At the end of this module you should be able to:
Table of contents
Worked Examples
A scalar quantity is one which can be completely specified by a number that expresses its magnitude in an appropriate unit.
Some examples are:
Scalar quantity SI unit
Mass kg
Length m
Area m2
A vector quantity requires a number (magnitude)
anddirection.
Some examples are:
Vector quantity SI unit
Velocity m/s
Acceleration m/s2
Force N (or kg m/s2)
Linear displacement and distance
The linear displacement is the length moved in a given direction - it is a vector quantity.
The magnitude of the displacement is the distance - a scalar
quantity.
The linear velocity is the rate of change of displacement with time. As displacement is a vector so velocity is a vector.
The magnitude of the velocity is speed. It is the rate of change of distance with time - hence it is a scalar.
If a body moves with uniform velocity then it must move in a fixed direction with constant speed.
The average speed of a body is the total distance moved divide by the total time taken.
Distance - time curve
A graph plotted for distance (s) against time (t), might look like that in Figure 1.1:
Figure 1.1: Distance-Time Curve
As speed is rate of change of distance with time, the slope, gradient, of the s/t curve is the speed.
Over the linear section OA of the curve the speed must be uniform.
Between A and B the gradient is becoming less and less, hence the body is slowing down.
At B the body is stopped (distance is not increasing) and remains at
rest between B and C.
The linear acceleration of a body is the rate of change of linear velocity with time. It is a vector.
If acceleration is uniform the speed must be increasing by equal amount in equal time intervals.
A car is travelling along a straight road at 13 m/s. It accelerates uniformly for 15 s until it is moving at 25 m/s
Solution
Speed - Time curve
A graph of speed (v) of a body plotted against time (t) might be as shown by the graph in figure 1.2:
Figure 1.2: Speed-Time Curve
As acceleration is rate of change of speed (v) with time (t), the slope, gradient, of the v/t curve is the speed.
In Figure 1.2 the gradient between A and B is increasing - hence acceleration - is increasing, between B and C it is constant, between C and D it is decreasing.
If in the small time interval dt, the speed is v. The distance covered in the time dt is
The total distance s travelled in the time interval between t1and t2 is the integral of this i.e.
This integral is the same as the area under the curve.
Thus the distance travelled in any time interval is the area under the
v/t
curve between the start and end time.
Equations for linear uniformly accelerated motion
If a body that is moving in a straight line and started with initial speed u undergoes a uniform acceleration a for a time t until its velocity is v, then the speed time curve would look like that in Figure 1.3:
Figure 1.3: Uniformly accelerated linear motion
Since a is uniform, its magnitude is
Since the distance done = average speed ´ time then
These four equations are the equations for linear, uniformly
accelerated motion. They all contain 4 unknowns, you must know three
before you can find the fourth.
A car starts from rest and accelerates in a straight line at 1.6 m/s2 for 10s.
a)
Initial speed u = 0.
Acceleration a = 1.6 m/s
Time t = 10s
Use Equation 1.1 to give v
b)
Use Equation 1.3 to give distance gone
c)
Initial speed u = 16 m/s
Final speed v = 0
Distance s = 20 m
Use Equation 1.4 to find acceleration
Deceleration is negative acceleration so
Deceleration = 6.4 m/s2
A hoist starts at ground level and accelerates as 1.2 m/s2 for 5s. It then moves with uniform speed for 10s and is finally brought to rest at the top of a building with a retardation (deceleration) of 2.0 m/s2.
a)
We can draw the speed - time curve shape as in Figure 1.4
Figure 1.4: Speed-Time curve shape for Worked Example 1.3
But to complete this we need to find the unknowns, the final speed u and the time of the retardation.
Initial speed u = 0.0
Acceleration a = 1.2 m/s2
Time t = 5 s
Use Equation 1.1 to give final speed
For the retardation time:
Initial speed u = 6.0 m/s
Final speed v = 0
Acceleration a = -2.0 m/s2 (NOTE: Negative sign as retardation)
Using Equation 1.1 to give time
We can now complete the speed time curve by putting in the values, as shown in Figure 1.5
Figure 1.5: Completed Speed-Time curve for Worked Example 1.3
b)
The total distance travelled is the area under the speed-time curve. i.e. the area of the trapezium OABC
Thus the height of the building is 84m
Free fall under gravity
When a body falls to earth freely (without any other forces involved) the acceleration is called acceleration due to gravity. It is often given the symbol g.
Provided that air resistance is negligible all bodies, heavy or light fall at the same acceleration. Although g varies very slightly at different points on he earth g = 9.81 m/s2 can be used in most calculations, and will be used in the rest of these notes. (To four decimal places g = 9.8142 m/s2.)
A worker drops a hammer from the top of a 60m high building. If the speed of sound in air is 340 m/s, how long does the worker have to shout down to warn colleagues (if his warning is to reach them before the hammer!) Neglect air resistance.
Solution:
The solution is, the length of time for the hammer to reach the ground minus the length of time it takes for the shouted warning to reach the workers on the ground.
For the hammer:
Initial speed u = 0
Acceleration a = 9.81 m/s2
Distance s = 60m
Use Equation 1.3 to give time.
For the shout
Difference in travel time between hammer and shout:
3.5 - 0.18 = 3.32 s
The warning must be shouted within 3.32 seconds of dropping the hammer.
Consider a point P moving along a line QP as shown in Figure 1.6
Figure 1.6: Angular motion
Let OX be a fixed line and q be the angle made with OP at some time t. The angular velocity w of P about O is the rate of change of qwith time in the sense of increasing q .
Hence
The SI unit for angular speed is radians per second, rad/s.
If the angular velocity of the point P in Figure 1.6 is changing with time, then the angular acceleration, a , of P is the rate of change of its angular velocity:
In SI units angular acceleration is measured in rad/s2.
If the angular acceleration is uniform, then when angular speed changes from w1 to w2in time t, its magnitude is
Relationship between linear and angular motion
Relationship between linear speed and angular speed
If a point P move round a circle of radius r with constant linear speed, v, (see Figure 1.7) then the angular speed, w, will be constant at
Figure 1.7: Circular Motion
Where t is the time to move from Q to P along the arc QP of the curve.
However, arc length QP is rqwhen q is measured in radians. Hence linear speed v is
Equation 1.10
Relationship between angular speed and frequency of rotation
Let P in Figure 1.7 rotate with constant frequency of n rev/s.
Since for each revolution turned the angle is 2prad, then the number of radians turned through per second is 2p n. This is the angular speed. So we can write
Relationship between linear acceleration and angular acceleration
By Equation 1.6 and Equation 1.10
and
as r is constant this can be written
and as is linear
acceleration a,
Uniform circular motion - (acceleration)
Consider again a point X moving round a circle, radius r, at uniform speed v.
Since the direction of motion is changing form instant to instant, the velocity is changing and hence point X has an acceleration.
Let AB represent the velocity vector when X is at point Q and AC represent
the velocity vector when X is at point P, as shown in Figure 1.8. The velocity
vectors at points Q and P are tangential to the circle of motion, sides
AB and AC are parallel to these vectors.
Figure 1.8: Angular Acceleration
The line BC represents the change in velocity as the point moves from Q to P.
The average acceleration of P is
length of arc QP = rq
So
time to travel from Q to P = rq / v
Also
To find the acceleration of P at Q, let the angle q/2 tend to zero, so
and the acceleration is given by
Hence, the acceleration of a point moving round a circle with radius r, at constant speed v is v2/r towards the centre of the circle.
A flywheel, diameter 1.1m, rotating at 1200 rev/min slows down at a constant rate to 900 rev/min in 30 s.
Find:
a)
From Equation 1.11 angular speed w= 2p n
Initial n = 1200 rev/min = 1200 /60 = 20 rev / s
b)
Final n = 900 rev/min = 900/60 = 15 rev/s
c)
Use Equation 1.7
d)
Equation 1.10 gives the linear speed, so initial speed on edge of flywheel is
The spin dryer in a washing machine is a cylinder with diameter 500mm. It spins at 900 rev/min.
Find
a)
Frequency of rotation in rev/s
n = 900/60 = 15 rev/s
From Equation 1.11
From Equation 1.10 the linear speed is
b)
From Equation 1.13
Note how large this is, it is 227 times the acceleration due to gravity!