius 40 m at a velocity of 8 m/s. Calculate
its angular velocity.
SOLUTION
v = ωr
ω = v/r = 8/40 = 0.2 rad/s
The tangential velocity of a point in angular velocity – in metric or imperial units like m/s or ft/s – can be calculated as
v = ω r (2b)
where
v = tangential velocity (m/s, ft/s, in/s)
r = distance from center to the point (m, ft, in)
Angular Moment – or Torque
Angular moment or torque can be expressed as:
T = α I (2f)
where
T = angular moment or torque (N m)
I = Moment of inertia (lbm ft2, kg m2)
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