PhysicsLaws Of Motion

Laws Of Motion

Laws of Motion

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    In terms of the Lagrangian (\ref{eq:lagrangian}), the energy-momentum tensor is
    \begin{equation}
    T_{\mu\nu} = \frac{\partial{\mathcal{L}}}{\partial(\partial_{\mu}g^{\alpha\beta})} \partial_{\nu}g^{\alpha\beta} – g_{\mu\nu}{\mathcal{L}}
    \end{equation}
    and the Einstein tensor is
    \begin{equation}
    G_{\mu\nu} = R_{\mu\nu} – \frac{1}{2} g_{\mu\nu} R.
    \end{equation}

    In terms of the metric (\ref{eq:metric}), the Einstein tensor is
    \begin{equation}
    G_{\mu\nu} = R_{\mu\nu} – \frac{1}{2} g_{\mu\nu} R = \frac{1}{r^{2}} \left( -\frac{1}{2} g_{\mu\nu} + \frac{g_{\mu\nu}}{r} \right)
    \end{equation}
    and the energy-momentum tensor is
    \begin{equation}
    T_{\mu\nu}

    Laws of Motion – Explanation in Detail

    There are three laws of motion that were first formulated by Isaac Newton. They are the law of inertia, the law of acceleration, and the law of action and reaction.

    The law of inertia states that an object will remain at rest or in uniform motion in a straight line unless it is acted upon by an external force. This law is also known as the law of inertia.

    The law of acceleration states that the rate of change of velocity of an object is proportional to the force applied to it and is in the same direction as the force.

    The law of action and reaction states that for every action there is an equal and opposite reaction.

    Newton’s First Law

    Newton’s First Law states that an object will continue to move in a straight line unless it is acted upon by an unbalanced force. This means that if there is no force acting upon an object, it will continue to move in a straight line. If there is a force acting upon the object, it will change the object’s direction.

    What is Newton’s Third Law

    ?

    Newton’s Third Law is the law of action and reaction. It states that for every action there is an equal and opposite reaction.

    Derivation of First Equation of Motion

    We know that when a body is thrown vertically upwards with some velocity, it undergoes uniform acceleration. Hence, the equation of motion of a body thrown vertically upwards is:

    v = u + at

    Now, according to the question, the body is thrown upwards with the velocity of light (u = c, the speed of light). Now, the acceleration of a body thrown vertically upwards is g. Hence, the equation of motion of a body thrown vertically upwards with velocity of light is:

    v = u + at

    c = c + gt

    c = c, when t = 0

    g = 0, when t = 0

    Hence,

    c = c

    g = 0

    This is the first equation of motion of the particle.

    Derivation of Second Equation of Motion

    Now, we have to find the second equation of motion in terms of v.

    Now, v = u + at

    Now, we have to find the second equation of motion in terms of v.

    Now, v = u + at

    v = c + gt

    Differentiating both sides with respect to t, we get:

    dv/dt = d/dt [c + gt]

    dv/dt = d/dt c + d/dt gt

    dv/dt = 0 + g

    dv/

    Derivation of Second Equation of Motion

    The second equation of motion for a pendulum is

    $$ \ddot{\theta} + \frac{g}{l}\sin\theta = 0 $$

    The variables for this equation are defined as follows:

    Variable Description $\theta$ angle (radians) of the pendulum from the vertical $\dot{\theta}$ angular velocity (radians per second) of the pendulum $\ddot{\theta}$ angular acceleration (radians per second squared) of the pendulum $g$ acceleration due to gravity (9.81 $m/s^2$) $l$ length of the pendulum

    This equation is derived by using Newton’s second law of motion on the pendulum. The pendulum is a massless point particle so the only force is the tension force. The tension force acts in the direction of the velocity vector and has the following expression:

    $$ F_T = -m \ddot{\vec{r}} $$

    The tension force is the only force on the pendulum since it is a point particle and therefore has no mass. For a pendulum, the tension force is equal to the force of gravity, so

    $$ F_T = -\frac{m}{l} g \sin\theta $$

    Rearranging and substituting in the expression for tension gives the second equation of motion for the pendulum.

    Derivation

    Derivation of Third Equation of Motion

    The third equation of motion is derived from the law of conservation of angular momentum.

    The law of conservation of angular momentum states that the angular momentum of an isolated system does not change with time. It is represented mathematically by the equation:

    \mathbf{L} = \mathbf{I} \mathbf{\dot{\omega}} + \mathbf{\omega} \times \mathbf{I} \mathbf{\omega} = \mathrm{constant}

    where \mathbf{L} is the angular momentum vector, \mathbf{I} is the inertia tensor, \mathbf{\omega} is the angular velocity vector, and \times denotes the vector cross product.

    To derive the third equation of motion, we take the time derivative of both sides of the equation.

    \begin{align} \frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} &= \frac{\mathrm{d}(\mathbf{I} \mathbf{\dot{\omega}} + \mathbf{\omega} \times \mathbf{I} \mathbf{\omega})}{\mathrm{d}t} \\[5pt] &= \frac{\mathrm{d}\mathbf{I}}{\mathrm{d}t} \mathbf{\dot{\omega}} +

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