Momentum Equation - Linear And Angular Conservation

Last edited: 2024-11-14 17:10:58

Linear and angular momentum has certain properties. Here we explore momentum's equations, rate of change, conservation, and link to torque and impulses.

Momentum Equation

The formula for momentum is

$$p = mv,$$

where $m$ is the mass of a moving object with velocity $v$. In vector form it becomes

$$\vec{p} = m \vec{v}.$$

Change in Momentum Formula

To calculate the change in momentum you have to use the change in velocity $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$ and plug it into the expression to get the change in momentum:

$$\Delta \vec{p} = m \Delta \vec{v}$$

Rate of Change of Momentum

To get the rate of change of momentum we will take the time derivative of the momentum equation:

$$\frac{d \vec{p}}{dt} = \frac{d}{dt} (m \vec{v}).$$

Since the mass is assumed to be constant the derivative only acts on the velocity and the derivative of velocity is acceleration ($\vec{a}$) so we get:

$$\frac{d \vec{p}}{dt} = m \vec{a} = \Sigma \vec{F}.$$

As you can see we get the expression for Newton's second law which says that the sum of all forces acting upon an object is equal to the mass times acceleration of the body.

Is Impulse The Change in Momentum?

To get the impulse we have to integrate the equation above over time. In other words, we have to integrate the forces acting on an object over time:

$$\int_{t_1}^{t_2} \Sigma \vec{F} dt = \int_{t_1}^{t_2} \frac{d \vec{p}}{dt} dt = \vec{p}_2 - \vec{p}_1 = \Delta \vec{p},$$

where $\vec{p}_1$ and $\vec{p}_2$ is the linear momentum at time $t_1$ and $t_2$ respectively. As can be seen, the impulse is the change in momentum.

Conservation of Momentum Equation

Observing the equation above, if the resultant force $\Sigma \vec{F}$ is zero then the momentum does not change and is conserved. Now consider two objects, $A$ and $B$, colliding. We know from Newton's third law that each action (force) has an equal opposite reaction. So the change in momentum for object $A$ is equal to the negative of the change of object $B$, $\Delta \vec{p}_1 = -\Delta \vec{p}_2$. So the total change for the system is

$$\Delta \vec{p} = \Delta \vec{p}_1 + \Delta \vec{p}_2 = 0$$

and the momentum is conserved.

Angular Momentum Equation

The formula for angular momentum $\vec{L}_O$ for an object with mass $m$ around origin $O$ is

$$\vec{L}_O = \vec{r} \times m \vec{v},$$

where $\vec{r}$ is the distance in vector form between the origin and the object and $\vec{v}$ is the object's velocity. If it is broken down into components then you get

$$L_x = m (v_z y - v_y z)$$ $$L_y = m (v_x z - v_z x)$$ $$L_z = m (v_y x - v_x y)$$

Angular Momentum Units

Looking at the equation above the angular momentum has the units $\text{kg} \cdot \text{m}^2/\text{s} = \text{N} \cdot \text{m} \cdot \text{s}$ and in US customary units that is lb-ft-sec.

Rate of Change of Angular Momentum

To get the rate of change of the angular momentum we have to take the time derivative of the equation above which gives us

$$\frac{d \vec{L}_O}{dt} = \frac{d \vec{r}}{dt} \times m \vec{v} + \vec{r} \times m \frac{d \vec{v}}{dt}$$ $$= \vec{v} \times m \vec{v} + \vec{r} \times m \vec{a}.$$

The cross product of parallel vectors is zero so $\vec{v} \times m \vec{v}$ is zero. Therefore we have the following equation for the angular momentum rate of change:

$$\frac{d \vec{L}_O}{dt} = \vec{r} \times m \vec{a}.$$

For the linear momentum, $\Sigma \vec{F}$ represented the resultant of all forces acting upon the object, and with Newton's law, we know that $\Sigma \vec{F} = m \vec{a}$. So for a total moment $\Sigma \vec{M}_O$ around the origin $O$ is given by

$$\Sigma \vec{M}_O = \vec{r} \times m \vec{a}.$$

As you can see the formula is the same as the one for the rate of change. In other words, the rate of change of angular momentum is equal to the total moment:

$$\Sigma \vec{M}_O = \frac{d \vec{L}_O}{dt}.$$

Relationship Between Torque And Angular Momentum

Since torque is a moment that causes rotation around an axis the formula above also gives us the formula for the torque around a fixed point $O$:

$$\vec{\tau} = \frac{d \vec{L}_O}{dt}.$$

Angular Impulse And Angular Momentum

Now that we know that the total moment acting upon an object is equal to the rate of change of the angular momentum we can obtain the effect of the moment on the angular momentum over a time interval using this formula:

$$\int_{t_1}^{t_2} \Sigma \vec{M}_O dt = \int_{t_1}^{t_2} \frac{d \vec{L}_O}{dt} dt$$ $$= [\vec{L}_O]_2 - [\vec{L}_O]_1 = \Delta \vec{L}_O.$$

The angular impulse is defined as the product of moment and time, so the formula above states that the total angular impulse is equal to the change in angular momentum.

Conservation of Angular Momentum Equation

Similar to the conservation of linear momentum if the resultant moment about a fixed point $O$ is zero over an interval of time then angular momentum $\vec{L}_O$ does not change and is conserved, as seen in the equation above. If two objects, $A$ and $B$, interact over an interval of time then we know from Newton's third law that each action (force) has an equal opposite reaction and this also applies to the moment, so $\Delta \vec{L}_A = -\Delta \vec{L}_B$. The total change in the angular momentum of the system is

$$\Delta \vec{L} = \Delta \vec{L}_A + \Delta \vec{L}_B = 0$$

and we can see that the total angular momentum is conserved.