7.21. In t second, the axis gradually becomes horizontal. (No figure was provided.) 7.28. EQUATIONS OF MOTION FOR PURE ROTATION When a rigid body rotates about a fixed axis perpendicular to the plane of the body at point O, the body's center of gravity G moves in a circular path of radius rG. \end{align}. \end{align} \end{align}, The torque on the rigid body about the axis of rotation is given by A block of mass m is attached to a light string that is wrapped around the rim of a uniform solid disc of radius R and mass M as in Fig. The use of a principal axis system greatly simplifies treatment of rigid-body rotation and exploits the powerful and elegant matrix algebra mentioned in appendix \(19.1\). The SI unit of the moment of inertia is kg\(\mathrm {m}^{2}\). On the other hand, Eq. We treat the whole system as a single point-like particle of mass m located at the center . The expressions for the kinetic energy of the object . Apply Newton's second law on the system to get What happens when a rigid object is rotating about a fixed axis? Abstract. The radius of said circle depends upon how far away that point particle is from the axis. &=(3m)\vec{a}. Thus the best solution for describing rotation of a rigid body is to use a rotation matrix that transforms from the stationary fixed frame to the instantaneous body-fixed frame for which the moment of inertia tensor can be evaluated. \begin{align} \label{dic:eqn:1} 5 we have seen that if the net external torque acting on a system of particles relative to an origin is zero then the total angular momentum of the system about that origin is conserved, In the case of a rigid object in pure rotational motion, if the component of the net external torque about the rotational axis (say the \(\mathrm {z}\)-axis) is zero then the component of angular momentum along that axis is conserved, i.e., if. Ans : When a rigid body is put into rotational motion, the amount of torque required to change the angular velocity of the body is called its rotational inertia. Table. When a body moves in a rotational motion around a given axis or a line, i.e., at a fixed distance and fixed orientation relative to the body, the body is rotating around the axis. Three particles A, B and C, each of mass $m$, are connected to each other by three massless rigid rods to form a rigid equilateral triangular body of side $l$. A rotating rigid object has an angular position given by \(\theta (t)=((0.3)t^{2}+(0.4)t^{3})\) rad. A body of mass m rotating about a fixed axis with angular velocity w will have a "rotational" kinetic energy of I w 2 where I = Moment of Inertia of the body. 7.9) and \(\theta \) is the angle between the position vector and the \(\mathrm {z}\)-axis. Three masses are connected by massless rods as in Fig. The motion of a body is controlled by certain variables, such as velocity, displacement, etc. \end{align} \begin{align} According to Euler's rotation theorem the rotation of a rigid body (or three-dimensional coordinate system with the fixed origin) is described by a single rotation about some axis. Find (in vector form) the linear velocity and acceleration of the point \(\mathrm {P}\) on the bar. As the top loses rotational kinetic energy due to friction, the top's rotation-axis precesses around a circle, as observed in the space-fixed frame. The kinematics and dynamics of rotation around a fixed axis of a rigid body are mathematically much simpler than those for free rotation of a rigid body; they are entirely analogous to those of linear motion along a single fixed direction, which is not true for free rotation of a rigid body. But what causes rotational motion? Springer, Cham. The centre of mass of the system (G) is at a distance $\mathrm{AG}={l}/{\sqrt{3}}$ from the hinge point A. The force responsible for rotational motion is called torque or the moment of the force. D) directed from the center of rotation toward G. 2. \nonumber A wheel of mass of 20 kg and radius of 0.75 \(\mathrm {m}\) is initially rotating at 120 rev/min. 7.5, we have, where \(r_{i}\) is the perpendicular distance from the particle to the axis of rotation. Rigid-body rotation can be broken into the following two classifications. Learn how to solve problems involving rigid bodies spinning around a fixed axis with animated examples. 0000002460 00000 n
Find the moment of inertia of an elliptical quadrant about the \(\mathrm {y}\)-axis (see Fig. Since A is a fixed point, the torque $\tau$ is related to the angular acceleration $\alpha$ by Thus, all particles have the same angular velocity and the same angular acceleration. Ans : Force is responsible for all motion that we observe in the physical world. Find the moment of inertia of a spherical shell of radius R and mass M about an axis passing through its center of mass. By contrast, in the stationary inertial frame the observables depend sensitively on the details of the rotational motion. This chapter extends the discussion to motion of finite-sized rigid bodies. Newton's first law of rotation In translation, a particle or particle like rigid body has constant linear velocity unless there is an external force being applied on it. Search: Opencv Rigid Transform. Salma Alrasheed . 7.11. For example, when observed in the stationary fixed frame, rapid rotation of a long thin cylindrical pencil about the longitudinal symmetry axis gives a time-averaged shape of the pencil that looks like a thin cylinder, whereas the time-averaged shape is a flat disk for rotation about an axis perpendicular to the symmetry axis of the pencil. Ans : Force is responsible for all motion that we observe in the physical world. As the rigid body rotates, a particle in the body will move through a distance s along its circular path (see Fig. 0000005734 00000 n
In contrast, when the torque acting on a body produces angular acceleration, it is called dynamic torque. a=\frac{2mg}{2m+M}.\nonumber Which of the sets can occur only if the rigid body rotates through more than 180? \begin{align} Consider a rigid body rotating about a fixed axis with an angular velocity $\omega$ and angular acceleration $\alpha$. When a rigid object rotates about a fixed axis all the points in the body have the same? However, for the general case of free rotation, the vector of angular velocity . 7.9, the direction of \(\mathrm {y}\) is perpendicular to the plane formed by \(\omega \) and \(\mathrm {R}\) where it can be verified using the right-hand rule. The kinematics and dynamics of rotation around a fixed axis of a rigid body are mathematically much simpler than those for free rotation of a rigid body; they are entirely analogous to those of linear motion along a single fixed direction, which is not true for free rotation of a rigid body. The body is set into rotational motion on the table about A with a constant angular velocity $\omega$. If a rigid object free to rotate about a fixed axis has a net external torque actingon it, the object undergoes an angular acceleration where The answer quick quiz 10.8 (b). Solution: - 199.241.137.45. 7.10. As seen from Fig. The rotational inertia of a body is affected by the mass and the distribution of the mass of the body with respect to the axis around which the body rotates. If its angular acceleration is given by \(\alpha =(4t)\,\mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}^{2}\) and if at \(t=0, \omega _{0}=0\), find the angular momentum of the sphere and the applied torque as a function of time. For a rigid body undergoing fixed axis rotation about the center of mass, our rotational equation of motion is similar to one we have already encountered for fixed axis rotation, ext = dLspin / dt . Ans : Angular velocity is the rate of change in angular displacement with respect to time. In the body-fixed frame, the ``vertical'' axis coincides with the top's axis of rotation (spin). 7.11 is valid only for a symmetrical homogeneous rigid object rotating about its symmetrical axis, where the angular momentum in the equation is the total angular momentum and it is directed along the axis of rotation. Every motion of a rigid body about a fixed point is a rotation about an axis through the fixed point. The common solutions to calculate an object's rigid body rotation need to use manually positioned references or track a single-point's rotation. F_h=(3m)a_c=\sqrt{3}m\omega^2 l. \nonumber For an arbitrarily shaped rigid body having a density , then the moment of inertia has to be calculated as an integral. As the body moves, the distance between the current and the initial position of the body changes. Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Differentiating the above equation with respect to t gives, Since ds/dt is the magnitude of the linear velocity of the particle and \(d\theta /dt\) is the angular velocity of the body we may write, Therefore, the farther the particle is from the rotational axis the greater its linear speed. Force is responsible for all motion that we observe in the physical world. Now consider another axis that is parallel to the first axis and that passes through a point \(\mathrm {P}\) as shown in Fig. For a rigid body which is a continuous system of particles, the sum is replaced by an integral. 7.16 shows a uniform thin rod of mass M and length L. Find the moment of inertia of the rod about an axis that is perpendicular to it and passing through: (a) the center of mass; (b) at one end; (c) at a distance of L/6 from one end. Applying Newtons second law in angular form to the disc gives, Since the acceleration of the block is equal to the (tangential) acceleration of a point at the rim of the disc we have. 0000001907 00000 n
0000002657 00000 n
Substitute $t=6$ sec in the expression for $\theta$ to get $\theta=36$ rad. 0000004127 00000 n
Solution. cm cm. If the rod is released from rest at an angle \(\theta =30^{\mathrm {o}}\) to the horizontal, find; (a) the initial angular acceleration of the rod when it is released; (b) the initial acceleration of a point at the end of the rod; (c) from conservation of energy find the angular speed of the rod at its lowest position (Neglect friction at the pivot). Here we are going to discuss Introduction to Rotational Kinematics of Rigid Body. The direction of the linear speed of the particles is always tangent to the path (as mentioned in Sect. 0000001452 00000 n
0000010219 00000 n
A rigid body has six degrees of freedom, three of translation and three of rotation. Rigid-body rotation can be broken into the following two classifications. Fixed-axis rotation describes the rotation around a . The pure rotational motion: The rigid body in such a motion rotates about a fixed axis that is perpendicular to a fixed plane. Find the moment of inertia of a uniform solid cylinder of radius R, length L and mass M about its axis of symmetry. 7.25. The disc rotates about a fixed point O. Legal. Let \(t_{1}=0, t_{2}=t, \omega _{1}=\omega _{\mathrm {o}}, \omega _{2}=\omega , \theta _{1}=\theta _{\mathrm {o}}\), and \(\theta _{2}=\theta .\) Because the angular acceleration is constant it follows that the angular velocity changes linearly with time and the average angular velocity is given by, Finally solving for t from Eq. Assuming that the string does not slip and that the disc rotates without friction, find: (a) the acceleration of the block; (b) the angular acceleration of the disc, and; (c) the tension in the string when the system is released from rest. \label{dic:eqn:2} Find the angular speed in radians per second of the earth about (a) its axis (b) the sun. A man stands on a platform that is free to rotate without friction about a vertical axis, Because the resultant external torque on the system is zero, it follows that the total angular momentum of the system is conserved. Therefore, it is necessary to treat the object as a system of particles. Its angular displacement is then given by, \(\triangle \theta \) is positive for counterclockwise rotations (increasing \(\theta \)) and negative for clockwise rotations (decreasing \(\theta \)). Note that this energy is not a new kind of energy; it is just the sum of the translational kinetic energies of the particles. \begin{align} Another example is a childs spinning top which has one point constrained to touch the ground but the orientation of the rotation axis is undefined. Provided by the Springer Nature SharedIt content-sharing initiative, Over 10 million scientific documents at your fingertips, Not logged in All lines on a rigid body on its plane of motion have the same angular velo. Rigid-body rotation features prominently in science, engineering, and sports. \alpha&=\frac{\mathrm{d}\omega}{\mathrm{d}t} \\ The axis referred to here is the rotation axis of the tensor . 7.3). You must there are over 200,000 words in our free online dictionary, but you are looking for one that's only in the Merriam-Webster Unabridged Dictionary. Fixed-axis rotation describes the rotation around a fixed axis of a rigid body; that is, an object that does not . The Zeroth law of thermodynamics states that any system which is isolated from the rest will evolve so as to maximize its own internal energy. This particle (at point P) will rotate in a circle of fixed radius r which represents the perpendicular distance from \(\mathrm {P}\) to the axis of rotation. So the shape of the rigid body must be specied, as well as the location of the rotation axis before the moment of inertia can be calculated. A man stands on a platform that is free to rotate without friction about a vertical axis as in Fig. 0000001628 00000 n
7.2 shows analogous equations in linear motion and rotational motion about a fixed axis. Abstract A rigid body has six degrees of freedom, three of translation and three of rotation. The description of rigid-body rotation is most easily handled by specifying the properties of the body in the rotating body-fixed coordinate frame whereas the observables are measured in the stationary inertial laboratory coordinate frame. The force $F$ acting on B causes an anticlockwise torque $\tau=Fl\sqrt{3}/2$ about the point A. since at \(t=0, \omega _{0}=0\) then \(c=0\) and, A uniform solid sphere rotating about an axis tangent to the sphere. and its angular acceleration is According to Newtons second law, all bodies tend to resist a change in their current state. In Example 7.8 find the angular momentum in each case. It will be shown that working in the body-fixed coordinate frame of a rotating body allows a description of the equations of motion in terms of the inertia tensor for a given point of the body, and that it is possible to rotate the body-fixed coordinate system into a principal axis system where the inertia tensor is diagonal. The tangential acceleration of the pulley at the point C is $\alpha R$. The quantities \(\theta , \omega \) and \(\alpha \) in pure rotational motion are the rotational analog of x,v and a in translational one-dimensional motion. 7.1 shows the linear/rotational analogous equations. From the workenergy theorem we have. The rotational inertia of a rigid body is an important concept as it helps us understand the amount of torque required to achieve a certain objective. A body can be constrained to rotate about an axis that has a fixed location and orientation relative to the body. Some bodies will translate and rotate at the same time, but many engineered systems have components that simply rotate about some fixed axis. This equation can also be written in component form since \(\mathbf {L}_{z}\) is parallel to \(\varvec{\omega }\), that is, Therefore, if a rigid body is rotating about a fixed axis (say the \(\mathrm {z}\)-axis), the component of the angular momentum along that axis is given by Eq. Let us denote the angle between the initial and current position of a rigid body as . Now the two angular velocities of rotation $\omega$ and $\Omega$ are not the same I suppose and there is no link between the two angular velocities, is that right? \label{fjc:eqn:3} I encourage you to solve this problem by energy method. If is the angular velocity of a rigid body, the angular acceleration of the body is given as =d/dt. The following open-ended questions, among others, were crafted to elicit students' thoughts about aspects of angular velocity of a rigid body. rigid body does not exist, it is a useful idealization. From conservation of energy we have \(K_{i}+U_{i}=K_{f}+U_{f}\). 5 ct 2 2 = ( o)2 + 2 c ( - o) o and o are the initial values of the body's angular Let $T$ be tension in the string, $a$ be the acceleration of the body of mass $m$, and $\alpha$ be the angular acceleration of the disc of mass $M$ and radius $R$ (see figure). To extend the particle model to the rigid-body model. `
>LL%>Z]"jZ0h2)c ZKk8yx F7 W
endstream
endobj
115 0 obj
397
endobj
92 0 obj
<<
/Type /Page
/Parent 85 0 R
/Resources 93 0 R
/Contents 97 0 R
/MediaBox [ 0 0 595 842 ]
/CropBox [ 0 0 595 842 ]
/Rotate 0
>>
endobj
93 0 obj
<<
/ProcSet [ /PDF /Text /ImageC ]
/Font << /TT2 95 0 R /TT4 100 0 R /TT6 99 0 R /TT8 102 0 R /TT10 107 0 R /TT11 105 0 R >>
/XObject << /Im1 113 0 R >>
/ExtGState << /GS1 109 0 R >>
/ColorSpace << /Cs6 94 0 R >>
>>
endobj
94 0 obj
[
/ICCBased 108 0 R
]
endobj
95 0 obj
<<
/Type /Font
/Subtype /TrueType
/FirstChar 32
/LastChar 148
/Widths [ 278 278 0 0 0 0 0 0 333 333 0 584 278 333 278 278 556 556 556 556
556 556 556 556 556 556 278 278 0 584 584 556 0 667 667 722 722
667 611 778 0 278 0 667 556 833 0 778 667 778 722 667 611 722 667
944 0 0 0 0 0 0 0 0 0 556 556 500 556 556 278 556 556 222 222 500
222 833 556 556 556 556 333 500 278 556 500 722 500 500 500 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 333 333 ]
/Encoding /WinAnsiEncoding
/BaseFont /Arial
/FontDescriptor 96 0 R
>>
endobj
96 0 obj
<<
/Type /FontDescriptor
/Ascent 905
/CapHeight 1041
/Descent -211
/Flags 32
/FontBBox [ -665 -325 2028 1006 ]
/FontName /Arial
/ItalicAngle 0
/StemV 94
/XHeight 747
>>
endobj
97 0 obj
<< /Length 1186 /Filter /FlateDecode >>
stream
Page ID 46089. If material is not included in the chapter's Creative Commons license and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. The rotational inertia of various rigid bodies of uniform density, Consider a rigid body rotating about a fixed axis (the \(\mathrm {z}\)-axis) with an angular speed \(\omega \) as shown in Fig. The speed at which the door opens can be controlled by the amount of force applied. Since all forces lie in the same plane the net torque is. Newton's second law gives a=\alpha R. Therefore, the total angular momentum of the rigid body along the \(\mathrm {z}\)-direction is. The angle of this position to the axis of rotation is taken as zero radians. Get answers to the most common queries related to the NEET UG Examination Preparation. Fig. Since one rotation (\(360^{\circ }\)) corresponds to \(\theta =2\pi r/r=2\pi \) rad, it follows that: Note that if the particle completes one revolution, \(\theta \) will not become zero again, it is then equal to \(2\pi \mathrm {r}\mathrm {a}\mathrm {d}\). \end{align} Find: (a) the rotational kinetic energy of the disc; (b) Suppose that the same disc rotate using a motor that delivers an instantaneous of power 0. TR=I_O\alpha=(MR^2/2)\alpha, \( \mathbf {L}_{i}\) can be analyzed to two components, \(\mathrm {a}\) component parallel to \(\varvec{\omega }\) written \((\mathbf {L}_{iz})\) and a component perpendicular to \(\varvec{\omega }\), \((\mathbf {L}_{i\perp })\). But we must first understand rotational motion and its nuances. View Answer. 7.7), thus, The instantaneous power delivered to rotate an object about a fixed axis is found from, Table. Thus, is the perpendicular distance between the two parallel axes. 1.9.1 \((d/dt(\mathbf {A}\times \mathbf {B})=\mathbf {A}\times d\mathbf {B}/dt+d\mathbf {A}/dt\times \mathbf {B})\) we have, Furthermore, the direction of \(\varvec{\alpha }\times \mathbf {R}\) is tangent to the circular path of the particle at any instant (see Fig. Another disc that is initially at rest is dropped on the first, the two will eventually rotate with the same angular speed due to friction between them. 7.13. The rotating motion is commonly referred to as "rotation about a fixed axis". Sample Problem. 0000004937 00000 n
Some bodies will translate and rotate at the same time, but many engineered systems have components that simply rotate about some fixed axis. They are related by 1 revolution = 2radians When a body rotates about a fixed axis, any point P in the body travels along a circular path. The rotational kinetic energy is the kinetic energy of rotation of a rotating rigid body or system of particles, and is given by K=12I2 K = 1 2 I 2 , where I is the moment of inertia, or "rotational mass" of the rigid body or system of particles. \end{align} \label{jpb:eqn:5} To understand rotation about a fixed axis. This is followed by a discussion of practical applications. \(2\mathrm {h}\mathrm {p}\), find in that case the torque applied to the disc. The average angular acceleration is defined as, The instantaneous angular acceleration is. \end{align} 7.13, \(\mathbf {L}_{i}\) is not parallel to \(\varvec{\omega }\). One radian is defined as the angle subtended by an arc of length that is equal to the radius of the circle. PubMedGoogle Scholar. Torque is described as the measure of any force that causes the rotation of an object about an axis. 7.12 gives the rotational inertia of various rigid bodies of uniform density. \end{align} 7.5) and therefore cannot be represented by a vector. 4oh5~ - 7.27). To simplify these problems, we define the translational and rotational motion of the body separately. 7.4. We should not ignore the fact that $\theta$ increases monotonically till pulley come to rest (see figure). Rotation of Rigid Bodies. A particle in rotational motion moves with an angular velocity. This body is placed on a horizontal frictionless table x-y plane) and is hinged to it at the point A, so that it can move without friction about the vertical axis through A (see figur). Angular acceleration also plays a role in the rotational inertia of a rigid body. about that axis. Write the expression for the same. The distance of the centre of mass from the axis of rotation increases or decreases the rotational inertia of a rigid body. (c) The acceleration of a point in the unwinding rope is the same as the acceleration of a point at the rim of the cylinder, i.e., (e) If the rope has moved a distance of lm, the angular displacement of the cylinder is, (f) The final angular speed when \(\theta =5\) rad is, That gives \(\omega =7.6 \; \mathrm {r}\mathrm {a}\mathrm {d}/\mathrm {s}\). 7.9). Variational Principles in Classical Mechanics (Cline), { "13.01:_Introduction_to_Rigid-body_Rotation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.02:_Rigid-body_Coordinates" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.03:_Rigid-body_Rotation_about_a_Body-Fixed_Point" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.04:_Inertia_Tensor" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.05:_Matrix_and_Tensor_Formulations_of_Rigid-Body_Rotation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.06:_Principal_Axis_System" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.07:_Diagonalize_the_Inertia_Tensor" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.08:_Parallel-Axis_Theorem" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.09:_Perpendicular-axis_Theorem_for_Plane_Laminae" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.10:_General_Properties_of_the_Inertia_Tensor" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.11:_Angular_Momentum_and_Angular_Velocity_Vectors" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.12:_Kinetic_Energy_of_Rotating_Rigid_Body" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.13:_Euler_Angles" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.14:_Angular_Velocity" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.15:_Kinetic_energy_in_terms_of_Euler_angular_velocities" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.16:_Rotational_Invariants" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.17:_Eulers_equations_of_motion_for_rigid-body_rotation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.18:_Lagrange_equations_of_motion_for_rigid-body_rotation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.19:_Hamiltonian_equations_of_motion_for_rigid-body_rotation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.20:_Torque-free_rotation_of_an_inertially-symmetric_rigid_rotor" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.21:_Torque-free_rotation_of_an_asymmetric_rigid_rotor" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.22:_Stability_of_torque-free_rotation_of_an_asymmetric_body" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.23:_Symmetric_rigid_rotor_subject_to_torque_about_a_fixed_point" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.24:_The_Rolling_Wheel" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.25:_Dynamic_balancing_of_wheels" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.26:_Rotation_of_Deformable_Bodies" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.E:_Rigid-body_Rotation_(Exercises)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13.S:_Rigid-body_Rotation_(Summary)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "01:_A_brief_History_of_Classical_Mechanics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "02:_Review_of_Newtonian_Mechanics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "03:_Linear_Oscillators" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "04:_Nonlinear_Systems_and_Chaos" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "05:_Calculus_of_Variations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "06:_Lagrangian_Dynamics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "07:_Symmetries_Invariance_and_the_Hamiltonian" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "08:_Hamiltonian_Mechanics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "09:_Hamilton\'s_Action_Principle" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "10:_Nonconservative_Systems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "11:_Conservative_two-body_Central_Forces" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "12:_Non-inertial_Reference_Frames" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "13:_Rigid-body_Rotation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "14:_Coupled_Linear_Oscillators" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "15:_Advanced_Hamiltonian_Mechanics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "16:_Analytical_Formulations_for_Continuous_Systems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "17:_Relativistic_Mechanics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "18:_The_Transition_to_Quantum_Physics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "19:_Mathematical_Methods_for_Classical_Mechanics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass226_0.b__1]()" }, 13.1: Introduction to Rigid-body Rotation, [ "article:topic", "authorname:dcline", "license:ccbyncsa", "showtoc:no", "principal axis system", "licenseversion:40", "source@http://classicalmechanics.lib.rochester.edu" ], https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FClassical_Mechanics%2FVariational_Principles_in_Classical_Mechanics_(Cline)%2F13%253A_Rigid-body_Rotation%2F13.01%253A_Introduction_to_Rigid-body_Rotation, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), source@http://classicalmechanics.lib.rochester.edu, status page at https://status.libretexts.org.
Polish Beet Soup Recipes, Meta Recruiter Phone Call, How To Create Directory In Internal Storage Android, Mountain Woods Bread Knife, Property Management Agreement With Owner, Ucf Nursing Transfer Requirements,
Polish Beet Soup Recipes, Meta Recruiter Phone Call, How To Create Directory In Internal Storage Android, Mountain Woods Bread Knife, Property Management Agreement With Owner, Ucf Nursing Transfer Requirements,