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    Overview - types of momentum

    Any moving object possesses both linear and angular momentum. Linear momentum \( \dot{\vec{p}} \) is an extensive property that quantifies the motion of a particle or a system proportional to the mass. Angular momentum \( \dot{\vec{L}} \) is an extensive property, proportional to the mass of the system, which applies to any object undergoing motion, particularly rotational motion about a point. Angular momentum is used to describe the torque on bodies in static and dynamic analyses of structures. Whereas linear momentum is parallel to the motion of an object, angular momentum is normal to an object’s motion.

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

    and

    $$ \dot{\vec{L}}=\vec{r} \times \dot{\vec{p}} = \vec{r} \times (\dot{m}\vec{v}) $$

    When a force acts on a system, it may produce a torque \( \vec{\tau} \), which is a measure of how a force changes the rotational motion of an object. Torque has both magnitude and direction and is calculated as the cross product of the position vector \( \vec{r} \) and the applied external force \( \vec{F} \):

    $$ \vec{\tau}=\vec{r} \times \vec{F} $$

    Conservation statements

    Linear and angular momentum are always conserved in the universe. Thus, momentum cannot be produced or destroyed in a system or in the universe. Recall that the Generation and Consumption terms describe the production and destruction of an extensive property in a system. In problems involving momentum, Generation and Consumption terms are eliminated from the accounting equation, which then reduces to the conservation equation.

    Linear Momentum

    $$ \sum_{in}\dot{\vec{p}}_{in} - \sum_{out}\dot{\vec{p}}_{out} + \sum{\vec{F}}=\frac{d\vec{p}^{sys}}{dt} $$

    Angular Momentum

    $$ \sum_{in}\dot{\vec{L}}_{in} - \sum_{out}\dot{\vec{L}}_{out} + \sum(\vec{r} \times \vec{F})=\frac{d\vec{L}^{sys}}{dt} $$

    Rigid Body Statics

    One common class of engineering problems involves the application of the conservation of momentum to closed, steady-state systems. In a closed (but not isolated) system, movement of momentum through bulk material transport across the system boundary does not occur. However, external forces may act on the system. For this situation, the differential form of the statement of the conservation of linear momentum is often used and is reduced from the closed, steady-state situation:

    $$ \sum \vec{F} = 0 $$

    additionally, this property is conserved across each dimension, such that

    $$ \sum \vec{F}_x = \sum \vec{F}_y = \sum \vec{F}_z= 0 $$

    Similarly, this property is conserved for angular momentum, such that

    $$ \sum(\vec{r} \times \vec{F})=0 $$

    and is similarly conserved across dimensions, such that

    $$ \sum(\vec{r} \times \vec{F})_x=\sum(\vec{r} \times \vec{F})_y=\sum(\vec{r} \times \vec{F})_z=0 $$

    Fluid Statics

    Another class of problems involves fluids that do not move, or static fluids. The viscosity of a static fluid does not influence its behavior, since it is not flowing. P is the pressure exerted on the system by the surroundings, z is the height in the z-direction, ρ is the density of the fluid, and g is the gravitational constant (magnitude only). An important result from above is that the pressure varies as a function of the position z within a system containing a static fluid.

    $$ \frac{dP}{dz}=-\rho g $$

    or

    Fluid Stastics One Dimension #hfp-eq
    $$ \Delta P = -\rho g \Delta z $$

    Ignoring air pressure:

    $$ \begin{aligned} \sum{F_y} &= 0 \\ P(z)A - W &= 0 \\ P(z)A - (\rho \ A \ z) \ g &= 0 \\ P(z) &= \rho \ g \ z \end{aligned} $$

    Warning: Don't confuse pressure with force! #force-pressure

    Pressure is a force distributed over an area (force per area) and therefore has units of Pascals (Pa). To calculate the resultant force from hydrostatic fluid pressure, you must use

    $$ F_R = P \ A $$

    where \( A \) is the area that the force is applied over.

    Collision systems

    Consider a system with no bulk mass transfer across the system boundary and no external forces. Such a system is isolated, and no momentum accumulates in it. Under these conditions, the integral formulation of the equation is commonly used. In these situations, the conservation of momentum equation is reduced to the following:

    Solution procedure of perfectly elastic collisions. #rec-ecp
    $$ \begin{aligned} m_1 \vec{v}_{1, i} + m_2 \vec{v}_{2, i} &= m_1 \vec{v}_{1, f} + m_2 \vec{v}_{2, f} \\ 0=\vec{p}_f^{sys}-\vec{p}_0^{sys} \end{aligned} $$

    where subscripts \(i, f\) denote initial and final velocities, respectively.

    type of collision definition
    elastic collisions collisions where there is no loss in kinetic energy
    inelastic collisions collisions where there are losses in kinetic energy, typically due to internal friction

    A relationship describing the elasticity of colliding objects, the coefficient of restitution (e), can be used to develop an additional equation to solve a problem. The coefficient is defined as the ratio of \( v_{separation} \), the difference in velocities after collision, and \( v_{approach} \), the difference in velocities before collision of two objects, A and B.

    Coefficient of Restitution. #rec-ecp
    $$ e = \frac{v_{separation}}{v_{approach}}= \frac{v_{B,f}-v_{A,f}}{v_{A,0}-v_{B,0}} $$
    range type of collision
    \(e = 0\) This corresponds to a perfectly inelastic collision. The velocity of both objects after the collision is the same
    \(0 < e < 1\) This corresponds to an inelastic collision that happens in the real world, where some kinetic energy is lost
    \(e = 1\) This corresponds to a perfectly elastic collision, where no kinetic energy is lost.
    \(e > 1\) This corresponds to a superelastic collision, where energy is gained/released, like a chemical reaction, a reduction of rotational energy.

    Unsteadystate momentum

    In an unsteady-state system, at least one variable describing the system (e.g., pressure, flow rate) changes with time. Unsteady-state systems gain or lose momentum with bulk material transfer or when external forces act on them; thus, the Accumulation term is always nonzero. For systems with no bulk transfer of material across the system boundary, the differential form of the conservation of linear momentum equation reduces to

    $$ \sum \vec{F}=\frac{d}{dt}(\vec{p}^{sys}) = m^{sys}\vec{a}^{sys} $$

    Mechanical Energy Accounting

    Mechanical energy involves the motion and displacement of fluids and bodies, as well as the forces that can change motion and displacement. Mechanical energy is the summation of the kinetic energy, potential energy, and work of a system. The mechanical energy equation accounts only for mechanical energy and its conversion to and from other types of energy.

    Movement of mass flowing as a fluid transfers mechanical energy into and out of a system in the forms of kinetic energy, potential energy, and flow work. When a fluid is in motion at a given velocity, the fluid possesses kinetic energy. The potential energy a fluid possesses results from its position in a gravitational field. Flow work is the energy required to push the fluid into and out of a system.

    The steady-state mechanical energy accounting equation is:

    $$ \dot{m} (\hat{E}_{P, in}-\hat{E}_{P, out})+\dot{m}(\hat{E}_{K, in}-\hat{E}_{K, out})-\dot{m}(\frac{P_{in}}{\rho_{in}}-\frac{P_{out}}{\rho_{out}}) + \dot{W}_{shaft} - \sum \dot{f} = 0 $$

    where \( \dot{m} \) is the mass flow rate, \( \hat{E}_P \) is the specific potential energy (potential energy per unit mass), \( \hat{E}_K \) is the specific kinetic energy (kinetic energy per unit mass), Pi and Pj are the pressures at the system boundary where mass flow enters and leaves, \( \rho \) is the density of the fluid, \( \dot{W}_{shaft} \) is the total nonflow and non-expansion work (i.e., shaft work), and \( \dot{f} \) is the total frictional losses.