Question

A system of particles with masses \(m_{\alpha}\) moves in a gravitational field \(\boldsymbol{g}\). Show that if \(\boldsymbol{F}_{\alpha}=-m_{\alpha} \boldsymbol{g}\) where \(\boldsymbol{g}\) is constant, then \(\sum_{\alpha} \boldsymbol{F}_{\alpha}=m \boldsymbol{g} \quad\) and \(\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha}=m \boldsymbol{c} \wedge \boldsymbol{g}\), where \(c\) is the position vector of the centre of mass and \(m\) is the total mass. Deduce that the effect of a uniform gravitational fieldon the total linear momentum and angular momentum is the same as that of a single force \(m g\) acting through the centre of mass. Show by counter-example that this is not true for a non-uniform field.

Short answer

Answer: No, the total linear momentum and total angular momentum relationships do not hold true for a non-uniform gravitational field, as shown by the counter-example in Step 4. In this case, the sum of forces differed from the expected -mg, and the sum of angular momenta was not equal to m * (position vector of centre of mass) x (gravitational field vector). These relationships only hold true for a uniform gravitational field.

Step-by-step solution

Define the given equation and calculate the sum of forces

We are given that \(\boldsymbol{F}_{\alpha} = -m_{\alpha}\boldsymbol{g}\), so we can first calculate the sum of forces, \(\sum_{\alpha} \boldsymbol{F}_{\alpha}\): $$\sum_{\alpha} \boldsymbol{F}_{\alpha} = \sum_{\alpha} (-m_{\alpha}\boldsymbol{g}) = -\boldsymbol{g} \sum_{\alpha} m_{\alpha}$$ Recall that the total mass of the system, \(m\), is just the sum of the individual particle masses, so \(m = \sum_{\alpha} m_{\alpha}\). Thus, we can substitute it into the above equation: $$\sum_{\alpha} \boldsymbol{F}_{\alpha} = -mg$$

Define the centre of mass and calculate the sum of angular momenta

Recall that the position vector of the centre of mass, \(\boldsymbol{c}\), is given by: $$\boldsymbol{c} = \frac{1}{m}\sum_{\alpha} m_{\alpha}\boldsymbol{r}_{\alpha}$$ Now, let's calculate the sum of angular momenta, \(\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha}\): $$\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha} = \sum_{\alpha} (\boldsymbol{r}_{\alpha} \wedge (-m_{\alpha}\boldsymbol{g})) = -\sum_{\alpha} m_{\alpha} (\boldsymbol{r}_{\alpha} \wedge \boldsymbol{g})$$ Now, we can factor out the constant \(\boldsymbol{g}\): $$-\sum_{\alpha} m_{\alpha} (\boldsymbol{r}_{\alpha} \wedge \boldsymbol{g}) = -\boldsymbol{g}\sum_{\alpha} m_{\alpha}\boldsymbol{r}_{\alpha}$$ Recall the equation for the centre of mass from before: $$\boldsymbol{c} = \frac{1}{m}\sum_{\alpha} m_{\alpha}\boldsymbol{r}_{\alpha}$$ Multiplying both sides by the total mass \(m\), we get: $$m\boldsymbol{c} = \sum_{\alpha} m_{\alpha}\boldsymbol{r}_{\alpha}$$ Substitute this into our equation for the sum of angular momenta: $$\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha} = -\boldsymbol{g}(m\boldsymbol{c})$$ So, $$\sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha} = m \boldsymbol{c} \wedge \boldsymbol{g}$$

Show the effect of a uniform gravitational field

From the previous steps, we have shown that the sum of forces equals \(-mg\) and the sum of angular momenta equals \(m \boldsymbol{c} \wedge \boldsymbol{g}\). This means that the effect of the uniform gravitational field on the total linear momentum and total angular momentum of the system of particles is the same as that of a single force, \(mg\), acting through the centre of mass.

Counter-example for a non-uniform field

To show that the same relationships do not hold true for a non-uniform field, let's consider a simple system with two particles, each particle having a mass equal to \(m_{1} = m_{2} = 1\) kg, and located at distances \(r_{1} = 0\) and \(r_{2} = 1\) m, respectively, from the origin. In a non-uniform field, the gravitational force acting on each particle can be represented by \(\boldsymbol{F}_{1} = -k(r_1) \hat{\boldsymbol{r}}_1 = -0\hat{\boldsymbol{r}}_1\) and \(\boldsymbol{F}_{2} = -k(r_2) \hat{\boldsymbol{r}}_2 = -k\hat{\boldsymbol{r}}_2\), where \(k\) is a constant. Thus, the sum of forces is: $$\boldsymbol{F}_{1} + \boldsymbol{F}_{2} = 0 - k\hat{\boldsymbol{r}}_2 \neq -mg$$ In this case, the sum of angular momenta is: $$\boldsymbol{r}_{1} \wedge \boldsymbol{F}_{1} + \boldsymbol{r}_{2} \wedge \boldsymbol{F}_{2} = \boldsymbol{r}_{1} \wedge 0 + \boldsymbol{r}_{2} \wedge (-k\hat{\boldsymbol{r}}_2) = -r_2 k \hat{\boldsymbol{r}}_3 \neq m\boldsymbol{c} \wedge \boldsymbol{g}$$ Therefore, for a non-uniform gravitational field, the total linear momentum and the total angular momentum relationships do not hold true, as shown by this counter-example.

Key concepts

Gravitational Field

The concept of a gravitational field is essential in understanding how objects interact with the forces of gravity surrounding them. A gravitational field can be thought of as an invisible force field that affects the motion of objects. This field is generated by masses and exerts forces on any mass within the field's influence.
In the problem we discussed, the gravitational field is denoted by \( \boldsymbol{g} \) and is assumed to be constant. This means that the strength and direction of gravity do not change over the space that the particles occupy, which is often an idealization used to simplify calculations.

• A uniform gravitational field applies the same force per unit mass everywhere within the field.
• This consistency allows us to predict the behavior of the particles affected by it.
• Understanding how forces such as \( \boldsymbol{F}_{\alpha} = -m_{\alpha} \boldsymbol{g} \) act helps us determine the resultant motion and dynamics of the system.

Gravitational fields are crucial in dynamics as they provide a simple model to test and understand various physical interactions.

Centre of Mass

The centre of mass is a geometric point in a body or system of particles where the entire mass could be considered to be concentrated. This concept is useful because it simplifies the description of the motion of the body or system, allowing us to describe it with fewer variables.
In our exercise, the centre of mass is expressed using the position vector \( \boldsymbol{c} \), which can be calculated using the formula:
\[ \boldsymbol{c} = \frac{1}{m}\sum_{\alpha} m_{\alpha} \boldsymbol{r}_{\alpha} \]
This equation shows how each particle's position and mass contribute to the overall centre of mass.

• The centre of mass moves as if all the system's mass and all external forces were concentrated at that point.
• This simplifies the effects of the gravitational field on the system, representing it as a single force \( m\boldsymbol{g} \) acting through the centre of mass.
• This representation helps us to predict how the system behaves dynamically as affected by external forces.

The centre of mass is an efficient tool for solving complex dynamics problems.

Particle Dynamics

Particle dynamics involves the study of forces and their effects on the motion of individual particles or systems of particles. It incorporates Newton's laws of motion, which are fundamental in describing how objects respond to external forces.
In our case, we are considering a system of particles that are influenced by a gravitational field. Each particle experiences a gravitational force \( \boldsymbol{F}_{\alpha} = -m_{\alpha} \boldsymbol{g} \).

• The first aspect is calculating the resultant force: the sum of all individual forces acting on particles in the system is calculated, leading us to \( \sum_{\alpha} \boldsymbol{F}_{\alpha} = m\boldsymbol{g} \).
• The second is understanding the rotational effects: determining the sum of angular momenta \( \sum_{\alpha} \boldsymbol{r}_{\alpha} \wedge \boldsymbol{F}_{\alpha} = m\boldsymbol{c} \wedge \boldsymbol{g} \).
• These calculations show how uniformly distributed forces affect the particle system.

Through particle dynamics, we gain insights into how systems like planets, stars, or engineered systems move and interact.

Uniform and Non-Uniform Fields

The distinction between uniform and non-uniform fields is important when analyzing the effects of gravitational forces.
A uniform gravitational field, like the one in our exercise, applies the same intensity across all points. This means it causes uniform acceleration on all particles, making the calculations and predictions straightforward, as seen with the total effects on linear and angular momentum.
A non-uniform field, however, has varying force intensities at different points, leading to complex effects on the system. The problem presents this with an example:

• Two particles in a non-uniform field experience different forces depending on their positions, illustrated by \( \boldsymbol{F}_{1} = 0 \) and \( \boldsymbol{F}_{2} = -k\hat{\boldsymbol{r}}_2 \).
• This non-uniformity leads to the sum of forces \( \boldsymbol{F}_{1} + \boldsymbol{F}_{2} \) not equaling \( -mg \) as it did in the uniform case.
• The calculation shows that the assumptions valid under a uniform field can fail in a non-uniform one, altering momentum results significantly.

Understanding the transition from uniform to non-uniform fields is crucial for addressing more complex real-world physics problems.