Angular Momentum of a Rigid Body
Angular momentum of a rigid body is a measure of its rotational motion. It is the product of the body's moment of inertia and its angular velocity. This quantity is conserved in the absence of external torques, making it a fundamental concept in the analysis of rotating systems in physics and engineering.
5 Key excerpts on "Angular Momentum of a Rigid Body"
eBook - ePub- Ronald J. Anderson(Author)
- 2020(Publication Date)
- Wiley(Publisher)
...This is because angular momentum is defined as the moment of linear momentum. Equation 2.35 defines angular momentum about the reference point due to the mass particle as, (2.46) where the linear momentum is defined to be, resulting in, (2.47) We assume that the absolute velocity of the reference point,, and the angular velocity of the rigid body,, are known. We can then write an expression for the velocity of particle (noting that the position vector has no rate of change of magnitude because both and are in the same rigid body and cannot move apart) as, (2.48) Referring to Figure 2.3, we can write expressions for the position of with respect to and the absolute velocity of point in the reference frame with unit vectors. Let the distance in the direction from point to point be. Similarly, we define and to be the distances in the and directions respectively. The position vector is then, Let the absolute velocity of have scalar components,, and. The velocity of is then, Further, let the angular velocity of the coordinate system be, The cross product can then be written as, and the velocity of particle is, (2.49) Equation 2.49 can be substituted into Equation 2.47 to give, after performing another cross product and gathering some terms, an expression for the angular momentum about due to particle as follows: (2.50) To get the total angular momentum vector about point, we write Equation 2.50 for every particle and add the resulting equations together to get, (2.51) or, (2.52) where the terms used in Equation 2.52 are defined as follows.. This is the total mass of the rigid body.. This summation is one component of that used to locate the center of mass of the body with respect to the reference point. is the ‐component of.. is the ‐component of.. is the ‐component of.. This term is a function of the spatial distribution of mass particles around the ‐axis (i.e. direction) passing through point. The term is always positive because of the sum of squares term...
eBook - ePubBiomechanics of Human Motion
Applications in the Martial Arts, Second Edition
- Emeric Arus, Ph.D.(Authors)
- 2017(Publication Date)
- CRC Press(Publisher)
...object. I = 1/3 m ⋅ w 2 Circular cone with the axis through the centers of x, y, and z axes and meets the thin edge point of the cone. The cone longitudinal axis is through the x line/axis. I = 3/10 m ⋅ r 2 Solid sphere about any diameter = 2 r. I = 2/5 m ⋅ r 2. Figure 9.4 demonstrates the parallel axis theorem. Looking at Figure 9.4, the black round spots represent CoM or CoG. Two other small circles (they are not black) also represent the moment of inertia of body parts about transverse axes through their CoM. The 0.28, 0.40, and 0.50 m represent the distances between the CoM of different body segments. Figure 9.3 shows in percentage the location of the mass centers of body segments. 9.7 ANGULAR MOMENTUM Recall that linear momentum is a vector quantity that has a magnitude and direction. A body or an object that has a motion has a momentum that measures the velocity and the quantity of that mass or body. To put into simpler words, the momentum is a measure of the force needed to start or stop a motion. Therefore, linear momentum (p) = m ⋅ v (N ⋅ s) or kg ⋅ m/s. In a similar way, the angular or rotary momentum H or (L) = moment of inertia times angular velocity. L = I ⋅ ω or kg ⋅ m 2 /s. The momentum of a diver who dives in a straight line is the equation p = m ⋅ v ; when the diver starts to rotate he has the equation L = I ⋅ ω. In contact sports such as karate, boxing, and so on, the linear momentum is closely related to the linear impulse (see Section 8.5); in the similar way in rotary motion, the rotary impulse (which will be described later) is also related to angular momentum. When a body rotates normally, it will not stop until a force intervenes in its rotation route? Let us take an example of a discus thrown. The discus will spin in the direction liberated by the thrower. The discus will rotate around its center (CoG) and also advances forward. The angular momentum L = I ⋅ ω (kg ⋅ m 2 /s)...
eBook - ePubIn Search of Divine Reality
Science as a Source of Inspiration
- Lothar Schäfer(Author)
- 1997(Publication Date)
- University of Arkansas Press(Publisher)
...Appendix 16 SOME TECHNICAL DETAILS CONCERNING BELL’S THEOREM Quantization of Angular Momenta VECTORS Among the dynamic physical attributes of things are those that are fully characterized by merely specifying an amount, or a magnitude, such as for example the energy or the mass of a system, and there are those that need both a magnitude and a direction to be fully defined, such as for example a force acting on a mass. In dealing with forces it is not only important to know how strong a given force is, but also the direction in which it acts. In physics, directed quantities that carry both a magnitude and a direction are called vector quantities. ANGULAR MOMENTUM In a closed system of moving objects, one of the constants of motion is linear momentum, the product mv of mass, m, and velocity, v. Since mv is a constant of motion, we say that in every mechanical process linear momentum is conserved. Similarly, when a particle with mass m is moving in a circle with radius r and its velocity is v, its motion is characterized by a quantity called angular momentum. Angular momentum is also a vector quantity and a constant of motion; that is, in every mechanical process angular momentum is conserved. For the particle moving in a circle the magnitude of angular momentum is given by the product mvr, mass times velocity times radius, and its direction is along the axis of rotation. It is seen from this formula that, the faster a greater mass moves in a larger circle, the greater its orbital angular momentum. Similarly, the faster a bigger spherical mass spins about an axis—like the earth is spinning about an axis through its poles—the greater its spin angular momentum. QUANTIZATION OF ANGULAR MOMENTUM It was one of the unexpected discoveries of this century that masses cannot revolve in an orbit or spin about an axis with arbitrary velocities; they can do so only at certain speeds. More precisely, angular momentum is quantized...
eBook - ePub- W. Bolton(Author)
- 2015(Publication Date)
- Routledge(Publisher)
...Chapter 27 Angular dynamics 27.1 Introduction Dynamics is the study of objects in motion. This chapter thus follows on from Chapter 5, where the terms and equations used in describing angular motion were introduced and Chapter 6 where the principles of dynamics were introduced. Here we now consider the torques and energy involved with angular motion. 27.1.1 Terms: torque and couple Figure 27.1 Couple If a force F is applied to the surface of a shaft of radius r (Figure 27.1), a reactive force R is set up which is equal in magnitude and opposite in direction to F, i.e. R 5 − F, and can be considered to act at O. This pair of oppositely directed, but equal in magnitude, forces which are not in the same straight line is called a couple. The turning moment of a couple is called the torque T and T = Fr. Figure 27.2 Rotation of a rigid body 27.2 Moment of inertia Consider a force F acting on a small element of mass δm of a rigid body (note that the symbol δ in front of a quantity is used to indicate that it is a small bit of the quantity), the element being a distance r from the axis of rotation at O (Figure 27.2). The torque T acting on the element is Fr and since F = δm × a we can express the torque as: torque T = δm × ar But a = rα, where α is the angular acceleration. Thus T = δm × r 2 α. We can express this as: torque T = Iα where I is termed the moment of inertia and for the small element of mass δm at radius r is given by: moment of inertia I = δm × r 2 The torque required to give a rotational acceleration for the entire rigid body will thus be the sum of the torques required to accelerate each element of mass in the body...
eBook - ePub- William Bolton(Author)
- 2012(Publication Date)
- Routledge(Publisher)
...Hence, calculate the coefficient. 4.6 Torque and angular motion Consider a force F acting on a small element of mass δ m of a rigid body, the element being a distance r from the axis of rotation at O (Figure 4.41). The torque T acting on the element is Fr and since F = δ m × a we can express the torque as: Figure 4.41 Rotation of a rigid body torque T = δ m × ar But a = ra, where a is the angular acceleration. Thus T = δ m × r 2 a. We can express this as: torque T = Ia [41] where I is termed the moment of inertia and for the small element of mass m at radius r is given by: moment of inertia I = δ m × r 2 [42] Equation [ 41 ] gives the torque required to give an angular acceleration to just an element of the mass of the rigid body. The torque required to give a rotational acceleration for the entire rigid body will thus be the sum of the torques required to accelerate each element of mass in the body and given by equation [ 41 ] when we consider the sum of the moments of inertia of the entire body. Thus: torque to give body angular acceleration T = Ia [43] where: moment of inertia of body I = ∫ r 2 d m [44] The moment of inertia of a body is the inertial property of a rotating body in much the same way as mass is the inertial property of a body for linear motion. An inertial property describes the inertia of the body that has to be overcome to get it moving. 4.6.1 Moment of inertia For a small mass at the end of a light pivoted arm with radius of rotation r, we can consider the entire mass of the body to be located at the same distance r and so the moment of inertia is given by equation [ 44 ] as: I = mr 2 [45] For a uniform disc of radius r and total mass m, the moment of inertia about an axis through the disc centre can be found by considering the disc to be composed of a number of rings (Figure 4.42). For a ring of thickness δ x and radius x, its area is effectively its circumference multiplied by the thickness and so is 2 πx δ x...
