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Revision:Dynamics

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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.

$\displaystyle F = f(t)$

By Newton's second law, F = ma , gives;

$\displaystyle ma = f(t)$

$\displaystyle a = \frac{\mathrm{d}v}{\mathrm{d}t}$

$\displaystyle m\frac{\mathrm{d}v}{\mathrm{d}t} = f(t)$

$\displaystyle mv = \int f(t) \mathrm{d}t + c$

Similarly, if is a function of the displacement, that is;

$\displaystyle F = g(x)$

then using Newton's second law gives;

$\displaystyle ma = g(x)$

$\displaystyle a = \frac{\mathrm{d}(\frac{1}{2}v^2)}{\mathrm{d}x}$

$\displaystyle \Rightarrow m\frac{\mathrm{d}(\frac{1}{2}v^2)}{\mathrm{d}x}= g(x)$

Integrating with respect to gives;

$\displaystyle \frac{1}{2}mv^2 = \int g(x) \mathrm{d}x + k$

Impulse and momentum

For a variable force f(t) , Newton's second law, f(t) = ma can be written;

$\displaystyle f(t) = m\frac{\mathrm{d}v}{\mathrm{d}t}$

Let the particle have a constant mass and velocity at time T_1 and a velocity at time T_2 . Then integrating with respect to over the integral from T_1 to T_2 gives;

$\displaystyle\int_{T_1}^{T_2} f(t) \mathrm{d}t = m \int_V^U 1 \mathrm{d}v$

$\displaystyle = m[v]_V^U$

$\displaystyle = m(V - U)$

$\displaystyle = mV - mU$

mV - mU is the change of momentum of the particle. The quantity

$\displaystyle\int_{T_1}^{T_2} f(t) \mathrm{d}t$

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 $= \displaystyle\int_{T_1}^{T_2} f(t) \mathrm{d}t$

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 $= F \times s$

$\displaystyle Fs = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

(where and are the initial and final speeds respectively of the particle)

$\displaystyle a = v\frac{\mathrm{d}v}{\mathrm{d}x}$

$\displaystyle f(x) = ma = mv\frac{\mathrm{d}v}{\mathrm{d}x}$ ---------- (1)

If the action of the force causes the particle to increase its speed from to while moving from the point x = x_1 to the point where x = x_2 then integrating (1) with respect to gives;

$\displaystyle\int_{x_1}^{x_2} f(x) \mathrm{d}x = m\int_U^V v \mathrm{d}v = m[\frac{1}{2}v^2]_U_V$

$= \frac{1}{2}mV^2 - \frac{1}{2}U^2$ ---------- (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 $= \displaystyle\int_{x_1}^{x_2} f(x) \mathrm{d}x$

3.2 - Newton's law of gravitation

The force of attraction between two bodies of masses M_1 and M_2 is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

$\displaystyle F = \frac{GM_1M_2}{d^2}$

is the constant of gravitation.

Units of

$\displaystyle F = \frac{GM_1M_2}{d^2}$

(where is a function of , the distance between the two bodies)

$\displaystyle \Rightarrow G = F\frac{d^2}{M_1M_2}$

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 M_e 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;

$\displaystyle F = \frac{GmM_e}{R^2}$

However, the force with which the earth attracts the particle is the weight of the particle, so;

$\displaystyle F = mg$

$\displaystyle \Rightarrow mg = \frac{GmM_e}{R^2}$

$\displaystyle \Rightarrow g = \frac{GM_e}{R^2}$

3.3 - Simple harmonic motion

Equation for acceleration:

$\displaystyle a = -\omega ^2 x$

Always get the acceleration of the oscillating object in this form to prove that the motion is simple harmonic.

Equation for speed:

$\displaystyle v^2 = \omega ^2(a^2 - x^2)$

$\displaystyle v_{max} = \omega a$

speed
$\omega = \frac{v}{a}$
amplitude of S.H.M
displacement moved by particle

Equations for displacement:

$\displaystyle x = a\sin(\omega t)$

Particle starts at centre of oscillation ( when )

$\displaystyle x = a\cos(\omega t)$

Particle starts at either end of oscillation ( when )

Equations for period of motion:

Period $\displaystyle = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{v_{max}}{a}} = \frac{2\pi a}{v_{max}}$

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.