Wednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #12Monday, Oct. 23, 2002Dr. JaehoonYu1. Rocket Propulsion2. Fundamentals on Rotation3. Rotational Kinematics4. Relationship Between Angular and Linear QuantitiesToday’s homework is homework #13 due 12:00pm, Wednesday, Nov. 6!!Wednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Announcements• 2ndTerm exam – Wednesday, Oct. 30, in the class– Covers chapters 6 – 10• Magda Cortez, David Hunt and Dhumil Patel, please come and see me after the classWednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3Motion 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 thetotal mass M is preserved, the velocity, total momentum, acceleration of the system areVelocity of the systemCMvTotal Momentum of the systemCMpAcceleration of the systemCMaExternal force exerting on the systemextF∑If net external force is 00=∑extFSystem’s momentum is conserved.What about the internal forces?dtrdCM==∑iirmMdtd 1∑=dtrdmMii1Mvmii∑=CMvM=MvmMii∑=∑= iivmtotipp ==∑dtvdCM==∑iivmMdtd 1∑=dtvdmMii1Mamii∑=CMaM=∑= iiamdtpdtot=dtpdtot=const=totpWednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Rocket PropulsionWhat is the biggest difference between ordinary vehicles and a rocket?Since there is no road to push against, rockets obtain propulsion from momentum conservation in the system consists of the rocket and gas from burnt fuel. ()vmMpi∆+=Initial momentum before burning fuelThe force that gives propulsion for normal vehicles is the friction between the surface of the road and the tire. The system in this case consists of the tire and the surface of the road. Newton’s 3rdlaw and the momentum conservation of an isolated system.M+∆mv()()gfvvmvvMp −∆+∆+=Final momentum after burning fuel and ejecting the gasMv+ ∆v∆mv−vgFrom momentum conservation()()gvvmvvM −∆+∆+Since dm is the same as –dM, one can obtain dvThrust is the force exerted on the rocket by the ejected gasThrustmvMv∆+=vM∆gmv∆=Mdmvg=vdfi∫ifvv −=∫=−=figMdMvfigMMv lndtdvM=dtdMvg=Wednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5Example 9.18Precisely the case we’ve discussed in the previous slide.A rocket moving in free space has a speed of 3.0x103m/s relative to the Earth. Its engines are turned on, and fuel is ejected in a direction opposite the rocket’s motion at a speed of 5.0x103m/s relative to rocket. A) What is the speed of the rocket relative to the Earth once its mass is reduced to one-half the mass before ignition?Find the thrust on the rocket if it burns fuel at the rate of 50kg/s?M+∆mvMv+ ∆v∆mv −vgfvSince the thrust is given proportional to the rate of mass change or the rate the fuel burns as given in the formuladtdMvdtdvMThurstg==Nskgsm75105.2/50/100.5×=××=One can obtain+=figiMMvv ln()sm/105.62ln100.5100.3333×=××+×=dtdMvThurstg=Wednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Fundamentals 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.o360Since the circumference of a circle is 2πr,The relationship between radian and degrees isθrPsOθrs=The arc length, or sergita, isrs=θTherefore the angle, θ, is . And the unit of the angle is in radian.rad 1rr /2π=π2=π2/360o= π/180o=Wednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Using what we have learned in the previous slide, how would you define the angular displacement?ifθθθ −=∆Angular Displacement, Velocity, and AccelerationHow about the average angular speed?tttifif∆∆=−−≡θθθωAnd the instantaneous angular speed?dtdttθθω =∆∆≡→∆lim0By the same token, the average angular accelerationtttifif∆∆=−−≡ωωωαAnd the instantaneous angular acceleration?dtdttωωα =∆∆≡→∆lim0When 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.θiθfWednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Rotational 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 acceleration, because these are the simplest motions in both cases.tifαωω +=Just like the case in linear motion, one can obtainAngular Speed under constant angular acceleration:Angular displacement under constant angular acceleration:221ttiifαωθθ ++=One can also obtain ()ififθθαωω −+= 222Wednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Example 10.1A wheel rotates with a constant angular acceleration pf 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 getsWhat 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.rad0.112=θ221tt αω +=( )200.250.32100.200.2 ×+×=rad0.11=.75.1.20.11revrev==πsrad/00.900.250.300.2=×+=( )rad8.2100.350.32100.300.223=×+×=θθ∆2θθ −=3rad8.10=.72.1.28.10revrev ==πWednesday, Oct. 23, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu10Relationship 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 motion components in its motion. What is the linear component of the motion you see?tvEvery particle (or masslet) in the object moves in a circle centered at the axis of rotation.riPθOxyvtLinear velocity along the tangential direction.How do we related this linear component of the motion with angular component?θrs=The arc-length is So the tangential speed
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