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UT Arlington PHYS 1443 - Motion with Resistive Forces

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1443-501 Spring 2002 Lecture #6Resistive Force Proportional to SpeedExample 6.11Air DragExample 6.4Analytical Method vs Numerical MethodThe Euler MethodUnit Conversion: Example 1.4Example 2.1Example 2.2Kinetic Equation of Motion in a Straight Line Under Constant AccelerationExample 2.12Example 3.12-dim Motion Under Constant AccelerationExample 4.1Example 4.5Uniform Circular MotionRelative Velocity and AccelerationExample 4.9ForceNewton’s LawsExample 5.1Example 5.4Example 5.12Example 6.8Example 6.91443-501 Spring 2002Lecture #6Dr. Jaehoon Yu1. Motion with Resistive Forces2. Resistive Force in Air at High Speed 3. Analytical and Numerical Methods4. Review Examples1st term exam on Monday Feb. 11, 2002, at 5:30pm, in the classroom!! Will cover chapters 1-6!!Will cover chapters 1-6!!Physics Clinic Hours Extened!!Physics Clinic Hours Extened!!Mon. – Thu. till 6pmMon. – Thu. till 6pmThursday: 6-9pm by appointments Thursday: 6-9pm by appointments ((e-mail to Andre Litvin [email protected]@yahoo.comMustafa [email protected]@hotmail.com ) )Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #62Resistive Force Proportional to Speed Since the resistive force is proportional to speed, we can write R=bvmLet’s consider that a ball of mass m is falling through a liquid.RmgvvmbgdtdvdtdvmmabvmgFFRFFyxg ;0This equation also tells you that0 when ,  vgvmbgdtdvThe above equation also tells us that as time goes on the speed increases and the acceleration decreases, eventually reaching 0. An object moving in a viscous medium will obtain speed to a certain speed (terminal speed) and then maintain the same speed without any more acceleration.What does this mean?What is the terminal speed in above case?bmgvvmbgdtdvt ;0How do the speed and acceleration depend on time?vmbgembbmgembbmgdtdvtgageembbmgdtdvatvebmgvtttmbtmbt11;0 when ;;0 when 0 ;1The time needed to reach 63.2% of the terminal speed is defined as the time constant, m/b.Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #63Example 6.11A small ball of mass 2.00g is released from rest in a large vessel filled with oil, where it experiences a resistive force proportional to its speed. The ball reaches a terminal speed of 5.00 cm/s. Determine the time constant  and the time it takes the ball to reach 90% of its terminal speed. mRmgvsskgkgbmskgsmsmkgvmgbbmgvtt332231010.5/392.01000.2/392.0/1000.5/80.91000.2Determine the time constant . Determine the time it takes the ball to reach 90% of its terminal speed.  msteeevvevebmgvtttttttt7.111010.530.230.21.0ln1.0 ;9.0119.0113Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #64Air Drag For objects moving through air at high speed, the resistive force is roughly proportional to the square of the speed.The magnitude of such resistive force221AvDRD: Drag Coefficient (dim.less): Density of Air A: Cross section of the objectLet’s analyze a falling object through the airmRmgvmgAvDmaFyy221gvmADay22ADmgvgvmADaty2 ;022ForceAccelerationTerminal SpeedFeb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #65Example 6.4A pitcher hurls a 0.145 kg baseball past a batter at 40.2m/s (=90 mi/h). Determine the drag coefficients of the air, density of air is 1.29kg/m3, the radius of the ball is 3.70cm, and the terminal velocity of the ball in the air is 43.0 m/s.Find the resistive force acting on the ball at this speed.    277.00.431030.429.180.9145.0221030.40370.02322322tAvmgDmrAUsing the formula for terminal speed NAvDR 24.12.401030.429.1277.02121232Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #66Analytical Method vs Numerical MethodAnalytical Method:The method of solving problems using cause and effect relationship and mathematical expressionsWhen solving for a motion of an object we follow the procedure:1. Find all the forces involved in the motion2. Compute the net force3. Compute the acceleration4. Integrate the acceleration in time to obtain velocity5. Integrate the velocity in time to obtain positionBut not all problems are analytically solvable due to complex conditions applied to the given motion, e.g. position dependent acceleration, etc.Numerical Method:The method of solving problems using approximation and computational tools, such as computer programs, stepping through infinitesimal intervals  Integration by numbers…Feb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #67The Euler MethodSimplest Numerical Method for solving differential equations, by approximating the derivatives as ratios of finite differences.Speed and the magnitude of Acceleration can be approximated in a small increment of time, t    ttvttvtvta     ttatvttv Thus the Speed at time t+t isIn the same manner, the position of an object and speed are approximated in a small time increment, t    ttxttxtxtvAnd the position at time t+t is              021022 ttvtxttxttattvtxttxtThe (t)2 is ignored in Euler method because it is closed to 0.We then use the approximated expressions to compute position at infinitesimal time interval, t, with computer to find the position of an object at any given timeFeb. 6, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #68Unit Conversion: Example 1.4•US and UK still use British Engineering units: foot, lbs, and seconds–1.0 in= 2.54 cm, 1ft=0.3048m=30.48cm–1m=39.37in=3.281ft~1yd, 1mi=1609m=1.609km–1lb=0.4535kg=453.5g, 1oz=28.35g=0.02835kg–Online unit converter: http://www.digitaldutch.com/unitconverter/•Example 1.4: Determine density in basic SI units (m,kg


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UT Arlington PHYS 1443 - Motion with Resistive Forces

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