Kinematics

It will help us in working examples if we summarise the relations which apply in kinematics, that is, the study of the movement of bodies irrespective of the forces acting upon them.

We shall consider only the two simple cases, those of uniform velocity and uniform acceleration.

Symbols and units will be as follows –

Time = t (sec)

Distance = s (metres)

Velocity (initial) = и (metres per sec) Velocity (final) = v (metres per sec)

Acceleration = a (metres per sec per sec)

Uniform velocity

If velocity is uniform at и metres per sec clearly

Distance travelled = Velocity X Time
or s = ut

Uniform acceleration

Final velocity = Initial velocity + Increase of velocity or v = и + at

Distance travelled = Initial velocity X Time

+ t Acceleration X Time squared

i. e. s = ut + at2

Final velocity squared = Initial velocity squared

+ 2 X Acceleration X Distance

or Vі = ur + las

With the aid of these simple formulae – all of which are founded on first prin­ciples – it is easy to work out problems of uniform velocity or uniform acceleration (see Examples 1.2 to 1.4).

EXAMPLE 1.2

If, during a take-off run an aeroplane starting from rest attains a velocity of 90 km/h in 10 seconds, what is the average acceleration?

SOLUTION Initial velocity и = 0 Final velocity v = 90 km/h = 25 m/s Time t = 10 sec a = ?

Since we are concerned with u, v, t and a, we use the formula v = и + at 25 =0 + 10d a = 25/10 = 2.5 m/s2

EXAMPLE 1.3

How far will the aeroplane of the previous example have travelled during the take-off run?

SOLUTION

и = 0, v = 25 m/s, t = 10 sec, a = 2.5 m/s2 To find s, we can either use the formula Final velocity squared = Initial velocity squared

+ 2 X Acceleration X Distance

s = ut + at2 = 0 + у X 2.5 X 102

= 125 m

or у1 = и2 + las 25 X 25 = 0 + 2 X 2.5 X s :.s = (25 X 15)1(1 X 2.5)

= 125 m

EXAMPLE 1.4

A bomb is dropped from an aeroplane which is in level flight at 200 knots at a height of 3500 m. Neglecting the effect of air resistance, how long will it be before the bomb strikes the ground, and how far horizontally before the target must the bomb be released?

SOLUTION

To find the time of fall we are concerned only with the vertical velocity, which was zero at release, с. и = 0

a = acceleration of gravity = 9.81 m/s2 s = vertical distance from aeroplane to ground = 3500 m t = ?

We need the formula connecting u, a, s and t, i. e. s = ut + at2

3500 = 0 + і X 9.81 X t2

t2 = (3500/9.81) X 2 = 713 .’. t = 27 sec (approx)

Since we are neglecting the effect of air resistance, the horizontal velocity of the bomb will, throughout the fall, remain the same as it was at the moment of release, i. e. the same as the velocity of the aeroplane, namely 200 knots or, converting into metres per second, (200 X 1852)73600 = 103 m/s (approx).

Therefore the distance that the bomb will travel forward during the falling time of 27 s will be 103 X 27 = 2781 m.

This, of course, is the distance before the target that the bomb must be released.

Note that in Example 1.4 we have neglected air resistance. Since we are interested in flying this may seem rather a silly thing to do, because we are only able to fly by making use of the same principles that are responsible for air resistance. In fact, too, the effects of air resistance on bombs are of vital importance and are always taken into account when bombing. But it is better to learn things in their most simple form first, then gradually to add the com­plications. As these complications are added we get nearer and nearer to the truth, but if we are faced with them all at once the picture becomes blurred and the fundamental principles involved fail to stand out clearly.

Other examples on kinematics will be found in Appendix 3, and the reader who is not familiar with examples of this type is advised to work through them.

Motion on curved paths

It has already been emphasised that bodies tend to continue in the same state of motion, and that this involves direction as well as speed. It is clear, there­fore, that if we wish to make a body change its motion by turning a comer or travelling on a curved path, we must apply a force to it in order to make it do so, and that this will apply even if the speed of the body does not change. This is a force exactly similar to the one that is required to accelerate an aircraft, that is to say: the force must be proportional to the mass of the body and to the acceleration which it is desired to produce. But what is the acceleration of a body that is going round a corner? Is there, in fact, any acceleration at all if the speed remains constant? And in what direction is the acceleration?

Let us deal with the last question first. There is another part of Newton’s second law which has not so far been mentioned, namely that the rate of change of momentum of the body will be in the direction of the applied force. If the mass of the body does not change as it goes round the corner the accel­eration must be in the direction of the force. But is there any acceleration if the speed does not change? Yes – because velocity is what we call a vector quan­tity, that is to say, it has both magnitude and direction, while speed has only magnitude. Thus if the direction of motion changes, the velocity changes even though the speed remains unaltered. But at what rate does the velocity change? – in other words, what is the acceleration? and in what direction is it?