講者： 許永昌 老師 1

Contents Rigid body Center of mass: r CM Rotational Energy Moment of inertia: I Mathematics: cross product Torque Properties Applications Rotation about a fixed axis Static Equilibrium Rolling Motion Angular Momentum Examples 2

Rigid Body ( 請預讀 P340~P345) 3 x y z x y z

Rotational Motion Top view from the tip of the axis of rotation: Right hand rule. Angular velocity: Exercise: 4

Center of Mass Benefits: Exercise: Find the center of mass of this object: 5

Homework Student Workbook: P12-1~P12-2 6

Rotational Energy & Moment of inertia ( 請預讀 P346~P350) 7

Parallel-axis Theorem (I=I CM +MR 2 ) Total kinetic energies of these two cases: 8 R K=? we get I=I CM +MR 2.

Exercises I CM of A cylinder. A thin rod. A sphere. Example 12.5 A 1.0-m-long, 200 g rod is hinged at one end and connected to a wall. It is held out horizontally, then released. What is the speed of the tip of the rod as it hits the wall? 9 v=?

Homework Student Workbook

Cross Product ( 請預讀 P368~P369) Geometrical definition: 11

Cross Product Properties: Anti-commutative: A  B=  B  A. Distributive: A  (  B+  C)=  A  B+  A  C. Based on the right-hand rule, we get: Exercise: If A=(1, 0, 0), B=(1, 1, 0). 12

Torque ( 請預讀 P370, P351~P356) Think of that, since Kinetic energy: Can we define If so, 13 FtFt

Torque Definition:  The torque contributed by F i exerted on particle i is It is dependent of the origin we chose.  i = r i F i sin  =d i F i =r i F i,t. d i : ( 力臂 ) Moment arm. Lever arm. 14 F i,t didi 顯然，在討論旋轉時， particle model 並不適用。 Particle model 適用於合力與動量的部分，而旋轉 則必須知道受力點與 origin 的相對位置。 顯然，在討論旋轉時， particle model 並不適用。 Particle model 適用於合力與動量的部分，而旋轉 則必須知道受力點與 origin 的相對位置。

Torque 15

Action (torque) 16

Is it possible for us to explain the equation r 1 F 1 =r 2 F 2 for a balance by Newton’s 2 nd and 3 rd Law directly? Hint: acceleration constraint & 17 r1r1 r2r2 F2F2 F1F1

Homework Student Workbook: 12.15, 12.17, 12.20, 12.22,

Applications ( 請預讀 P357~P367) 19

Rotation about a fixed axis ( 請預讀 P357~P359) E.g. rope and pulley. Kinematics: Dynamics:. 20 TranslationalRotational velocity vt=rvt=r acceleration at=r.at=r. r

Exercise The acceleration of box m 1. Frictionless. Ignore drag. Massless string. The rope turns on the pulley without slipping. 21 I m1m1 m2m2

Static Equilibrium ( 請預讀 P360~P363) The condition for a rigid body to be in static equilibrium is both Exercise: problem (P381) A 3.0-m-long rigid beam with a mass of 100 kg is supported at each end. An 80 kg student stands 2.0 m from support I. How much upward force does each support exert on the beam? 22 12

Rolling Motion ( 請預讀 P364~P367) 23 P 2R  Inertial reference frame P RR Viewed from its center of mass RR

Exercise The acceleration of this cylinder. I= ½MR 2. No slipping. Friction? 24 

Homework Student Workbook 12.25, 12.29,

Angular Momentum ( 請預讀 P371~P375) Angular Momentum: Properties: For a rigid body 2. I is a tensor in general. 26 riri LiLi

Exercises The change of the diver’s angular velocity when he extend his legs and arms. Rotating bike’s wheel and rotating coins. Questions: The motion of this wheel or coin. Frictionless? Which point you chose to be as the origin, and why? 27 I CM1 I CM2 v=Rv=R

Homework Student Workbook P12-11~P12-12 Textbook 請製作卡片 28

Summary 29 Kinematics Single particleMany particlesRotation Positionriri r CM  i m i r i /M =s/r=s/r Mass (moment of inertia for rotation) mimi MimiMimi Iimiri2.Iimiri2. Velocityv i =dr i /dt v CM  =dr CM /dt  =d  /dt Momentumpi=mivipi=mivi P CM =Mv CM L i =r i  p i L tot =I  (*) Accelerationa i =dv i /dt a CM  =dv CM /dt  =d  /dt Force (torque)F net on i F net on system =F ext Newton’s 2 nd Law Kinetic Energy K i = ½m i v i 2.K tot = ½Mv CM 2 +K micro. K rot = ½ I  2. New