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What is Newton's second law

  1. Jul 24, 2014 #1
    Definition/Summary

    Net force = rate of change of momentum (= mass times acceleration of the centre of mass, if the mass is constant)

    Net impulse = change of momentum

    These are vector equations, so they apply to each direction individually

    The rotational versions are:
    net torque = rate of change of angular momentum
    net impulsive torque = change of angular momentum

    Equations

    Newton's second law:

    [tex]\sum_{\text{all}}\mathbf{F}\,=\,\frac{d(m\mathbf{v}_{c.o.m.})}{dt}[/tex]

    If m is constant:

    [tex]\sum_{\text{all}}\mathbf{F}\,=\,m\frac{d(\mathbf{v}_{c.o.m.})}{dt}\,=\,m \mathbf{a}_{c.o.m.}[/tex]

    Torque version:

    [tex]\sum_{\text{all}}\mathbf{\tau}\,=\,\frac{d\mathbf{\mathcal{L}}}{dt}[/tex]

    Torque version for rigid body:

    [tex]\sum_{\text{all}}\mathbf{\tau}\,=\,I\mathbf{\alpha}[/tex]

    Extended explanation

    Net force = mass times acceleration:

    This only applies when the mass is constant.

    For example, it does not apply to a rocket, whose mass is continually reduced by the burning of fuel.

    "Acceleration" means the acceleration of the centre of mass: this even applies when the body is rotating: the sum of the (linear) forces equals the mass times the (linear) acceleration of the c.o.m.:

    [tex]\sum\mathbf{F}\ =\ m\,\mathbf{a}_{c.o.m.}[/tex]​

    Standard (non-impulsive) version:

    Newton's second law states that the total vector sum of the forces on a body is equal to the rate of change of its momentum. This formulation is valid when mass changes, and is even valid at relativistic speeds.​

    [tex]\sum\mathbf{F}\ =\ \frac{d(m\mathbf{v})}{dt}[/tex]​
    Component version:

    Force and momentum are vectors, so Newton's second law is a vector equation.

    That means that it applies to the components in any direction, and so the law can also be written:

    In any direction, the total sum of the components of the forces on a body in that direction is equal to the rate of change of the component of its momentum in that direction.​

    [tex]\sum\mathbf{F}_k\ =\ \frac{d(m\mathbf{v}_k)}{dt}[/tex]​
    Torque:

    The vector nature of the equation also means that its cross-product with any fixed vector gives another vector equation:

    Net torque on a point particle = rate of change of angular momentum:​

    [tex]\mathbf{\tau}_{total}\equiv\mathbf{r} \times \sum\mathbf{F}=\frac{d}{dt}(\mathbf{r} \times \mathbf{p})[/tex]​

    Combining this with the strong version of Newton's third law (action is equal and opposite to reaction, and in the same line) gives:

    Net torque on a rigid body = rate of change of angular momentum:​

    [tex]\mathbf{\tau}_{total}\equiv\sum\mathbf{r} \times \mathbf{F}=\frac{d}{dt}\left(\int\mathbf{r} \times d\mathbf{p}\right)= \frac{d \mathbf{\mathcal{L}}}{dt}[/tex]​

    This is also a vector equation, of course, and so we may take one component at a time, to give the scalar equation:

    Net torque about any axis = rate of change of angular momentum about that axis.​

    Impulsive version:

    Impulse is the integral of force times time.

    By comparison, work done is the integral of force times distance.

    Sometimes (for example, when a bat hits a ball), the force changes quickly, and it is difficult to measure it, but it is easy to measure the overall effect of the force. The impulse is that overall effect.

    Newton's second law states that the total vector sum of the impulses on a body is equal to the change of its momentum.​

    Collisions, and conservation of momentum:

    A collision or explosion is impulsive, and there are no external impulsive forces in a collision or explosion (in particular, the sum of the external forces is zero :wink:), so the impulsive version of the law states:

    Total momentum immediately before = total momentum immediately after:​

    [tex]\sum m_i\,u_i\,=\,\sum m_i\,v_i[/tex]​

    In other words: conservation of momentum applies to all collisions or explosions.

    By comparison, conservation of energy applies only to elastic collisions. Most collisions are not elastic.

    Static problems:

    When nothing moves, the momentum is constant, and so Newton's second law states:

    The total sum of the forces on a body in equilibrium is zero.​

    This applies separately to each body in a system, or to the external forces on any combination of bodies.

    When there are only three forces, a vector triangle may be used.

    Sliding problems (normal components):

    When one body slides against another along a common flat surface, the acceleration normal (perpendicular) to that surface is zero.

    If the surface is curved, the acceleration normal to the surface is not zero, but is centripetal towards the instantaneous centre of curvature of the surface (note: this is a matter of geometry, not physics :wink:).

    Accordingly, Newton's second law for normal components states:

    The total sum of the components of the normal forces on a sliding body on a flat surface is zero, and on a curved surface equals the mass times the centripetal acceleration.​

    String problems:

    When a number of bodies are connected by a string (an inextensible "massless" cable), in a straight line, all have the same (vector) acceleration, and so the tension in the string can be ignored (as an internal force), and the bodies can be treated together as a single body with the same total mass and acceleration and subject only to the external forces.

    When the string goes over a "massless" pulley, so that the string is in two parts in different directions, and the tensions either side of the pulley are the same, the bodies still all have the same scalar acceleration, and the bodies can be treated together as a single body with the same total mass and acceleration and subject only to the components of the external forces along the string.

    When the string goes over a massive pulley, so that the string is in two parts in different directions, and the tensions either side of the pulley are different, the previous paragraph applies except that the pulley must be treated as a body with the same acceleration and with an effective mass equal to the moment of inertia divided by the radius squared: [itex]m_{eff}\ =\ I/r^2[/itex] (for a uniform cylindrical pulley, that is half the actual mass: [itex]m_{eff}\ =\ m/2[/itex]).

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
     
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