Particles kinetics
Kinetics is a branch of mechanics that examines the forces and torques responsible for the motion of objects. Unlike kinematics, which describes motion without addressing its causes, kinetics analyzes the interactions that result in changes in velocity and acceleration. By considering factors such as mass, force, and energy, this section provides essential tools for solving problems related to force-driven motion, enabling a deeper understanding of dynamic systems in engineering and physics.
Newton’s and Euler’s laws
We already encountered the first law when working on statics here.
Newton’s Second Law of Motion states:
The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship is commonly represented by the formula F = ma, where F is the net force, m is the mass, and a is the acceleration.
We are not considering at this stage relativistic effects, so the mass is constant.
Finally, Newton’s Third Law of Motion states:
For every action, there is an equal and opposite reaction.
This means that if one object exerts a force on a second object, the second object exerts an equal and opposite force back on the first.
Euler’s first law
Consider a system of N particles, each with mass m_i and subjected to an external force \mathbf{F}_i. Let \mathbf{f}_{ij} denote the internal force exerted by particle j on particle i.
For each particle i, Newton’s second law gives:
m_i \frac{\mathrm d\mathbf{v}_i}{\mathrm dt} = \mathbf{F}_i + \sum_{j \neq i} \mathbf{f}_{ij}
where \mathbf{F}_i is the external force on particle i and \sum_{j \neq i} \mathbf{f}_{ij} is the sum of internal forces acting on particle i due to all other particles j in the system.
Summing Newton’s second law over all particles in the system:
\sum_{i=1}^{N} m_i \frac{\mathrm d\mathbf{v}_i}{\mathrm dt} = \sum_{i=1}^{N} \mathbf{F}_i + \sum_{i=1}^{N} \sum_{j \neq i} \mathbf{f}_{ij}
According to Newton’s third law, the internal forces between particles i and j are equal in magnitude and opposite in direction, so \mathbf{f}_{ij} = -\mathbf{f}_{ji}. Therefore, when summing over all particles, the internal forces cancel out:
\sum_{i=1}^{N} \sum_{j \neq i} \mathbf{f}_{ij} = 0
We then have:
\sum_{i=1}^{N} m_i \frac{\mathrm d\mathbf{v}_i}{\mathrm dt} = \sum_{i=1}^{N} \mathbf{F}_i
The center of mass \mathbf{R} of the system is defined by:
\mathbf{R} = \frac{1}{M} \sum_{i=1}^{N} m_i \mathbf{r}_i
where M = \sum_{i=1}^{N} m_i is the total mass of the system, and \mathbf{r}_i is the position vector of the i^{th} particle.
Taking the second derivative of \mathbf{R} with respect to time gives the acceleration of the center of mass:
\frac{\mathrm d^2 \mathbf{R}}{\mathrm dt^2} = \frac{1}{M} \sum_{i=1}^{N} m_i \frac{\mathrm d^2 \mathbf{r}_i}{\mathrm dt^2} = \frac{1}{M} \sum_{i=1}^{N} \mathbf{F}_i
The equation above shows that the acceleration of the center of mass \frac{\mathrm d^2 \mathbf{R}}{\mathrm dt^2} is determined only by the total external force \sum_{i=1}^{N} \mathbf{F}_i acting on the system, independent of the internal forces.
The motion of the center of mass describes the overall motion of the system of particles as if all mass were concentrated at \mathbf{R} and subjected to the net external force.
To transition from a discrete sum over particles to a continuous distribution, we consider a continuous mass distribution. For this, we introduce the mass density \rho(\mathbf{r}), representing the mass per unit volume at a point \mathbf{r} within the body.
The center of mass \mathbf{R} for a continuous distribution of mass is given by:
\mathbf{R} = \frac{1}{M} \int_V \mathbf{r} \, \rho(\mathbf{r}) \, dV
where M = \int_V \rho(\mathbf{r}) \, dV is the total mass of the body, and V is the volume occupied by the mass distribution.
Taking the time derivative of the center of mass position \mathbf{R} gives the velocity of the center of mass:
\mathbf{v}_C = \frac{\mathrm d\mathbf{R}}{\mathrm dt} = \frac{1}{M} \int_V \mathbf{v} \, \rho(\mathbf{r}) \, dV
where \mathbf{v} = \frac{\mathrm d\mathbf{r}}{\mathrm dt} is the velocity at a point \mathbf{r} in the body.
Taking another time derivative, we obtain the acceleration of the center of mass:
\mathbf{a}_C = \frac{\mathrm d\mathbf{V}}{\mathrm dt} = \frac{1}{M} \int_V \mathbf{a} \, \rho(\mathbf{r}) \, dV
where \mathbf{a} = \frac{\mathrm d\mathbf{v}}{\mathrm dt} is the local acceleration at \mathbf{r}.
For a continuous distribution, the force acting on each infinitesimal mass element \rho(\mathbf{r}) \, dV is given by:
d\mathbf{F} = \rho(\mathbf{r}) \mathbf{a} \, dV
Summing up (integrating) over the entire volume, we get the total force:
\int_V d\mathbf{F} = \int_V \rho(\mathbf{r}) \mathbf{a} \, dV
According to Newton’s second law for the entire body, the net external force \mathbf{F}_{\text{ext}} acting on the body can be written as:
\mathbf{F}_{\text{ext}} = M \mathbf{a}_C = \int_V \rho(\mathbf{r}) \mathbf{a} \, dV
This result:
\mathbf{F}_{\text{ext}} = M \mathbf{a}_C
is known as Euler’s First Law for a continuous body. It states that the total external force acting on a continuous mass distribution is equal to the total mass M times the acceleration of the center of mass \mathbf{A}. This is analogous to Newton’s second law applied to the entire body, confirming that the center of mass motion fully describes the translational motion of the entire continuous system.
Euler’s second law
Newton’s Second Law states that the sum of the forces acting on a particle is equal to the time derivative of its linear momentum:
\sum \mathbf{F} = \frac{\mathrm d\mathbf{p}}{\mathrm dt}
where \mathbf{p} = m\mathbf{v} is the linear momentum of the particle. This applies to individual particles, which only have linear momentum. For a system of particles, we extend this concept.
Consider a system of particles observed from an inertial reference frame, denoted as F. Let the position vector \mathbf{r}_i represent the location of the i^{th} particle relative to the origin of the inertial reference frame. Crossing \mathbf{r}_i with both sides of Newton’s Second Law gives:
\mathbf{r}_i \times \sum \mathbf{F}_i = \mathbf{r}_i \times \frac{\mathrm d\mathbf{p}_i}{\mathrm dt}
The left-hand side, \mathbf{r}_i \times \mathbf{F}_i, defines the moment of force (torque) about the origin:
\sum \mathbf{M}_O = \sum \mathbf{r}_i \times \mathbf{F}_i
The right-hand side becomes the time derivative of the angular momentum \mathbf{L}_O about the origin:
\sum \mathbf{M}_O = \frac{\mathrm d\mathbf{L}_O}{\mathrm dt}
where \mathbf{L}_O = \sum \mathbf{r}_i \times \mathbf{p}_i = \sum \mathbf{r}_i \times m_i \mathbf{v}_i. This is known as the moment of momentum equation, relating the net torque about the origin to the rate of change of angular momentum about the origin.
The angular momentum \mathbf{L}_P of a system of particles about an arbitrary point P is derived from the definition of angular momentum and the principle of relative motion.
For a system of N particles, the angular momentum about a point P is the sum of the angular momenta of all individual particles about that point:
\mathbf{L}_P = \sum_{i=1}^N \mathbf{r}_{Pi} \times m_i \mathbf{v}_i
Substitute \mathbf{r}_{Pi} = (\mathbf{r}_i - \mathbf{r}_P) into the expression for \mathbf{L}_P:
\mathbf{L}_P = \sum_{i=1}^N \left( (\mathbf{r}_i - \mathbf{r}_P) \times m_i \mathbf{v}_i \right)
Distribute the cross product:
\mathbf{L}_P = \sum_{i=1}^N (\mathbf{r}_i \times m_i \mathbf{v}_i) - \sum_{i=1}^N (\mathbf{r}_P \times m_i \mathbf{v}_i)
The first term, \sum_{i=1}^N (\mathbf{r}_i \times m_i \mathbf{v}_i), is the angular momentum about the origin O. For simplicity, we rewrite this relative to the mass center C.
The second term, \sum_{i=1}^N (\mathbf{r}_P \times m_i \mathbf{v}_i), simplifies because \mathbf{r}_P is constant relative to all particles, allowing us to factor it out:
\sum_{i=1}^N (\mathbf{r}_P \times m_i \mathbf{v}_i) = \mathbf{r}_{PC} \times \sum_{i=1}^N m_i \mathbf{v}_i
The angular momentum \mathbf{L}_P about an arbitrary point P is then:
\mathbf{L}_P = \mathbf{L}_C + \mathbf{r}_{PC} \times \mathbf{p}_C
Returning to the system of particles referenced from a fixed point O in an inertial frame, the sum of moments about O is:
\sum \mathbf{M}_O = \frac{\mathrm d\mathbf{L}_O}{\mathrm dt} = \sum_{i=1}^N \mathbf{r}_i \times \mathbf{F}_i
Using the product rule, the time derivative of angular momentum can be expanded as needed to analyze rotational dynamics. These principles form the foundation of Euler’s Second Law, which describes the relationship between torques and angular momentum in a rigid body or system of particles.
From Newton’s Second Law, the force \mathbf{F}_i acting on a particle is related to its acceleration:
\mathbf{F}_i = m_i \mathbf{a}_i
Substituting this into the moment equation gives:
\sum \mathbf{M}_O = \sum_{i=1}^N \mathbf{r}_i \times m_i \mathbf{a}_i
The angular momentum \mathbf{L}_O of the system about O is defined as:
\mathbf{L}_O = \sum_{i=1}^N \mathbf{r}_i \times m_i \mathbf{v}_i
Taking the time derivative of \mathbf{L}_O:
\frac{\mathrm d\mathbf{L}_O}{\mathrm dt} = \sum_{i=1}^N \frac{\mathrm d}{\mathrm dt} \left( \mathbf{r}_i \times m_i \mathbf{v}_i \right)
Using the product rule:
\frac{\mathrm d\mathbf{L}_O}{\mathrm dt} = \sum_{i=1}^N \left[ \frac{\mathrm d\mathbf{r}_i}{\mathrm dt} \times m_i \mathbf{v}_i + \mathbf{r}_i \times \frac{\mathrm d(m_i \mathbf{v}_i)}{\mathrm dt} \right]
The time derivative of the position vector \frac{\mathrm d\mathbf{r}_i}{\mathrm dt} is the velocity \mathbf{v}_i:
\frac{\mathrm d\mathbf{r}_i}{\mathrm dt} = \mathbf{v}_i
Substituting:
\frac{\mathrm d\mathbf{L}_O}{\mathrm dt} = \sum_{i=1}^N \left( \mathbf{v}_i \times m_i \mathbf{v}_i + \mathbf{r}_i \times m_i \mathbf{a}_i \right)
The term \mathbf{v}_i \times m_i \mathbf{v}_i vanishes because the cross product of any vector with itself is zero:
\mathbf{v}_i \times m_i \mathbf{v}_i = 0
Therefore:
\frac{\mathrm d\mathbf{L}_O}{\mathrm dt} = \sum_{i=1}^N \mathbf{r}_i \times m_i \mathbf{a}_i
Substituting this back into the moment equation, we have:
\sum \mathbf{M}_O = \frac{\mathrm d\mathbf{L}_O}{\mathrm dt}
This is Euler’s Second Law, which states that the sum of the moments about a fixed point O in an inertial reference frame is equal to the time derivative of the angular momentum about O. This equation, often referred to as the Moment Equation, highlights the relationship between torques and the rate of change of angular momentum in a system of particles.
To derive Euler’s Second Law about the mass center, we follow a similar process as for a fixed point O, but now referencing all quantities relative to the mass center C.
The angular momentum of a system of N particles about the mass center C is defined as:
\mathbf{L}_C = \sum_{i=1}^N \mathbf{r}_{iC} \times m_i \mathbf{v}_{iC}
The velocity of the i^{th} particle relative to C is given by:
\mathbf{v}_{iC} = \mathbf{v}_i - \mathbf{v}_C
Taking the time derivative of \mathbf{L}_C:
\frac{\mathrm d\mathbf{L}_C}{\mathrm dt} = \sum_{i=1}^N \frac{\mathrm d}{\mathrm dt} \left( \mathbf{r}_{iC} \times m_i \mathbf{v}_{iC} \right)
Using the product rule:
\frac{\mathrm d\mathbf{L}_C}{\mathrm dt} = \sum_{i=1}^N \left[ \frac{\mathrm d\mathbf{r}_{iC}}{\mathrm dt} \times m_i \mathbf{v}_{iC} + \mathbf{r}_{iC} \times \frac{\mathrm d(m_i \mathbf{v}_{iC})}{\mathrm dt} \right]
The time derivative of \mathbf{r}_{iC} is the relative velocity \mathbf{v}_{iC}:
\frac{\mathrm d\mathbf{r}_{iC}}{\mathrm dt} = \mathbf{v}_{iC}
Substituting this gives:
\frac{\mathrm d\mathbf{L}_C}{\mathrm dt} = \sum_{i=1}^N \left( \mathbf{v}_{iC} \times m_i \mathbf{v}_{iC} + \mathbf{r}_{iC} \times m_i \mathbf{a}_{iC} \right)
The term \mathbf{v}_{iC} \times m_i \mathbf{v}_{iC} vanishes because the cross product of a vector with itself is zero:
\mathbf{v}_{iC} \times m_i \mathbf{v}_{iC} = 0
Therefore:
\frac{\mathrm d\mathbf{L}_C}{\mathrm dt} = \sum_{i=1}^N \mathbf{r}_{iC} \times m_i \mathbf{a}_{iC}
Using Newton’s Second Law, the acceleration \mathbf{a}_{iC} is related to the force \mathbf{F}_{iC} acting on the i^{th} particle relative to the center of mass:
\mathbf{F}_{iC} = m_i \mathbf{a}_{iC}
Substituting this into the expression for \frac{\mathrm d\mathbf{L}_C}{\mathrm dt}:
\frac{\mathrm d\mathbf{L}_C}{\mathrm dt} = \sum_{i=1}^N \mathbf{r}_{iC} \times \mathbf{F}_{iC}
The right-hand side represents the sum of the moments about the center of mass C. Denoting this as \sum \mathbf{M}_C, we have:
\sum \mathbf{M}_C = \frac{\mathrm d\mathbf{L}_C}{\mathrm dt}
This equation describes how the moments about the center of mass are related to the rate of change of angular momentum about the center of mass. It is the Moment Equation for the mass center.
The relationship between the sum of the moments and the time derivative of angular momentum is only valid under specific conditions. This equivalence holds for a point fixed in an inertial reference frame or for the mass center of a system. It does not generally apply for an arbitrary point P. For such arbitrary points, the relationship must account for additional terms related to the motion of the reference point.
The sum of the moments about an arbitrary point P, denoted \sum \mathbf{M}_P, does not directly equal the time derivative of the angular momentum about P, \frac{\mathrm d\mathbf{L}_P}{\mathrm dt}. Instead, it relates to a balance of equivalent force systems. Specifically, the moments about P balance with the moments about the mass center C and an additional term involving the cross product of the position vector and the linear acceleration of the mass center.
For the mass center C, the sum of the moments about C, \sum \mathbf{M}_C, is:
\sum \mathbf{M}_C = \frac{\mathrm d\mathbf{L}_C}{\mathrm dt}
where \mathbf{L}_C = \sum_{i=1}^N \mathbf{r}_{iC} \times m_i \mathbf{v}_{iC} is the angular momentum about the mass center.
The angular momentum about an arbitrary point P, \mathbf{L}_P, can be expressed as:
\mathbf{L}_P = \mathbf{L}_C + \mathbf{r}_{PC} \times M \mathbf{v}_C
Taking the time derivative of \mathbf{L}_P:
\frac{\mathrm d\mathbf{L}_P}{\mathrm dt} = \frac{\mathrm d\mathbf{L}_C}{\mathrm dt} + \frac{\mathrm d}{\mathrm dt} \left( \mathbf{r}_{PC} \times M \mathbf{v}_C \right)
Using the product rule on the second term:
\frac{\mathrm d\mathbf{L}_P}{\mathrm dt} = \frac{\mathrm d\mathbf{L}_C}{\mathrm dt} + \frac{\mathrm d\mathbf{r}_{PC}}{\mathrm dt} \times M \mathbf{v}_C + \mathbf{r}_{PC} \times \frac{\mathrm d(M \mathbf{v}_C)}{\mathrm dt}
The time derivative of \mathbf{r}_{PC} is the relative velocity \mathbf{v}_{PC}:
\frac{\mathrm d\mathbf{r}_{PC}}{\mathrm dt} = \mathbf{v}_{PC}
The time derivative of M \mathbf{v}_C is the total force \mathbf{F} = M \mathbf{a}_C, where \mathbf{a}_C is the acceleration of the mass center.
Substituting these into the equation:
\frac{\mathrm d\mathbf{L}_P}{\mathrm dt} = \frac{\mathrm d\mathbf{L}_C}{\mathrm dt} + \mathbf{v}_{PC} \times M \mathbf{v}_C + \mathbf{r}_{PC} \times M \mathbf{a}_C
Since \mathbf{v}_{PC} \times M \mathbf{v}_C = 0 (the cross product of parallel vectors is zero):
\frac{\mathrm d\mathbf{L}_P}{\mathrm dt} = \frac{\mathrm d\mathbf{L}_C}{\mathrm dt} + \mathbf{r}_{PC} \times M \mathbf{a}_C
The sum of the moments about P is related to the sum of the moments about C and the additional term \mathbf{r}_{PC} \times M \mathbf{a}_C:
\sum \mathbf{M}_P = \sum \mathbf{M}_C + \mathbf{r}_{PC} \times M \mathbf{a}_C
Substituting \sum \mathbf{M}_C = \frac{\mathrm d\mathbf{L}_C}{\mathrm dt}:
\sum \mathbf{M}_P = \frac{\mathrm d\mathbf{L}_C}{\mathrm dt} + \mathbf{r}_{PC} \times M \mathbf{a}_C
This shows that the moments about P depend on the moments about the mass center C, the angular momentum about C, and the term involving the cross product of the position vector \mathbf{r}_{PC} and the linear acceleration M \mathbf{a}_C.
Kinetic diagram
To solve for the equation of the body, we can use two diagrams: one we already used in statics, the free body diagram (FBD), which represent the external forces action on the body. At this stage we assume that the line of action of all the forces are concurrent, so we are looking at translational motion; at a later stage we will be also looking at rotational motion and therefore the force will be non-concurrent.
The Free Body Diagram (FBD) was sufficient in statics because, in statics, we deal only with bodies at rest or in equilibrium. This means that the sum of forces and moments on the body is zero, and there is no acceleration. Since we only need to ensure equilibrium (i.e., \sum F = 0 and \sum M = 0), there’s no need to consider inertial forces or moments.
In dynamics, in addition to the Free Body Diagram (FBD) that captures the external forces, we introduce a Kinetic Diagram (KD). The KD is essential in dynamics because it includes the inertial forces and moments, representing the body’s resistance to changes in motion. While the FBD addresses the actual forces acting on the object, the KD represents the effects of acceleration due to the body’s mass.
For translational motion, Newton’s Second Law, F = ma, governs the relationship between forces and the acceleration of the center of mass. In the kinetic diagram, each mass in the body has an inertia force, \mathbf{F} = - m \mathbf{a}, opposing the direction of acceleration. By summing the forces in the FBD and accounting for these inertial effects in the KD, we form equations of motion that can solve for unknown accelerations or forces.
Together, the FBD and KD yield the complete dynamic analysis for bodies in motion.
Particle on a curve
It is possible to consider a particle moving along the outside of a curve surface and find the condition that ensure that the particle will stay in contact with the surface.
Using the FBF, we can draw the three forces acting on the body: the external force \mathbf F, the normal reaction \mathbf N, and the weight m \mathbf g. Since the particle moved on a curved surface we can decompose the motion along the normal and tangential components. For the forces, that gives:
\begin{aligned} F_T & = F - mg \sin(\alpha) \\ F_N & = N - mg \cos(\alpha) \end{aligned}
The KD gives:
\begin{aligned} & m a_T \\ & - m a_N = - m \frac{|v|^2}{\rho} \end{aligned}
In it now possible to write the equations of equilibrium. While for statics the forces where equal to zero, now the forces are equal to the kinetic vectors:
\begin{aligned} m a_T & = F - mg \sin(\alpha) \\ - m a_N & = N - mg \cos(\alpha) \end{aligned}
The particle will stay on the curve if the normal force is positive (N \geq 0):
N = mg\cos(\alpha) - m a_N = mg\cos(\alpha) - m \frac{|v|^2}{\rho} \ge 0
which means:
v \leq \sqrt{\rho g \cos(\alpha)}