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## Main Question or Discussion Point

**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*: this

**of the centre of mass***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 ), 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 ).

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]).

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