# Classical Mechanics. Luis Anchordoqui

Size: px
Start display at page:

Transcription

1 1 Rigid Body Motion Inertia Tensor Rotational Kinetic Energy Principal Axes of Rotation Steiner s Theorem Euler s Equations for a Rigid Body Eulerian Angles

2 Review of Fundamental Equations 2 Rigid body collection of a large number of small mass elements which all maintain a fixed spatial relationship with respect to one another has mass m i instantaneous position vector r i If there are N elements and the ith element: the equation of motion of the ith element is written d 2 j i r i m i dt 2 = j=1,n f ij + F i f ij internal force exerted on the ith element by the jth element F i the external force acting on the ith element Internal forces f ij stresses which develop within the body to ensure its various elements maintain a constant spatial relationship with respect to one another Of course f ij = f ji by Newton s third law The external forces represent forces which originate outside the body

3 Review of Fundamental Equations (cont d) 3 Recall that summing over all mass elements M d2 r cm dt 2 = F M = i=1,n m i the total mass r cm position vector of the center of mass F = i=1,n F i total external force Recall that the center of mass of a rigid body moves under the action of the external forces as a point particle whose mass is identical with that of the body Likewise d L dt = τ L = i=1,n m i r i d r i /dt total angular momentum of the body τ = i=1,n r i F i the total external torque

4 Moment of Inertia Tensor 4 Consider a rigid body rotating with fixed angular velocity ω about an axis which passes through the origin Let r i be the position vector of the ith mass element whose mass is m i This position vector to precesses about the axis of rotation with angular velocity ω d r i dt = ω r i This equation specifies the velocity v i = d r i /dt of each mass element as the body rotates with fixed angular velocity ω about an axis passing through the origin The total angular momentum of the body (about the origin) is written L = m i r i d r i dt = m i r i ( ω r i ) = [ ] m i r 2 i ω ( r i ω) r i i=1,n i=1,n i=1,n

5 Moment of Inertia Tensor (cont d) 5 The ith component reads L i = α = α = j m α ω i x 2 α,k x α,i x α,j ω j k ( ) m α ω j δ ij x 2 α,k ω j x α,i x α,j j k ( ) ω j m α δ ij x 2 α,k x α,i x α,j α k j L i = j I ij ω j

6 Moment of Inertia Tensor (cont d) 6 The previous equation can be written in a matrix form L x I xx I xy I xz ω x L y = I yx I yy I yz ω y L z I zx I zy I zz ω z I xx = i=1,n I yy = i=1,n I zz = i=1,n I xy = I yx = (y 2 i + z 2 i ) m i = (x 2 i + z 2 i ) m i = (x 2 i + y 2 i ) m i = i=1,n I yz = I zy = i=1,n I xz = I zx = i=1,n x i y i m i = y i z i m i = x i z i m i = (y 2 + z 2 ) dm (x 2 + z 2 ) dm (x 2 + y 2 ) dm x y dm y z dm x z dm

7 Moment of Inertia Tensor (cont d) 7 I xx moment of inertia about the x-axis I yy moment of inertia about the y-axis I zz moment of inertia about the z-axis I ij the ij product of inertia i j The matrix of the I ij values is known as the moment of inertia tensor a sum over separate mass elements Each component of the moment of inertia tensor can be written as an integral over infinitesimal mass elements In the integrals dm = ρdv ρ is the mass density dv a volume element The total angular momentum of the body can be written more succinctly as L = I ω ( ) L and ω are both column vectors and I is the matrix of the I ij values Note that I is a real symmetric matrix I ij = I ij and I ji = I ij Although the above results were obtained assuming a fixed angular velocity they remain valid at each instant in time even if the angular velocity varies

8 Rotational Kinetic Energy 8 The instantaneous rotational kinetic energy of a rotating rigid body is T = 1 2 i=1,n T = 1 2 i=1,n m i ( d ri dt ) 2 m i ( ω r i ) ( ω r i ) = 1 2 ω T = 1 ω L 2 or equivalently T = 1 2 ωt I ω i=1,n m i r i ( ω r i ) T = 1 2 ω T row vector of the Cartesian components (ω x, ω y, ω z ) which is the transpose (denoted T ) of the column vector ω when written in component form ( Ixx ω 2 x + I yy ω 2 y + I zzω 2 z + 2I xy ω x ω y + 2I yz ω y ω z + 2 I xz ω x ω z )

9 Matrix Theory 9 denotes a complex conjugate T denotes a transpose vector x eigenvector of A It is time to review a little matrix theory If A is a real symmetric matrix of dimension n A = A and A T = A Consider the matrix equation Ax = λx the associated number λ eigenvalue of A Properties of the eigenvectors and eigenvalues of a real symmetric matrix the matrix equation can be rearranged to give (A λ1)x = 0 where 1 is the unit matrix

10 Matrix Theory (cont d) 10 Matrix Equation set of n homogeneous simultaneous algebraic equations for the n components of x such a set of equations only has a non-trivial solution when the determinant of the associated matrix is set to zero a necessary condition for the set of equations to have a non-trivial solution is A λ1 = 0 The above formula is essentially an nth-order polynomial equation for λ We know that such an equation has n (possibly complex) roots conclusion there are n eigenvalues and n associated eigenvectors of the n-d matrix A

11 Matrix Theory (cont d) 11 theorem The n eigenvalues and eigenvectors of the real symmetric matrix A are all real Ax i = λ i x i ( ) taking the transpose and complex conjugate x T i A = λ i x T i ( ) where x i and λ i are the ith eigenvector and eigenvalue of A respectively Left multiplying Eq. ( ) by x i T we obtain λ i = λ i x T i x T i Ax i = λ i x T i x i Right multiplying ( ) by x i we get x T i Ax i = λ i x T i The difference of the previous two equations yields x i (λ i λ i )x T i x i = 0 x i (which is x i x i in vector notation) is positive definite λ i is real it follows that x i is real

12 Matrix Theory (cont d) 12 theorem 2 eigenvectors corresponding to 2 different eigenvalues are mutually orthogonal Let Ax i = λ i x i and Ax j = λ j x j λ i λ j Taking the transpose of the first equation and right multiplying by x j and left multiplying the second equation by x T i we obtain x T i Ax j = λ i x T i x j and x T i Ax j = λ j x T i x j Taking the difference of these equations we get (λ i λ j )x T i x j = 0 Since by hypothesis λ i λ j it follows that x T i x j = 0 In vector notation x i x j = 0 the eigenvectors x i and x j are mutually orthogonal

13 Matrix Theory (cont d) 13 If λ i = λ j = λ we cannot conclude that x T i x j = 0 by the above argument nevertheless it is easily seen that any linear combination of x i and x j is an eigenvector of A with eigenvalue λ it is possible to define two new eigenvectors of A which are mutually orthogonal (with the eigenvalue λ) For example ( ) x x i = x i and x T j = x j i x j x T i x x i i This can be generalized to any number of eigenvalues which take the same value conclusion A real symmetric n-dimensional matrix possesses n real eigenvalues with n associated real eigenvectors which are or can be chosen to be mutually orthogonal

14 Principal Axes of Rotation 14 The moment of inertia tensor I takes the form of a real symmetric three-dimensional matrix from the matrix theory which we have just reviewed the moment of inertia tensor possesses 3 mutually orthogonal eigenvectors which are associated with 3 real eigenvalues Let the ith eigenvector be denoted ˆω i and the ith eigenvalue λ i (which can be normalized to be a unit vector) I ˆω i = λ i ˆω i i = 1, 2, 3 ( ) The directions of the 3 mutually orthogonal unit vectors ˆω i define the 3 so-called principal axes of rotation of the rigid body These axes are special because when the body rotates about one of them (i.e., when ω is parallel to one of them) the angular momentum vector L becomes parallel to the angular velocity vector ω This can be seen from a comparison of Eq. ( ) and Eq. ( )

15 Principal Axes of Rotation (cont d) 15 Next reorient the Cartesian coordinate axes so the they coincide with the mutually orthogonal principal axes of rotation In this new reference frame the eigenvectors of I are the unit vectors e x e y e z and the eigenvalues are the moments of inertia about these axes I xx I yy I zz These latter quantities are referred to as the principal moments of inertia Note that the products of inertia are all zero in the new reference frame In this frame the moment of inertia tensor takes the form of a diagonal matrix I = I xx I yy I zz

16 Principal Axes of Rotation (cont d) 16 (I) verify that: homework e x e y e z are indeed the eigenvectors of the matrix I = I xx I yy I zz with the eigenvalues I xx I yy I zz (II) verify that: L = I ω is indeed parallel to ω whenever ω is directed along e x, e y, or e z

17 Principal Axes of Rotation (cont d) 17 When expressed in the new coordinate system, Eq. ( ) yields L = (I xx ω x, I yy ω y, I zz ω z ) the rotational kinetic energy becomes T = 1 ( Ixx ωx I yy ωy 2 + I zz ωz 2 ) conclusion There are many great simplifications to be had by choosing a coordinate system whose axes coincide with the principal axes of rotation of the rigid body But how do we determine the directions of the principal axes in practice?

18 Principal Axes of Rotation (cont d) 18 In general we have to solve the eigenvalue equation I ˆω = λ ˆω or I xx λ I xy I xz I yx I yy λ I yz I zx I zy I zz λ cosα cosβ cosγ = ˆω = (cosα, cos β, cosγ) cos 2 α + cos 2 β + cos 2 γ = 1 α is the angle the unit eigenvector subtends with the x-axis β the angle it subtends with the y-axis γ the angle it subtends with the z-axis Unfortunately the analytic solution of the above matrix equation is generally quite difficult Fortunately sometimes the rigid body under investigation possesses some kind of symmetry so that at least one principal axis can be found by inspection the other two principal axes can be determined as follows

19 Principal Axes of Rotation (cont d) 19 If the z-axis is known to be a principal axes in some coordinate system the two products of inertia I xz and I yz are zero otherwise (0, 0, 1) would not be an eigenvector The other two principal axes must lie in the x-y plane cosγ = 0 cosβ = sinα, since cos 2 α + cos 2 β + cos 2 γ = 1 The first two rows in the matrix equation thus reduce to (I xx λ) cosα + I xy sinα = 0 ( ) I xy cosα + (I yy λ) sin α = 0 Eliminating λ between the above two equations, we obtain I xy (1 tan 2 α) = (I xx I yy ) tanα ( )

20 Principal Axes of Rotation (cont d) 20 Now tan(2 α) 2 tanα/(1 tan 2 α) Eq. ( ) yields tan(2 α) = 2 I xy I xx I yy ( ) There are two values of α which satisfy the above equation (lying between π/2 and π/2) These specify the angles α which the two mutually orthogonal principal axes in the x-y plane make with the x-axis we have now determined the directions of all three principal axes Incidentally once we have determined the orientation angle α of a principal axis we can substitute back into Eq. ( ) to obtain the corresponding principal moment of inertia λ

21 Principal Axes of Rotation (cont d) 21 Consider a uniform rectangular lamina of mass m and sides a and b which lies in the x-y plane Suppose that the axis of rotation passes through the origin (i.e., through a corner of the lamina) Since z = 0 throughout the lamina I xz = I yz = 0 the z-axis is a principal axis

22 Principal Axes of Rotation (cont d) 22 After some straightforward integration I xx = 1 3 m b2 I yy = 1 3 m a2 I xy = 1 4 m a b from Eq. ( ) α = 1 ( ) 3 a b 2 tan 1 2 a 2 b 2 which specifies the orientation of the two principal axes which lie in the x-y plane For the special case where a = b α = π/4 and α = 3π/4 i.e., the two in-plane principal axes of a square lamina (at a corner) are parallel to the two diagonals of the lamina

23 Steiner s Theorem 23 For the kinetic energy to be separable into translational and rotational portions choose a body coordinate system whose origin is the c.m. of the body For certain geometrical shapes it may not be always convinient to compute the elements of the inertia tensor using such a coordinate system Consider some other set of coordinate axis X i (also fixed with respect to the body) having the same orientation that x i -axes but with origin Q that does not corresponds with the c.m. origin O (origin Q may be located either outside or within the body) The elements of the inertia tensor relative to X i -axes can be written as J ij = α ( ) m α δ ij Xα,k 2 X αi X α,j k

24 Steiner s Theorem (cont d) 24 If the vector connecting Q and O is a R = a + r with components X i = a i + x i J ij = α The tensor element J ij then becomes [ ] m α δ ij (x α,k + a k ) 2 (x α,j + a i )(x α,i + a j ) k

25 Steiner s Theorem (cont d) 25 J ij = α + α [ ] m α δ ij x 2 α,k x α,ix α,j k [ ] m α δ ij (2x α,k a k + a 2 k) (a i x α,j + a j x α,i + a i a j ) k Identifying the first summation as I ij and regrouping J ij = I ij + [ ] m α δ ij a 2 k a ia j α k + [ ] m α 2δ ij (x α,k a k a i x α,j a j x α,i ) α k Each term in the last sumation involves a term of the form α m αx α,k but because O is located at the c.m. α m αr α = 0 or for the kth component α m αx α,k = 0 all such terms vanish

26 Steiner s Theorem (cont d) 26 J ij = I ij + α [ ] m α δ ij a 2 k a ia j k α m α = M k a2 k a2 solving for I ij we have the result I ij = J ij M(a 2 δ ij a i a j ) which allows the calculation of the elements I ij of the desired inertia tensor (with origin at c.m.) once those with respect to X i -axes are known The second term in the right hand side is the inertia tensor referred to the origin Q for a point mass M The above equation is the general form of Steiner s parallel-axis theorem ( )

27 Euler s Equations for a Rigid Body 27 The fundamental equation of motion of a rotating body τ = d L dt is only valid in an inertial frame However we have seen that L is most simply expressed in a frame of reference whose axes are aligned along the principal axes of rotation of the body Such a frame of reference rotates with the body and is therefore non-inertial it is helpful to define two Cartesian coordinate systems with the same origins coordinates x, y, z denote the fixed inertial frame coordinates x, y, z co-rotates with the body in such a manner that the x -, y -, and z -axes are always pointing along its principal axes of rotation Since this body frame co-rotates with the body its instantaneous angular velocity is the same as that of the body d L dt = d L dt + ω L

28 Euler s Equations for a Rigid Body (cont d) 28 Recall that d/dt is the time derivative in the fixed frame d/dt the time derivative in the body frame The equation of motion of a rotating body can be re-written as τ = d L dt + ω L (ℵ) In the body frame let τ = (τ x, τ y, τ z ) and ω = (ω x, ω y, ω z ) L = (I x x ω x, I y y ω y, I z z ω z ) I x x I y y I z z are the principal moments of inertia

29 Euler s Equations for a Rigid Body (cont d) 29 In the body frame the components of Eq. (ℵ) yield τ x = I x x ω x (I y y I z z ) ω y ω z τ y = I y y ω y (I z z I x x ) ω z ω x τ z = I z z ω z (I x x I y y ) ω x ω y = d/dt We have made use of the fact that the moments of inertia of a rigid body are constant in time in the co-rotating body frame The above equations are known as Euler s equations

30 Euler s Equations for a Rigid Body (cont d) 30 Consider a rigid body which is constrained to rotate about a fixed axis with constant angular velocity ω x = ω y = ω z = 0 Euler s equations reduce to τ x = (I y y I z z ) ω y ω z τ y = (I z z I x x ) ω z ω x τ z = (I x x I y y ) ω x ω y These equations specify the components of the steady torque exerted on the body by the constraining supports (in the body frame) The steady angular momentum is written (in the body frame) L = (I x x ω x, I y y ω y, I z z ω z )

31 Euler s Equations for a Rigid Body (cont d) 31 It is easily seen that τ = ω L the angular velocity vector The torque is perpendicular to the angular momentum vector If the axis of rotation is a principal axis 2 of the 3 components of ω are zero (in the body frame) all 3 components of the torque are zero To make the body rotate steadily about a principal axis zero external torque is required

32 Euler s Equations for a Rigid Body (cont d) 32 Suppose that the body is freely rotating there are no external torques Let the body be also rotationally symmetric about the z -axis I x x = I y y = I and I z z = I (of course in general I I ) Euler s equations yield I dω x dt + (I I ) ω z ω y = 0 ( ) dω y I (I I ) ω z ω x = 0 ( ) dt dω z = 0 ω z is a constant of motion dt Equation ( ) and ( ) can be written dω x dt + Ω ω y = 0 and dω y dt Ω ω x = 0 Ω = (I /I 1) ω z

33 Euler s Equations for a Rigid Body (cont d) 33 It is easily seen that the solution to the ( ) and ( ) system of equations is ω x = ω cos(ω t) ω y = ω sin(ω t) ω is a constant The projection of ω onto the x -y plane has the fixed length ω and rotates steadily about the z -axis with angular velocity Ω the length of the angular velocity vector ω = (ω 2 x + ω2 y + ω2 z )1/2 is a constant of the motion The angular velocity vector makes some constant angle α with the z -axis which implies that ω z = ω cosα and ω = ω sinα

34 Euler s Equations for a Rigid Body (cont d) 34 The components of the angular velocity vector are ω x = ω sinα cos(ω t) ω y = ω sinα sin(ω t) ω z = ω cosα where ( ) I Ω = ω cosα 1 I (Ⅎ) In the body frame the angular velocity vector precesses about the symmetry axis (i.e., the z -axis) with the angular frequency Ω The components of the angular momentum vector are L x = I ω sinα cos(ω t) L y = I ω sinα sin(ω t) L z = I ω cosα In the body frame the angular momentum vector is also of constant length and precesses about the symmetry axis with the angular frequency Ω

35 Euler s Equations for a Rigid Body (cont d) 35 The angular momentum vector makes a constant angle θ with the symmetry axis tanθ = I I tanα (ð) Note that the angular momentum vector the angular velocity vector all lie in the same plane the symmetry axis In other words e z L ω = 0 The angular momentum vector lies between the angular velocity vector and the symmetry axis (i.e., θ < α) for a flattened (or oblate) body (i.e., I < I ) The angular velocity vector lies between the angular momentum vector and the symmetry axis (i.e., θ > α) for an elongated (or prolate) body (i.e., I > I )

36 Eulerian Angles 36 We have seen how we can solve Euler s equations to determine the properties of a rotating body in the co-rotating body frame next we investigate how to determine the same properties in the inertial fixed frame The fixed frame and the body frame share the same origin we can transform from one to the other by means of an appropriate rotation of our vector space If we restrict ourselves to rotations about 1 of the Cartesian coordinate axes 3 successive rotations are required to transform the fixed into the body frame There are many different ways to combined 3 successive rotations Next we describe the most widely used method which is due to Euler

37 Eulerian Angles (cont d) 37 We start in the fixed frame x e x which has coordinates y and unit vectors e y z e z Our first rotation is counterclockwise through an angle φ about the z-axis (looking down the axis) End in the new frame x e x which has coordinates y and unit vectors e y z e z The transformation of coordinates can be represented by x cosφ sinφ 0 y = sinφ cosφ 0 x y z z The angular velocity vector associated with φ has the magnitude φ and is directed along e z along the axis of rotation ω φ = φe z φ is the precession rate about the e z axis as seen in the fixed frame

38 Eulerian Angles (cont d) 38 The second rotation is counterclockwise through an angle θ about the x -axis (looking down the axis) the new frame x e x will have coordinates y and unit vectors e y z e z The transformation of coordinates can be represented by x x y = 0 cosθ sinθ y z 0 sinθ cosθ z The angular velocity vector associated with θ has the magnitude θ and is directed along e x along the axis of rotation ω θ = θ e x

39 Eulerian Angles (cont d) 39 The third rotation is counterclockwise through an angle ψ about the z -axis (looking down the axis) the new frame is the body frame x e x which has coordinates y and unit vectors e y z e z The transformation of coordinates can be represented by x cosψ sinψ 0 x y = sinψ cosψ 0 y z z The angular velocity vector associated with ψ has the magnitude ψ and is directed along e z along the axis of rotation e z = e z because the third rotation is about e z ω ψ = ψ e z ψ is minus the precession rate about the e z axis as seen in the body frame

40 Eulerian Angles (cont d) 40 The full transformation between the fixed frame and the body frame is complicated however the following results can easily be verified e z = sin ψ sinθ e x + cosψ sinθ e y + cosθ e z ( ) e x = cosψ e x sinψ e y It follows from Eq. ( ) that e z e z = cosθ θ is the angle of inclination between the z- and z -axes The total angular velocity is found to be ω = ω φ + ω θ + ω ψ ω x = sinψ sinθ φ + cosψ θ (ℶ) ω y = cosψ sinθ φ sinψ θ (ℸ) ω z (ג) = cosθ φ + ψ

41 Eulerian Angles (cont d) 41 φ The angles θ are termed Eulerian angles ψ Each has a clear physical interpretation φ is the angle of precession about the e z axis in the fixed frame ψ is minus the angle of precession about the e z axis in the body frame θ is the angle of inclination between the e z and e z axes (ℶ) Using Eqs. (ℸ) (ג) We can express the components of the angular velocity vector ω in the body frame entirely in terms of the Eulerian angles and their time derivatives

42 Warning! 42 The designation of the Euler angles is not universally agreed upon (even the manner in which they are generated) some care must be taken in comparing results from different sources If we identify the fixed system with x and the body system with x

43 Eulerian Angles (cont d) 43 Consider a freely rotating body which is rotationally symmetric about one axis In the absence of an external torque L is a constant of motion Let L point along the z-axis We saw that: L subtends a constant angle θ with the axis of symmetry (i.e., with the z -axis) the time derivative of the Eulerian angle θ is zero We also saw that: the angular momentum vector the axis of symmetry are co-planar the angular velocity vector Consider an instant in time at which all of these vectors lie in the y -z plane this implies that ω x = 0 ω subtends a constant angle α with the symmetry axis ω y = ω sinα and ω z = ω cosα

44 Eulerian Angles (cont d) 44 Eq. (ℶ) yields ψ = 0 Eq. (ℸ) yields ω sinα = sinθ φ (κ) This can be combined with Eq. (ð) to give φ = ω [ 1 + ( I 2 I 2 1 ) cos 2 α] 1/2 ( ) Finally Eqs. (ג) (ð) and (κ) yields ( ψ = ω cosα cosθ φ = ω cosα 1 tanα ) = ω cosα tanθ A comparison of the above equation with Eq. (Ⅎ) gives ψ = Ω ( 1 I ) I ψ is minus the precession rate in the body frame (of the angular momentum and angular velocity vectors) φ is the precession rate in the fixed frame (of the angular velocity vector and the symmetry axis)

45 Eulerian Angles (cont d) 45 It is known that the Earth s axis of rotation is slightly inclined to its symmetry axis (which passes through the two poles) the angle α is approximately 0.2 seconds of an arc It is also known that the ratio of the moments of inertia is about I I = as determined from the Earth s oblateness from (Ⅎ) as viewed on Earth the precession rate of the angular velocity vector about the symmetry axis is Ω = ω giving a precession period of T = 2π Ω = 305 days (of course 2π/ω = 1 day) The observed period of precession is about 440 days The disagreement between theory and observation is attributed to the fact that the Earth is not perfectly rigid

46 Eulerian Angles (cont d) 46 The (theroretical) precession rate of the Earth s symmetry axis is given by Eq. ( ) (as viewed from space) φ = ω The associated precession period is T = 2π φ = days The free precession of the Earth s symmetry axis in space is superimposed on a much slower precession with a period of about 26,000 years due to the small gravitational torque exerted on Earth by Sun and Moon because of the Earth s slight oblateness

### Rotational Motion. Chapter 4. P. J. Grandinetti. Sep. 1, Chem P. J. Grandinetti (Chem. 4300) Rotational Motion Sep.

Rotational Motion Chapter 4 P. J. Grandinetti Chem. 4300 Sep. 1, 2017 P. J. Grandinetti (Chem. 4300) Rotational Motion Sep. 1, 2017 1 / 76 Angular Momentum The angular momentum of a particle with respect

### Lecture 37: Principal Axes, Translations, and Eulerian Angles

Lecture 37: Principal Axes, Translations, and Eulerian Angles When Can We Find Principal Axes? We can always write down the cubic equation that one must solve to determine the principal moments But if

### Phys 7221 Homework # 8

Phys 71 Homework # 8 Gabriela González November 15, 6 Derivation 5-6: Torque free symmetric top In a torque free, symmetric top, with I x = I y = I, the angular velocity vector ω in body coordinates with

### Artificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik. Robot Dynamics. Dr.-Ing. John Nassour J.

Artificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik Robot Dynamics Dr.-Ing. John Nassour 25.1.218 J.Nassour 1 Introduction Dynamics concerns the motion of bodies Includes Kinematics

### CP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS. Prof. N. Harnew University of Oxford TT 2017

CP1 REVISION LECTURE 3 INTRODUCTION TO CLASSICAL MECHANICS Prof. N. Harnew University of Oxford TT 2017 1 OUTLINE : CP1 REVISION LECTURE 3 : INTRODUCTION TO CLASSICAL MECHANICS 1. Angular velocity and

### Lecture 38: Equations of Rigid-Body Motion

Lecture 38: Equations of Rigid-Body Motion It s going to be easiest to find the equations of motion for the object in the body frame i.e., the frame where the axes are principal axes In general, we can

### 9 Kinetics of 3D rigid bodies - rotating frames

9 Kinetics of 3D rigid bodies - rotating frames 9. Consider the two gears depicted in the figure. The gear B of radius R B is fixed to the ground, while the gear A of mass m A and radius R A turns freely

### Lecture 38: Equations of Rigid-Body Motion

Lecture 38: Equations of Rigid-Body Motion It s going to be easiest to find the equations of motion for the object in the body frame i.e., the frame where the axes are principal axes In general, we can

### Rotational & Rigid-Body Mechanics. Lectures 3+4

Rotational & Rigid-Body Mechanics Lectures 3+4 Rotational Motion So far: point objects moving through a trajectory. Next: moving actual dimensional objects and rotating them. 2 Circular Motion - Definitions

### Physics 312, Winter 2007, Practice Final

Physics 312, Winter 2007, Practice Final Time: Two hours Answer one of Question 1 or Question 2 plus one of Question 3 or Question 4 plus one of Question 5 or Question 6. Each question carries equal weight.

### Part 8: Rigid Body Dynamics

Document that contains homework problems. Comment out the solutions when printing off for students. Part 8: Rigid Body Dynamics Problem 1. Inertia review Find the moment of inertia for a thin uniform rod

### General Physics I. Lecture 10: Rolling Motion and Angular Momentum.

General Physics I Lecture 10: Rolling Motion and Angular Momentum Prof. WAN, Xin (万歆) 万歆 ) xinwan@zju.edu.cn http://zimp.zju.edu.cn/~xinwan/ Outline Rolling motion of a rigid object: center-of-mass motion

### A set of N particles forms a rigid body if the distance between any 2 particles is fixed:

Chapter Rigid Body Dynamics.1 Coordinates of a Rigid Body A set of N particles forms a rigid body if the distance between any particles is fixed: r ij r i r j = c ij = constant. (.1) Given these constraints,

### 12. Rigid Body Dynamics I

University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 015 1. Rigid Body Dynamics I Gerhard Müller University of Rhode Island, gmuller@uri.edu Creative Commons License

### Rigid Body Rotation. Speaker: Xiaolei Chen Advisor: Prof. Xiaolin Li. Department of Applied Mathematics and Statistics Stony Brook University (SUNY)

Rigid Body Rotation Speaker: Xiaolei Chen Advisor: Prof. Xiaolin Li Department of Applied Mathematics and Statistics Stony Brook University (SUNY) Content Introduction Angular Velocity Angular Momentum

### In this section of notes, we look at the calculation of forces and torques for a manipulator in two settings:

Introduction Up to this point we have considered only the kinematics of a manipulator. That is, only the specification of motion without regard to the forces and torques required to cause motion In this

### 1/30. Rigid Body Rotations. Dave Frank

. 1/3 Rigid Body Rotations Dave Frank A Point Particle and Fundamental Quantities z 2/3 m v ω r y x Angular Velocity v = dr dt = ω r Kinetic Energy K = 1 2 mv2 Momentum p = mv Rigid Bodies We treat a rigid

### Stress, Strain, Mohr s Circle

Stress, Strain, Mohr s Circle The fundamental quantities in solid mechanics are stresses and strains. In accordance with the continuum mechanics assumption, the molecular structure of materials is neglected

### 6. 3D Kinematics DE2-EA 2.1: M4DE. Dr Connor Myant

DE2-EA 2.1: M4DE Dr Connor Myant 6. 3D Kinematics Comments and corrections to connor.myant@imperial.ac.uk Lecture resources may be found on Blackboard and at http://connormyant.com Contents Three-Dimensional

### Spacecraft Dynamics and Control

Spacecraft Dynamics and Control Matthew M. Peet Arizona State University Lecture 16: Euler s Equations Attitude Dynamics In this Lecture we will cover: The Problem of Attitude Stabilization Actuators Newton

### Properties of the stress tensor

Appendix C Properties of the stress tensor Some of the basic properties of the stress tensor and traction vector are reviewed in the following. C.1 The traction vector Let us assume that the state of stress

### If the symmetry axes of a uniform symmetric body coincide with the coordinate axes, the products of inertia (Ixy etc.

Prof. O. B. Wright, Autumn 007 Mechanics Lecture 9 More on rigid bodies, coupled vibrations Principal axes of the inertia tensor If the symmetry axes of a uniform symmetric body coincide with the coordinate

### Classical Mechanics III (8.09) Fall 2014 Assignment 3

Classical Mechanics III (8.09) Fall 2014 Assignment 3 Massachusetts Institute of Technology Physics Department Due September 29, 2014 September 22, 2014 6:00pm Announcements This week we continue our discussion

### Manipulator Dynamics 2. Instructor: Jacob Rosen Advanced Robotic - MAE 263D - Department of Mechanical & Aerospace Engineering - UCLA

Manipulator Dynamics 2 Forward Dynamics Problem Given: Joint torques and links geometry, mass, inertia, friction Compute: Angular acceleration of the links (solve differential equations) Solution Dynamic

### UAV Coordinate Frames and Rigid Body Dynamics

Brigham Young University BYU ScholarsArchive All Faculty Publications 24-- UAV oordinate Frames and Rigid Body Dynamics Randal Beard beard@byu.edu Follow this and additional works at: https://scholarsarchive.byu.edu/facpub

### Physics 106b/196b Problem Set 9 Due Jan 19, 2007

Physics 06b/96b Problem Set 9 Due Jan 9, 2007 Version 3: January 8, 2007 This problem set focuses on dynamics in rotating coordinate systems (Section 5.2), with some additional early material on dynamics

### Physics 106a, Caltech 4 December, Lecture 18: Examples on Rigid Body Dynamics. Rotating rectangle. Heavy symmetric top

Physics 106a, Caltech 4 December, 2018 Lecture 18: Examples on Rigid Body Dynamics I go through a number of examples illustrating the methods of solving rigid body dynamics. In most cases, the problem

### Chapter 6: Momentum Analysis

6-1 Introduction 6-2Newton s Law and Conservation of Momentum 6-3 Choosing a Control Volume 6-4 Forces Acting on a Control Volume 6-5Linear Momentum Equation 6-6 Angular Momentum 6-7 The Second Law of

### V (r,t) = i ˆ u( x, y,z,t) + ˆ j v( x, y,z,t) + k ˆ w( x, y, z,t)

IV. DIFFERENTIAL RELATIONS FOR A FLUID PARTICLE This chapter presents the development and application of the basic differential equations of fluid motion. Simplifications in the general equations and common

### Astronomy 6570 Physics of the Planets. Precession: Free and Forced

Astronomy 6570 Physics of the Planets Precession: Free and Forced Planetary Precession We have seen above how information concerning the distribution of density within a planet (in particular, the polar

### identify appropriate degrees of freedom and coordinates for a rigid body;

Chapter 5 Rotational motion A rigid body is defined as a body (or collection of particles) where all mass points stay at the same relative distances at all times. This can be a continuous body, or a collection

### Torque and Rotation Lecture 7

Torque and Rotation Lecture 7 ˆ In this lecture we finally move beyond a simple particle in our mechanical analysis of motion. ˆ Now we consider the so-called rigid body. Essentially, a particle with extension

### PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2)

PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2) We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane. As discussed in Chapter 16, when

### Chapter 11. Angular Momentum

Chapter 11 Angular Momentum Angular Momentum Angular momentum plays a key role in rotational dynamics. There is a principle of conservation of angular momentum. In analogy to the principle of conservation

### Lagrange Multipliers

Optimization with Constraints As long as algebra and geometry have been separated, their progress have been slow and their uses limited; but when these two sciences have been united, they have lent each

### Rigid bodies - general theory

Rigid bodies - general theory Kinetic Energy: based on FW-26 Consider a system on N particles with all their relative separations fixed: it has 3 translational and 3 rotational degrees of freedom. Motion

### 19. Principal Stresses

19. Principal Stresses I Main Topics A Cauchy s formula B Principal stresses (eigenvectors and eigenvalues) C Example 10/24/18 GG303 1 19. Principal Stresses hkp://hvo.wr.usgs.gov/kilauea/update/images.html

### Physics 351 Wednesday, April 4, 2018

Physics 351 Wednesday, April 4, 2018 Pick up worksheet (!) on your way in. HW10 due this Friday. I tried (!) to make it short. For HW help, Bill is in DRL 3N6 Wed 4 7pm. Grace is in DRL 2C2 Thu 5:30 8:30pm.

### Dynamics. 1 Copyright c 2015 Roderic Grupen

Dynamics The branch of physics that treats the action of force on bodies in motion or at rest; kinetics, kinematics, and statics, collectively. Websters dictionary Outline Conservation of Momentum Inertia

### Lesson Rigid Body Dynamics

Lesson 8 Rigid Body Dynamics Lesson 8 Outline Problem definition and motivations Dynamics of rigid bodies The equation of unconstrained motion (ODE) User and time control Demos / tools / libs Rigid Body

### Physics 351, Spring 2015, Final Exam.

Physics 351, Spring 2015, Final Exam. This closed-book exam has (only) 25% weight in your course grade. You can use one sheet of your own hand-written notes. Please show your work on these pages. The back

### Rigid Body Dynamics: Kinematics and Kinetics. Rigid Body Dynamics K. Craig 1

Rigid Body Dynamics: Kinematics and Kinetics Rigid Body Dynamics K. Craig 1 Topics Introduction to Dynamics Basic Concepts Problem Solving Procedure Kinematics of a Rigid Body Essential Example Problem

### Rigid body dynamics. Basilio Bona. DAUIN - Politecnico di Torino. October 2013

Rigid body dynamics Basilio Bona DAUIN - Politecnico di Torino October 2013 Basilio Bona (DAUIN - Politecnico di Torino) Rigid body dynamics October 2013 1 / 16 Multiple point-mass bodies Each mass is

### REVIEW - Vectors. Vectors. Vector Algebra. Multiplication by a scalar

J. Peraire Dynamics 16.07 Fall 2004 Version 1.1 REVIEW - Vectors By using vectors and defining appropriate operations between them, physical laws can often be written in a simple form. Since we will making

### Physics 351, Spring 2018, Homework #9. Due at start of class, Friday, March 30, 2018

Physics 351, Spring 218, Homework #9. Due at start of class, Friday, March 3, 218 Please write your name on the LAST PAGE of your homework submission, so that we don t notice whose paper we re grading

### Chapter 6: Momentum Analysis of Flow Systems

Chapter 6: Momentum Analysis of Flow Systems Introduction Fluid flow problems can be analyzed using one of three basic approaches: differential, experimental, and integral (or control volume). In Chap.

### GG612 Lecture 3. Outline

GG61 Lecture 3 Strain and Stress Should complete infinitesimal strain by adding rota>on. Outline Matrix Opera+ons Strain 1 General concepts Homogeneous strain 3 Matrix representa>ons 4 Squares of line

### CH.4. STRESS. Continuum Mechanics Course (MMC)

CH.4. STRESS Continuum Mechanics Course (MMC) Overview Forces Acting on a Continuum Body Cauchy s Postulates Stress Tensor Stress Tensor Components Scientific Notation Engineering Notation Sign Criterion

### PEAT SEISMOLOGY Lecture 2: Continuum mechanics

PEAT8002 - SEISMOLOGY Lecture 2: Continuum mechanics Nick Rawlinson Research School of Earth Sciences Australian National University Strain Strain is the formal description of the change in shape of a

### Rotational motion of a rigid body spinning around a rotational axis ˆn;

Physics 106a, Caltech 15 November, 2018 Lecture 14: Rotations The motion of solid bodies So far, we have been studying the motion of point particles, which are essentially just translational. Bodies with

### MECH 5312 Solid Mechanics II. Dr. Calvin M. Stewart Department of Mechanical Engineering The University of Texas at El Paso

MECH 5312 Solid Mechanics II Dr. Calvin M. Stewart Department of Mechanical Engineering The University of Texas at El Paso Table of Contents Preliminary Math Concept of Stress Stress Components Equilibrium

### SOLUTIONS, PROBLEM SET 11

SOLUTIONS, PROBLEM SET 11 1 In this problem we investigate the Lagrangian formulation of dynamics in a rotating frame. Consider a frame of reference which we will consider to be inertial. Suppose that

### Rigid body simulation. Once we consider an object with spatial extent, particle system simulation is no longer sufficient

Rigid body dynamics Rigid body simulation Once we consider an object with spatial extent, particle system simulation is no longer sufficient Rigid body simulation Unconstrained system no contact Constrained

### Robot Control Basics CS 685

Robot Control Basics CS 685 Control basics Use some concepts from control theory to understand and learn how to control robots Control Theory general field studies control and understanding of behavior

### PLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION

PLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION I. Moment of Inertia: Since a body has a definite size and shape, an applied nonconcurrent force system may cause the body to both translate and rotate.

### Problem Set 2 Due Tuesday, September 27, ; p : 0. (b) Construct a representation using five d orbitals that sit on the origin as a basis: 1

Problem Set 2 Due Tuesday, September 27, 211 Problems from Carter: Chapter 2: 2a-d,g,h,j 2.6, 2.9; Chapter 3: 1a-d,f,g 3.3, 3.6, 3.7 Additional problems: (1) Consider the D 4 point group and use a coordinate

### Lecture 41: Highlights

Lecture 41: Highlights The goal of this lecture is to remind you of some of the key points that we ve covered this semester Note that this is not the complete set of topics that may appear on the final

### PHYS 705: Classical Mechanics. Euler s Equations

1 PHYS 705: Classical Mechanics Euler s Equations 2 Euler s Equations (set up) We have seen how to describe the kinematic properties of a rigid body. Now, we would like to get equations of motion for it.

### Introduction to Seismology Spring 2008

MIT OpenCourseWare http://ocw.mit.edu 12.510 Introduction to Seismology Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Stress and Strain

### Unit IV State of stress in Three Dimensions

Unit IV State of stress in Three Dimensions State of stress in Three Dimensions References Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi 2004. Rattan.S.S., "Strength

### M E 320 Professor John M. Cimbala Lecture 10

M E 320 Professor John M. Cimbala Lecture 10 Today, we will: Finish our example problem rates of motion and deformation of fluid particles Discuss the Reynolds Transport Theorem (RTT) Show how the RTT

### CHAPTER 4 Stress Transformation

CHAPTER 4 Stress Transformation ANALYSIS OF STRESS For this topic, the stresses to be considered are not on the perpendicular and parallel planes only but also on other inclined planes. A P a a b b P z

### Chapter 1. Continuum mechanics review. 1.1 Definitions and nomenclature

Chapter 1 Continuum mechanics review We will assume some familiarity with continuum mechanics as discussed in the context of an introductory geodynamics course; a good reference for such problems is Turcotte

### 2 so that the (time-dependent) torque, as measured in space-xed coordinates, is ~N = ma2! 2 (^x cos!t +^y sin!t) 10 = a p 5 2 ^z ~ F 2 ~F 1 = a p 5^z

1 PHYS 321 Homework Assignment #9 Due: Friday, 22 November 2002 (4 probs) 1. Problem 7-14, B&O p. 280. The situation is as shown above. We take the body-xed axes ^x 0 and ^y 0 to lie in the plane of the

### Differential Kinematics

Differential Kinematics Relations between motion (velocity) in joint space and motion (linear/angular velocity) in task space (e.g., Cartesian space) Instantaneous velocity mappings can be obtained through

### Physics 339 Euler Angles & Free Precession November 2017 Hamilton s Revenge

Physics 339 Euler Angles & Free Precession November 217 Hamilton s Revenge As described in the textbook, Euler Angles are a way to specify the configuration of a 3d object. Starting from a fixed configuration

### 27. Euler s Equations

27 Euler s Equations Michael Fowler Introduction We ve just seen that by specifying the rotational direction and the angular phase of a rotating body using Euler s angles, we can write the Lagrangian in

### Lecture II: Rigid-Body Physics

Rigid-Body Motion Previously: Point dimensionless objects moving through a trajectory. Today: Objects with dimensions, moving as one piece. 2 Rigid-Body Kinematics Objects as sets of points. Relative distances

### Rotation. Rotational Variables

Rotation Rigid Bodies Rotation variables Constant angular acceleration Rotational KE Rotational Inertia Rotational Variables Rotation of a rigid body About a fixed rotation axis. Rigid Body an object that

### Jitter and Basic Requirements of the Reaction Wheel Assembly in the Attitude Control System

Jitter and Basic Requirements of the Reaction Wheel Assembly in the Attitude Control System Lulu Liu August, 7 1 Brief Introduction Photometric precision is a major concern in this space mission. A pointing

### Chapter 9 Notes. x cm =

Chapter 9 Notes Chapter 8 begins the discussion of rigid bodies, a system of particles with fixed relative positions. Previously we have dealt with translation of a particle: if a rigid body does not rotate

### Physics 351 Friday, April 7, 2017

Physics 351 Friday, April 7, 2017 Turn in HW11 today. Handing out HW12 due next Friday. The only homework you ll do on last weekend s Chapter 11 (coupled oscillators, normal modes, etc.) is solving the

### Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter.

Chapter 3 Mechanical Systems A. Bazoune 3.1 INRODUCION Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter. 3. MECHANICAL ELEMENS Any mechanical system consists

### Leonhard Euler ( September 1783)

LEONHARD EULER (1707 - SEPTEMBER 1783) Leonhard Euler (1707 - September 1783) BEYOND EQUATIONS Leonhard Euler was born in Basle, Switzerland; he was in fact a born mathematician, who went on to become

### Rotational Kinematics and Dynamics. UCVTS AIT Physics

Rotational Kinematics and Dynamics UCVTS AIT Physics Angular Position Axis of rotation is the center of the disc Choose a fixed reference line Point P is at a fixed distance r from the origin Angular Position,

### CSE 167: Introduction to Computer Graphics Lecture #2: Linear Algebra Primer

CSE 167: Introduction to Computer Graphics Lecture #2: Linear Algebra Primer Jürgen P. Schulze, Ph.D. University of California, San Diego Fall Quarter 2016 Announcements Monday October 3: Discussion Assignment

### 3 2 6 Solve the initial value problem u ( t) 3. a- If A has eigenvalues λ =, λ = 1 and corresponding eigenvectors 1

Math Problem a- If A has eigenvalues λ =, λ = 1 and corresponding eigenvectors 1 3 6 Solve the initial value problem u ( t) = Au( t) with u (0) =. 3 1 u 1 =, u 1 3 = b- True or false and why 1. if A is

### Tensor Visualization. CSC 7443: Scientific Information Visualization

Tensor Visualization Tensor data A tensor is a multivariate quantity Scalar is a tensor of rank zero s = s(x,y,z) Vector is a tensor of rank one v = (v x,v y,v z ) For a symmetric tensor of rank 2, its

### Exercise 1: Inertia moment of a simple pendulum

Exercise : Inertia moment of a simple pendulum A simple pendulum is represented in Figure. Three reference frames are introduced: R is the fixed/inertial RF, with origin in the rotation center and i along

### ROBOTICS: ADVANCED CONCEPTS & ANALYSIS

ROBOTICS: ADVANCED CONCEPTS & ANALYSIS MODULE 5 VELOCITY AND STATIC ANALYSIS OF MANIPULATORS Ashitava Ghosal 1 1 Department of Mechanical Engineering & Centre for Product Design and Manufacture Indian

### PHY 5246: Theoretical Dynamics, Fall Assignment # 9, Solutions. y CM (θ = 0) = 2 ρ m

PHY 546: Theoretical Dnamics, Fall 5 Assignment # 9, Solutions Graded Problems Problem (.a) l l/ l/ CM θ x In order to find the equation of motion of the triangle, we need to write the Lagrangian, with

### Chapter 5. The Differential Forms of the Fundamental Laws

Chapter 5 The Differential Forms of the Fundamental Laws 1 5.1 Introduction Two primary methods in deriving the differential forms of fundamental laws: Gauss s Theorem: Allows area integrals of the equations

### Chapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics

Chapter 1: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Translational vs Rotational / / 1/ m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv / / 1/ I

### Chapter 10. Rotation of a Rigid Object about a Fixed Axis

Chapter 10 Rotation of a Rigid Object about a Fixed Axis Angular Position Axis of rotation is the center of the disc Choose a fixed reference line. Point P is at a fixed distance r from the origin. A small

### Getting started: CFD notation

PDE of p-th order Getting started: CFD notation f ( u,x, t, u x 1,..., u x n, u, 2 u x 1 x 2,..., p u p ) = 0 scalar unknowns u = u(x, t), x R n, t R, n = 1,2,3 vector unknowns v = v(x, t), v R m, m =

### 16. Rotational Dynamics

6. Rotational Dynamics A Overview In this unit we will address examples that combine both translational and rotational motion. We will find that we will need both Newton s second law and the rotational

### 1 Stress and Strain. Introduction

1 Stress and Strain Introduction This book is concerned with the mechanical behavior of materials. The term mechanical behavior refers to the response of materials to forces. Under load, a material may

### Noninertial Reference Frames

Chapter 12 Non Reference Frames 12.1 Accelerated Coordinate Systems A reference frame which is fixed with respect to a rotating rigid is not. The parade example of this is an observer fixed on the surface

### Chapter 4: Fluid Kinematics

Overview Fluid kinematics deals with the motion of fluids without considering the forces and moments which create the motion. Items discussed in this Chapter. Material derivative and its relationship to

### CSE 167: Introduction to Computer Graphics Lecture #2: Linear Algebra Primer

CSE 167: Introduction to Computer Graphics Lecture #2: Linear Algebra Primer Jürgen P. Schulze, Ph.D. University of California, San Diego Spring Quarter 2016 Announcements Project 1 due next Friday at

### Mechanics Physics 151

Mechanics Phsics 151 Lecture 8 Rigid Bod Motion (Chapter 4) What We Did Last Time! Discussed scattering problem! Foundation for all experimental phsics! Defined and calculated cross sections! Differential

### Revision of the physical backgrounds of Earth s rotation

Revision of the physical backgrounds of Earth s rotation L. Völgyesi Budapest University of Technology and Economics, H-151 Budapest, Müegyetem rkp. 3. Hungary e-mail: lvolgyesi@epito.bme.hu bstract. Rotation

### arxiv: v1 [math.ds] 18 Nov 2008

arxiv:0811.2889v1 [math.ds] 18 Nov 2008 Abstract Quaternions And Dynamics Basile Graf basile.graf@epfl.ch February, 2007 We give a simple and self contained introduction to quaternions and their practical

### Lecture 9 - Rotational Dynamics

Lecture 9 - Rotational Dynamics A Puzzle... Angular momentum is a 3D vector, and changing its direction produces a torque τ = dl. An important application in our daily lives is that bicycles don t fall

### Notes on multivariable calculus

Notes on multivariable calculus Jonathan Wise February 2, 2010 1 Review of trigonometry Trigonometry is essentially the study of the relationship between polar coordinates and Cartesian coordinates in

### 14. Rotational Kinematics and Moment of Inertia

14. Rotational Kinematics and Moment of nertia A) Overview n this unit we will introduce rotational motion. n particular, we will introduce the angular kinematic variables that are used to describe the

### Lecture AC-1. Aircraft Dynamics. Copy right 2003 by Jon at h an H ow

Lecture AC-1 Aircraft Dynamics Copy right 23 by Jon at h an H ow 1 Spring 23 16.61 AC 1 2 Aircraft Dynamics First note that it is possible to develop a very good approximation of a key motion of an aircraft

### Properties of surfaces II: Second moment of area

Properties of surfaces II: Second moment of area Just as we have discussing first moment of an area and its relation with problems in mechanics, we will now describe second moment and product of area of