Thursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu1PHYS 1443 – Section 001Lecture #13Thursday, June 22, 2006Dr. Jaehoon Yu• CM and the Center of Gravity• Fundamentals on Rotational Motion• Rotational Kinematics• Relationship between angular and linear quantities• Rolling Motion of a Rigid Body• Torque• Torque and Vector Product• Moment of InertiaThursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu2Announcements• Reading assignments– CH. 11.6, 11.8, 11.9 and 11.10• Last quiz next Wednesday– Early in the class– Covers Ch. 10 – what we cover next TuesdayThursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu3The net effect of these small gravitational forces is equivalent to a single force acting on a point (Center of Gravity) with mass M.Center of Mass and Center of GravityThe center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry, if object’s mass is evenly distributed throughout the body.Center of GravityHow do you think you can determine the CM of objects that are not symmetric?gFG∆miCMAxis of symmetryOne can use gravity to locate CM.1. Hang the object by one point and draw a vertical line following a plum-bob.2. Hang the object by another point and do the same.3. The point where the two lines meet is the CM. ∆migSince a rigid object can be considered as a collection of small masses, one can see the total gravitational force exerted on the object as What does this equation tell you?iiF=∑Giimg=∆∑GMg=GThe CoG is the point in an object as if all the gravitational force is acting on!Thursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu4Motion of a Group of ParticlesWe’ve learned that the CM of a system can represent the motion of a system. Therefore, for an isolated system of many particles in which the total mass M is preserved, the velocity, total momentum, acceleration of the system areVelocity of the systemCMvGTotal Momentum of the systemCMpGAcceleration of the systemCMaGExternal force exerting on the systemextF∑GIf net external force is 00extF=∑GSystem’s momentum is conserved.What about the internal forces?CMdrdt=G1iidmrdt M⎛⎞=⎜⎟⎝⎠∑G1iidrmMdt=∑GiimvM=∑GCMMv=GiimvMM=∑Giimv=∑Gto tpp==∑GGCMdvdt=G1iidmvdt M⎛⎞=⎜⎟⎝⎠∑G1iidvmMdt=∑GiimaM=∑GCMMa=Giima=∑Gtotdpdt=Gto tdpdt=Gconsttotp=GThursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu5Therefore the angle, θ, is . And the unit of the angle is in radian. It is dimensionless!!Fundamentals on RotationLinear motions can be described as the motion of the center of mass with all the mass of the object concentrated on it. Is this still true for rotational motions?No, because different parts of the object have different linear velocities and accelerations.Consider a motion of a rigid body – an object that does not change its shape – rotating about the axis protruding out of the slide. One radian is the angle swept by an arc length equal to the radius of the arc.D360Since the circumference of a circle is 2πr,The relationship between radian and degrees isl =The arc length islRθ=rad 1rr /2π=π2=π2/360D=π/180D=57.3≅D180 3.14≅DRθThursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu6Example 10 – 1 A particular bird’s eyes can barely distinguish objects that subtend an angle no smaller than about 3x10-4rad. (a) How many degrees is this? (b) How small an object can the bird just distinguish when flying at a height of 100m? (a) One radian is 360o/2π. Thus4310rad−×l=()4310rad−=××()360 2 radπD0.017=D(b) Since l=rθ and for small angle arc length is approximately the same as the chord length.rθ=4100 3 10mrad−××=2310 3mcm−×=Thursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu7=fωRotational KinematicsThe first type of motion we have learned in linear kinematics was under a constant acceleration. We will learn about the rotational motion under constant angular acceleration, because these are the simplest motions in both cases.Just like the case in linear motion, one can obtainAngular Speed under constant angular acceleration:Angular displacement under constant angular acceleration:=fθOne can also obtain =2fωiω+iθ+()ifiθθαω−+ 22tαitω+212tαThursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu8Using what we have learned in the previous slide, how would you define the angular displacement?=∆θAngular Displacement, Velocity, and AccelerationHow about the average angular speed?≡ωAnd the instantaneous angular speed?≡ωBy the same token, the average angular acceleration≡αAnd the instantaneous angular acceleration?≡αWhen rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and angular acceleration.θiθfifθθ−=−−ififttθθt∆∆θ=∆∆→∆ttθlim0dtdθ=−−ififttωωt∆∆ω=∆∆→∆ttωlim0dtdωUnit? rad/sUnit? rad/sUnit? rad/s2Unit? rad/s2Thursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu9Example for Rotational KinematicsA wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what angle does the wheel rotate in 2.00s?ifθθ−Using the angular displacement formula in the previous slide, one getstω=()200.250.32100.200.2 ×+×=rad0.11=.75.1.20.11revrev ==π212tα+Thursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu10Example for Rotational Kinematics cnt’dWhat is the angular speed at t=2.00s?tifαωω+=Using the angular speed and acceleration relationshipFind the angle through which the wheel rotates between t=2.00 s and t=3.00 s.2tθ==srad/00.900.250.300.2=×+=3tθ==θ∆2θθ−=3rad8.10=.72.1.28.10revrev ==πifθθ−221ttαω+=Using the angular kinematic formulaAt t=2.00sAt t=3.00sAngular displacement2.00 2.00×13.50 2.002+×11.0rad=2.00 3.00×()213.50 3.002+×21.8rad=Thursday, June 22, 2006 PHYS 1443-001, Summer 2006Dr. Jaehoon Yu11Relationship Between Angular and Linear QuantitiesWhat do we know about a rigid object that rotates about a fixed axis of rotation?When a point rotates, it has both the linear and angular components in its motion. What is the linear component of the motion you see?vEvery particle (or masslet) in the object moves in a circle centered at the axis of rotation.Linear velocity along the tangential direction.How do we related this linear
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