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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 F 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 F 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 x 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 m and velocity U at time T_1 and a velocity V at time T_2. Then integrating with respect to t 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 F acts on a particle of mass m and moves it through a distance s 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 u and v 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 U to V while moving from the point x = x_1 to the point where x = x_2 then integrating (1) with respect to x 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}


G is the constant of gravitation.


Units of G

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

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


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


  • F is measured in Newtons,
  • d in metres
  • M_1, M_2 in kilograms.


So, units for G are;

Nm²kg-² = kgms-²m²kg-²

= m³/kgs²

Hence G is 6.67 x 10^-11 m³/kgs²


Relationship between G and g

Consider a particle of mass m which is at rest on the surface of the earth.

Suppose the mass of the earth is M_e kg and its radius is R. 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

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


Equations for displacement:

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

  • Particle starts at centre of oscillation (x = 0 when t = 0)


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

  • Particle starts at either end of oscillation (x = a when t = 0)


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 P resting on a smooth horizontal surface and attached to one end of an elastic spring whose other end is fixed to a point O on the surface. If P is pulled aside in the direction OP and then released the spring will be stretched and so will exert a tension on P resulting in an acceleration.


\displaystyle F = ma

\displaystyle -T = ma

\displaystyle \frac{-\lambda x}{L} = ma

\displaystyle a = \frac{-\lambda x}{mL} ---------- (1)


Equation (1) is in the form a = -\omega ^2x

\displaystyle \Rightarrow \omega^2 = \frac{\lambda}{mL}


Since period \displaystyle = \frac{2\pi}{\omega}

The period of the particle's oscillations is \displaystyle 2\pi \sqrt{\frac{mL}{\lambda}}


String is only taut during S.H.M. When the string goes slack, T = 0 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;

\displaystyle T = mg

\displaystyle T = \frac{\lambda e}{L}

\displaystyle \frac{\lambda e}{L} = mg

\displaystyle e = \frac{mgL}{\lambda}


Hence, the total extension when the particle is displaced a distance x from the equilibrium position is;


\displaystyle e + x = \frac{mgL}{\lambda} + x


Tension at this point is;


\displaystyle T = \lambda \frac{(\frac{mgl}{\lambda} + x)}{L} = mg + \frac{\lambda x}{L}


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


\displaystyle mg - T = ma

\displaystyle mg - [mg + \frac{\lambda x}{L}] = ma

\displaystyle ma = \frac{-\lambda x}{L}

\displaystyle a = \frac{-\lambda x}{mL}


String becomes slack when x = -e and the particle will then move freely under gravity until it falls back to x = -e again.


Comments

Originally written by Widowmaker on TSR forums.

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