TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Dynamics
These notes are based on the requirements of the M3 A Level mathematics module.
3.1 - Newton's laws for a particle moving in a straight line when the applied force is variable
Consider a particle of mass m moving under the influence of a force 
which is a function of time.
By Newton's second law, 
, gives;
Similarly, if is a function of the displacement, that is;
then using Newton's second law gives;
Integrating with respect to 
gives;
Impulse and momentum
For a variable force 
, Newton's second law, 
can be written;
Let the particle have a constant mass 
and velocity 
at time 
and a velocity 
at time 
. Then integrating with respect to 
over the integral from to gives;
![\displaystyle = m[v]_V^U](http://www.thestudentroom.co.uk/latexrender/pictures/cd/cd254079098c118cbcd8a14df7e172bd.png "\displaystyle = m[v]_V^U")
is the change of momentum of the particle. The quantity
is the impulse of the variable force on the particle.
Thus the equation;
Impulse = change of momentum
is still valid but for a variable force the impulse is given by;
Impulse 
Work and energy
When a constant force acts on a particle of mass and moves it through a distance 
in the direction of the force, the work done by the force is defined by;
Work done 
(where 
and 
are the initial and final speeds respectively of the particle)
---------- (1)
If the action of the force causes the particle to increase its speed from to while moving from the point 
to the point where 
then integrating (1) with respect to gives;
![\displaystyle\int_{x_1}^{x_2} f(x) \mathrm{d}x = m\int_U^V v \mathrm{d}v = m[\frac{1}{2}v^2]_U_V](http://www.thestudentroom.co.uk/latexrender/pictures/13/1372c94a2a6b2b7096b768fc61362269.png "\displaystyle\int_{x_1}^{x_2} f(x) \mathrm{d}x = m\int_U^V v \mathrm{d}v = m[\frac{1}{2}v^2]_U_V")
---------- (2)
The RHS of (2) is the increase in the K.E. of the particle and the LHS is the work done by the variable force.
Therefore: work done 
3.2 - Newton's law of gravitation
The force of attraction between two bodies of masses 
and 
is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
is the constant of gravitation.
Units of
(where is a function of 
, the distance between the two bodies)
- is measured in Newtons,
- in metres
in kilograms.
So, units for are;
Nm²kg-² = kgms-²m²kg-²
- = m³/kgs²
Hence G is 6.67 x 10^-11 m³/kgs²
Relationship between and 
Consider a particle of mass which is at rest on the surface of the earth.
Suppose the mass of the earth is 
kg and its radius is 
. By Newton's law of gravitation the magnitude of the force of attraction between the particle and the earth is;
However, the force with which the earth attracts the particle is the weight of the particle, so;
3.3 - Simple harmonic motion
Equation for acceleration:
Always get the acceleration of the oscillating object in this form to prove that the motion is simple harmonic.
Equation for speed:
speed
amplitude of S.H.M
displacement moved by particle
Equations for displacement:
- Particle starts at centre of oscillation (
when 
)
- Particle starts at either end of oscillation (
when )
Equations for period of motion:
Period 
3.4 - Horizontal oscillations of a particle attached to the end of an elastic string or spring
Consider a particle 
resting on a smooth horizontal surface and attached to one end of an elastic spring whose other end is fixed to a point 
on the surface. If is pulled aside in the direction 
and then released the spring will be stretched and so will exert a tension on resulting in an acceleration.
---------- (1)
Equation (1) is in the form 
Since period 
The period of the particle's oscillations is 
String is only taut during S.H.M. When the string goes slack, 
and the particle moves at a constant speed.
So using time = distance/speed you can calculate the length of time for which the string is slack etc.
3.5 - Vertical oscillations of a particle attached to the end of an elastic string or spring
When the particle is in equilibrium;
Hence, the total extension when the particle is displaced a distance x from the equilibrium position is;
Tension at this point is;
Using Newton's second law, , gives;
![\displaystyle mg - [mg + \frac{\lambda x}{L}] = ma](http://www.thestudentroom.co.uk/latexrender/pictures/75/75ebc22881ef3e2fe8f6ac9252c5b1fc.png "\displaystyle mg - [mg + \frac{\lambda x}{L}] = ma")
String becomes slack when 
and the particle will then move freely under gravity until it falls back to again.
Comments
Originally written by Widowmaker on TSR forums.
