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