UNIVERSITY PHYSICS I CHAPTER 10 Chapter 10 Spin and orbital motion Rotation: All around us: wheels, skaters, ballet, gymnasts helicopter, rotors, mobile engines, CD disks, Atomic world: electrons“spin”,“ orbit Universe: planets spin and orbiting the sun galaxies spin,… Chapter 4 kinematics Chapter 10 dynam
1 Chapter 10 Spin and orbital motion Rotation: All around us: wheels, skaters, ballet, gymnasts, helicopter, rotors, mobile engines, CD disks, … Atomic world: electrons— “spin”, “orbit”. Universe: planets spin and orbiting the sun, galaxies spin, … Chapter 4 kinematics Chapter 10 dynamics
s10.1 Some concepts about rotation 1. Spin--describe rotational motion of a system about an axis through its center of mass 2. Rigid body-a system composed of many pointlike particles that maintain fixed distances from each other at all time each particle of the spinning rigid body system executes circular motion about the axis through the center of mass 3. Orbital motion-the center of the mass of system is moving in space from a perspective t a particular reference frame. 810.1 Some concepts about rotation The motion of the center of mass must not be circular ●
2 §10.1 Some concepts about rotation 1. Spin—describe rotational motion of a system about an axis through its center of mass. 2. Rigid body—a system composed of many pointlike particles that maintain fixed distances from each other at all time. each particle of the spinning rigid body system executes circular motion about the axis through the center of mass. 3. Orbital motion—the center of the mass of the system is moving in space from a perspective of a particular reference frame. The motion of the center of mass must not be circular. §10.1 Some concepts about rotation
s10.1 Some concepts about rotation 4. The orbital angular momentum of a particle Define:L=r×p=rxm Magnitude L L=rmvsin8=Pr=rp Direction: Perpendicular to the plane containing the 6∴ r and p 810.1 Some concepts about rotation Notice O L is measured with respect to the origin at O; ②unit:kgm2/s; 3 whatever the path or trajectory of a particle is straightline. curved path, closed orbital ath 5. The angular momentum of the circular orbital motion of a particle (a) Angular momentum
3 4. The orbital angular momentum of a particle Define: L r p r mv r r r r r = × = × x y z m θ r r p r o ⊥r L r ⊥ p Magnitude: L = rmvsinθ = p⊥r = r⊥ p Direction: r p r r and Perpendicular to the plane containing the §10.1 Some concepts about rotation Notice: 1 is measured with respect to the origin at O; 2 unit: kg·m2/s; 3 whatever the path or trajectory of a particle is straightline , curved path, closed orbital path, …. L r 5. The angular momentum of the circular orbital motion of a particle (a) Angular momentum §10.1 Some concepts about rotation
s10.1 Some concepts about rotation L=rxp=r×m L= myr V=xI v=⑦r then mor=mro L=mro (b)Moment of inertia of a particle define = mr then L=mro=lo 810.2 The time rate of change of angular momentum and torque 1. The time rate of change of angular momentum for a single particle L=r×P P) X D+rx dt dt dt d p=×p=vxmv=0 F dt dL =rX d rxF dt total
4 L mvr L r p r mv = = × = × r r r r r v r v r ω ω = = × r r r r r p r o m ω r ω ω 2 2 then L = m r = mr (b) Moment of inertia of a particle ω r r 2 L = mr §10.1 Some concepts about rotation 2 I = mr ω ω r r r L = mr = I 2 define then §10.2 The time rate of change of angular momentum and torque 1. The time rate of change of angular momentum for a single particle t p p r t r r p t t L d d d d ( ) d d d d r r r r r r r = × = × + × L r p r r r = × total d d d d 0 d d r F t p r t L p v p v mv t r r r r r r r r r r r r Q ∴ = × = × × = × = × = r r F r θ o m
810.2 The time rate of change of angular momentum and torque 2. torque Define: T total=/ F total Magnitude t=rF sin 0=n F=rF1=d' Direction: Perpendicular to the plane containing the r and F Unit of the torque: n'm r is the position vector of the point of application of the force with respect to the chosen origin 8 10.2 The time rate of change of angular momentum and torque Discussion: F 6=0or丌, If 3 F=0 total total =0 F cross the O
5 Define: total Ftotal r r r r τ = × 2. torque is the position vector of the point of application of the force with respect to the chosen origin. r r Unit of the torque: N·m Magnitude: = = ⊥ = ⊥ τ rF sinθ r F rF Direction: Perpendicular to the plane containing the r F r r and r r F r θ o m r⊥ = d F⊥ §10.2 The time rate of change of angular momentum and torque r r F r θ o m r⊥ = d F⊥ If cross the , 0, 0 or , total F O Fr r = θ = π 0 τ total = r §10.2 The time rate of change of angular momentum and torque Discussion:
8 10.2 The time rate of change of angular momentum and torque 3. Dynamics of circular orbital motion of a single particle L=mr20=lo r×F dt d d total (o= dr Example 1: P3 10.5 Can not be used Example 2: P433 10.6 in noncircular orbital motion 8 10.3 The angular momentum of a system of particles and moment of inertia of rigid body 1. The angular momentum of a system of particles L=∑L=∑x=∑x CM +r! P L=∑Gw+xm ×∑m+∑xm(m+) =xm+mm+∑m可 6
6 3. Dynamics of circular orbital motion of a single particle ω ω r r r Q L = mr = I 2 total total d d d d τ r r r r r r = × = r × F = t p r t L Example 1: P433 10.5 Example 2: P433 10.6 α ω τ ω r r r r I t I I t ∴ = = = d d ( ) d d total Can not be used in noncircular orbital motion. §10.2 The time rate of change of angular momentum and torque §10.3 The angular momentum of a system of particles and moment of inertia of rigid body 1. The angular momentum of a system of particles = ∑ = ∑ × = ∑ × ii i i i i i i i L L r p r m v r r r r r r ⎩ ⎨ ⎧ = + ′ = + ′ i CM i i CM i v v v r r r r r r r r r Q θ pi r o rCM r ir r mi ir r C ′ ( ) ( ) i i i i CM i i i CM i i i i CM i i i CM i i i i i i CM i r m v r m v r m v r m v r m v v L r r m v = × + ′× + ′× ′ = × + ′× + ′ = + ′ × ∑ ∑ ∑ ∑ ∑ ∑ r r r r r r r r r r r r r r r ∴
810.3 The angular momentum of a system of particles and moment of inertia of rigid body First term: FcM x2m, ",=FCM XMVCM Second term ∑水mFCM=∑mx下m= Mcmv=0 le position vector of center of mass with respect to the center of mass Third term: ∑ r×mv is the vector sum of angular momentum of all particles with respect to the center of mass. 33中mHm出e then L=×Mc+∑可xm可 orbital 十 2. Spin angular momentum of oN a rigid body about a axis through the center of mass ∑xm可 =0X ⊥ i∥ 十 ∴Lm=∑m+)x(D×n)
7 ×∑ = × i rCM mivi rCM MvCM r r r r First term: ∑ ′× = ∑ ′× = ′ × = 0 CM CM CM i i CM i i i i r m v m r v Mr v r r r r r r Second term: The position vector of center of mass with respect to the center of mass i i i ir m v r r Third term: ∑ ′× ′ is the vector sum of angular momentum of all particles with respect to the center of mass. §10.3 The angular momentum of a system of particles and moment of inertia of rigid body Lorbital Lspin L r Mv r m vi i i CM CM i r r r r r r r = + = × +∑ ′× ′ then 2. Spin angular momentum of a rigid body about a axis through the center of mass i v ′ r o ω r z o′ mi ir′ r ⊥ ′ ir r i // r′ r i i i i L r m v r r r Q = ∑ ′× ′ spin ⊥ ⊥ ′ = × ′ ′ = ′ + ′ i i i i i v r r r r r r r r r r ω // ( ) ( ) spin // ⊥ ⊥ ∴ =∑ ′ + ′ × × ′ i i i i i L m r r r r r r r r ω §10.3 The angular momentum of a system of particles and moment of inertia of rigid body
810.3 The angular momentum of a system of particles and moment of inertia of rigid body ∑m而X(x) +∑m×(xh) n..uol iL +∑mh The vector is an involved vector summation. The rotation of an oddly shaped object about any axis of rotation is beyond the scope of this course 8 10.3 The angular momentum of a system of particles and moment of inertia of rigid body 3. The moment of inertia or rotational inertia of a rigid body about a fixed axis through 乡 center of mass ∑mhX(x)=∑mrO1 O fOthe rigid body is symmetry r about the axis; ② the axis is fixed. This term has no effect. Then ∑ (a×r1) ∑m 8
8 §10.3 The angular momentum of a system of particles and moment of inertia of rigid body ∑ ∑ ∑ ∑ ⊥ ⊥ ⊥ ⊥ ⊥ + ′ = − ′ + ′ × × ′ = ′ × × ′ i i i i i i i i i i i i i i i m r m r r m r r L m r r ω ω ω ω r r r r r r r r r 2 // spin // ( ) ( ) i v ′ r o ω r z o′ mi ir′ r ⊥ ′ ir r i // r′ r The vector is an involved vector summation. The rotation of an oddly shaped object about any axis of rotation is beyond the scope of this course. i v ′ r o ω z r o′ mi ir′ r ⊥ ′ ir r i // r′ r 3. The moment of inertia or rotational inertia of a rigid body about a fixed axis through center of mass ∑ ⊥ ∑ ⊥ ′ × × ′ = − ′ i i i i i i i i m r r m r r r r r r // (ω ) ω If 1the rigid body is symmetry about the axis; 2the axis is fixed. This term has no effect. Then ∑ ∑ ⊥ ⊥ ⊥ = ′ = ′ × × ′ i i i i i i i m r L m r r ω ω r r r r r 2 spin ( ) §10.3 The angular momentum of a system of particles and moment of inertia of rigid body
810.3 The angular momentum of a system of particles and moment of inertia of rigid body Define:cw=∑mh This is the moment of inertia or rotational inertia of a rigid body about a fixed axis through center of mass The angular momentum 0 MO=C∑m s10.3 The angular momentum of a system of particles and moment of f inerti a of rigid body 4. The moment of inertia of various rigid bodies (a) point particle r-a distance from the axis of rotation (b) Collection of point particles I=∑ ri1 --the perpendicular distance of each mass m from the axis of rotation (c) Rigid body of distributive mass 几L d
9 i v ′ r o ω z r o′ mi ir′ r ⊥ ′ ir r i // r′ r ∑ ⊥ = ′ i CM i i I m r 2 Define: This is the moment of inertia or rotational inertia of a rigid body about a fixed axis through center of mass The spin angular momentum of a rigid body ω ω r r r ( ) 2 spin = = ∑ ⊥ i CM i i L I m r §10.3 The angular momentum of a system of particles and moment of inertia of rigid body 4. The moment of inertia of various rigid bodies (a) Point particle 2 I = mr r –a distance from the axis of rotation (b) Collection of point particles = ∑ ⊥ i i I m ri 2 --the perpendicular distance of each mass mi from the axis of rotation ri⊥ §10.3 The angular momentum of a system of particles and moment of inertia of rigid body (c) Rigid body of distributive mass I r dm2 = ∫ ⊥
810.3 The angular momentum of a system of particles and moment of inertia of rigid body -the perpendicular distance of each mass dm from the axis of rotation The element of mass: adl linear density: a dm= ds surface density: o pdv volum density P 8 10.3 The angular momentum of a system of particles and moment of inertia of rigid body Examplel: 5 particles are connected by 4 light staffs as shown in figure. Find the moment of the system with respect to the axis through point A, and perpendicular to the paper plane. Solution:I=∑m ●4 I=2m12+3m(2) n +(4m+5m)(√2l) =32ml 5m 10
10 --the perpendicular distance of each mass dm from the axis of rotation r⊥ dm = λdl linear density：λ σdS surface density：σ ρdV volum density：ρ The element of mass: §10.3 The angular momentum of a system of particles and moment of inertia of rigid body l l l l A m 2m 3m 4m 5m 2 2 2 2 32 (4 5 )( 2 ) 2 3 (2 ) ml m m l I ml m l = + + = + Example1: 5 particles are connected by 4 light staffs as shown in figure. Find the moment of the system with respect to the axis through point A, and perpendicular to the paper plane. Solution: 2 = ∑ i⊥ i i I m r §10.3 The angular momentum of a system of particles and moment of inertia of rigid body
