康涅狄格州大学：《普通物理》课程PPT教学课件（英文版）Lecture 19 Today's Agenda

Physics 121: Lecture 19 Today's Agenda Announcements Homework 8: due Friday Nov 11@6: 00 PM Chap.7:#7,22,28,33,35,44,45,50,54,61,and65 Today's topics Kinetic energy rotation Rolling motion Angular momentum Fluid and solids Physics 121: Lecture 19, Pg

Physics 121: Lecture 19, Pg 1 Physics 121: Lecture 19 Today’s Agenda Announcements Homework 8: due Friday Nov. 11 @ 6:00 PM. Chap. 7: # 7, 22, 28, 33, 35, 44, 45, 50, 54, 61, and 65. Today’s topics Kinetic energy & rotation Rolling motion Angular momentum Fluid and solids

Summary (with comparison to 1-D kinematics) Angular Linear a= constant a= constant V=Vo +at 0=00+Oot+ al And for a point at a distance R from the rotation axis X=RO V=OR a=ar Physics 121: Lecture 19, Pg 2

Physics 121: Lecture 19, Pg 2 Summary (with comparison to 1-D kinematics) Angular Linear  = constant  = 0 +0 +  1 2 2 t t a = constant v = v + at 0 x = x + v t + at 0 0 1 2 2 And for a point at a distance R from the rotation axis: x = R v = R a = R

Rotation Kinetic Energy ■■■ The kinetic energy of a rotating system looks similar to that of a point particle Point Particle Rotating System v is"linear velocity o is angular velocity m is the mass I is the moment of inertia about the rotation axis mi ri Physics 121: Lecture 19, Pg 3

Physics 121: Lecture 19, Pg 3 Rotation & Kinetic Energy... The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System I = m r i i i 2 v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis

Rolling Motion Now consider a cylinder rolling at a constant speed CM CM The cylinder is rotating about CM and its CM is moving at constant speed (VcM). Thus its total kinetic energy is given by OT 2 CMO+o NANCM Physics 121: Lecture 19, Pg 4

Physics 121: Lecture 19, Pg 4 Rolling Motion Now consider a cylinder rolling at a constant speed. KTOT = CM + MVCM 1 2 1 2 2 2 I  VCM CM The cylinder is rotating about CM and its CM is moving at constant speed (vCM). Thus its total kinetic energy is given by :

Rolling Motion Consider again a cylinder rolling at a constant speed CM CM At any instant the cylinder is rotating about point P. Its kinetic energy is given by its rotational energy about that point TOt 1/2 P Physics 121: Lecture 19, Pg 5

Physics 121: Lecture 19, Pg 5 Rolling Motion Consider again a cylinder rolling at a constant speed. VCM P Q CM  At any instant the cylinder is rotating about point P. Its kinetic energy is given by its rotational energy about that point. KTOT = 1/2 IP 2

Rolling Motion We can find Ip using the parallel axis theorem CM CM Ip=ICM+ MR2 Ko=1/2(cM+MR2)2 Kor=12lcM02+12M(R202)=1/2lcM02+1/2MvM2! TOT CM0+M∥ Physics 121: Lecture 19, Pg 6

Physics 121: Lecture 19, Pg 6 Rolling Motion We can find IP using the parallel axis theorem KTOT = CM + MVCM 1 2 1 2 2 2 I  VCM P Q CM  IP = ICM + MR2 KTOT = 1/2 (ICM + MR2 ) 2 KTOT = 1/2 ICM 2 + 1/2 M (R22 ) = 1/2 ICM 2 + 1/2 M vCM 2 !

Rolling Motion Cylinders of different /rolling down an inclined plane 0 △K=-AU=Mgh R 0 M K=O TOT CM +-M CM V=0 Physics 121: Lecture 19, Pg7

Physics 121: Lecture 19, Pg 7 Rolling Motion Cylinders of different I rolling down an inclined plane: h v = 0  = 0 K = 0 R K = - U = Mgh v = R M KTOT = CM + MVCM 1 2 1 2 2 2 I 

Rolling If there is no slipping(due to friction 2V Where= OR In the lab reference frame In the cm reference frame Physics 121: Lecture 19, Pg 8

Physics 121: Lecture 19, Pg 8 Rolling... If there is no slipping (due to friction): v 2v In the lab reference frame v In the CM reference frame v Where v = R 

Rolling Kot =lIcmo2+lMvM use y= or and /=cMR2, hoop: C disk C=1/2 sphere: C=2/5 C etc So C+1Mv2= Mgh V=√2qh C+1 The rolling speed is always lower than in the case of simple Sliding since the kinetic energy is shared between CM motion and rotation Physics 121: Lecture 19, Pg 9

Physics 121: Lecture 19, Pg 9 Rolling... Use v= R and I = cMR2 . So: ( 1)Mv Mgh 2 1 2 + = 1 1 v 2gh + = The rolling speed is always lower than in the case of simple Sliding since the kinetic energy is shared between CM motion and rotation. hoop: c=1 disk: c=1/2 sphere: c=2/5 etc... c c c c KTOT = CM + MVCM 1 2 1 2 2 2 I 

Example: Rolling Motion A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane Ball has radius r 0 M Physics 121: Lecture 19, Pg 10

Physics 121: Lecture 19, Pg 10 Example : Rolling Motion A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? M  h M v ? Ball has radius R