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PHYS 1443 – Section 003 Lecture #7 Monday, Oct. 7, 2002 Dr. Jaehoon Yu 1. 2. 3. 4. 5. 6. Numerical Modeling in Particle Dynamics (Euler Method) Work & Scalar product of vectors Kinectic Energy Power Potential Energy • Gravitational Potential Energy • Elastic Potential Energy Conservative Forces and Mechanical Energy Conservation Today’s homework is homework #8, due 12:00pm, next Monday!! Monday, Oct. 7, 2002 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 1 Announcements • Term Exam – Grading is completed • Maximum Score: 85 • Numerical Average: 51.7 Very good!!! • One person missed the exam without a prior approval – Can look at your exam after the class – All scores are relative based on the curve • One worst after the adjustment will be dropped – Exam constitutes only 50% of the total • Do your homework well • Come to the class and do well with quizzes Monday, Oct. 7, 2002 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 2 First Term Exam Distributions Gaussian Mean: 51 This is what I will use to adjust term exam scores Monday, Oct. 7, 2002 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 3 Resistive Force Proportional to Speed Since the resistive force is proportional to speed, we can write R=bv Let’s consider that a ball of mass m is falling through a liquid. R v m F F mg g F R x Fy mg bv ma m dv dt 0 dv b g v dt m This equation also tells you that dv b g v g , when v 0 dt m The above equation also tells us that as time goes on the speed increases and the acceleration decreases, eventually reaching 0. What does this mean? 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 is the terminal speed in above case? How do the speed and acceleration depend on time? dv b mg g v 0; vt dt m b Monday, Oct. 7, 2002 mg bt 1 e m ; v 0 when t 0; b dv mg b bt m t a e ge t ; a g when t 0; dt b m dv mg b t t mg b b t e 1 1 e t g v dt b m b m m v PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu The time needed to reach 63.2% of the terminal speed is defined as the time constant, tm/b. 4 Example 6.11 A 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 t and the time it takes the ball to reach 90% of its terminal speed. vt R v m Determine the time constant t. mg Determine the time it takes the ball to reach 90% of its terminal speed. mg b mg 2.00 10 3 kg 9.80m / s 2 b 0.392kg / s vt 5.00 10 2 m / s m 2.00 103 kg t 5.10 103 s b 0.392kg / s v mg t t 1 e b v 1 e t t t t 0.9vt vt 1 e t 1 e t t 0.9; e t t 0.1 t t ln 0.1 2.30t 2.30 5.10 103 11.7ms Monday, Oct. 7, 2002 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 5 Numerical Modeling in Particle Dynamics • The method we have been using to solve for particle dynamics is called Analytical Method Solve motions using differential equations – – – – • Use Newton’s second law for net force in the motion Use net force to determine acceleration, a=SF/m Use the acceleration to determine velocity, dv/dt=a Use the velocity to determine position, dx/dt=v Some motions are too complicated to solve analytically Numeric method or Euler method to describe the motion – Differential equations are divided in small increments of time or position – Acceleration is determined from net force: a ( x, v , t ) F ( x , v, t ) m – Determine velocity using a(x,v,t): v( x, t ) v( x, t t ) a( x, v, t )t – Determine position using a and v: xt x(t t ) v( x, t )t Compute the quantities at every small increments of time t and plot position, velocity, or acceleration as a function of time to describe the motion. Monday, Oct. 7, 2002 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 6 Work Done by a Constant Force Work in physics is done only when a sum of forces exerted on an object made a motion to the object. F M y q FN M Free Body Diagram F q d x FG M g Which force did the work? Force F How much work did it do? W F d Fd cosq What does this mean? Monday, Oct. 7, 2002 Unit? N m J (for Joule) Physical work is done only by the component of of the force along the movement of the object. PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu Work is energy transfer!! 7 Example 7.1 A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F=50.0N at an angle of 30.0o with East. Calculate the work done by the force on the vacuum cleaner as the vacuum cleaner is displaced by 3.00m to East. F M 30o M W F d F d cosq W 50.0 3.00 cos 30 130 J d Does work depend on mass of the object being worked on? Why don’t I see the mass term in the work at all then? Monday, Oct. 7, 2002 Yes It is reflected in the force. If the object has smaller mass, its would take less force to move it the same distance as the heavier object. So it would take less work. Which makes perfect sense, doesn’t it? PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 8 Scalar Product of Two Vectors • Product of magnitude of the two vectors and the cosine of the angle between them A B A B cosq • Operation is commutative A B A B cosq B A cosq B A • Operation follows distribution A B C A B AC law of multiplication • Scalar products of Unit Vectors i i j j k k 1 i j j k k i 0 • How does scalar product look in terms of components? A Ax i Ay j Az k B Bx i By j Bz k A B Ax i Ay j Az k Bx i By j Bz k A B Ax Bx Ay B y Az Bz Monday, Oct. 7, 2002 C Cx i C y j Cz k A B Ax Bx i i Ay B y j j Az Bz k k Angle between the vectors PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu cos q Ax Bx Ay B y Az Bz AB 9 Example 7.3 A particle moving in the xy plane undergoes a displacement d=(2.0i+3.0j)m as a constant force F=(5.0i+2.0j) N acts on the particle. a) Calculate the magnitude of the displacement and that of the force. Y d F X d d x2 d y2 2.0 3.0 3.6m 2 F Fx2 Fy2 2 5.02 2.02 5.4 N b) Calculate the work done by the force F. W F d 2.0 5.0 i i 3.0 2.0 j j 10 6 16( J ) 2.0 i 3.0 j 5.0 i 2.0 j Can you do this using the magnitudes and the angle between d and F? W F d F d cosq Monday, Oct. 7, 2002 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 10 Work Done by Varying Force • If the force depends on position of particle during the motion – one must consider work segment in small segment of the position where the force can be considered constant W Fx x – Then add them all up throughout the entire motion (xi xf) xf W Fx x xi In the limit where x0 xf lim x 0 Fx x Fx dx W xf xi xi – If more than one force is acting, the net work is done by the net force W (net ) xf xi F dx ix One of the forces depends on position is force by a spring The work done by the spring force is W 0 xmax Monday, Oct. 7, 2002 Fs dx 0 xmax PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu Fs kx kxdx 1 kx2 2 11 Kinetic Energy and Work-Kinetic Energy Theorem • Some problems are hard to solve using Newton’s second law – If forces exerting on the object during the motion are so complicated – Relate the work done on the object by the net force to the change of the speed of the object M SF vi M vf Work The work on the object by the net force SF is W d Displacement Suppose net force SF was exerted on an object for displacement d to increase its speed from vi to vf. d F d mad cos 0 mad 1 v f vi t 2 Acceleration v f vi 1 1 1 v f vi t mv2f mvi2 W mad m 2 2 t 2 a v f vi t Kinetic KE 1 mv2 2 Energy 1 2 1 2 W mv f mvi KE f KEi KE The work done by the net force caused Work change of object’s kinetic energy. 2 2 Monday, Oct. 7, 2002 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 12 Example 7.7 A 6.0kg block initially at rest is pulled to East along a horizontal, frictionless surface by a constant horizontal force of 12N. Find the speed of the block after it has moved 3.0m. M F M vi=0 vf Work done by the force F is W F d F d cosq 12 3.0 cos 0 36J d From the work-kinetic energy theorem, we know Since initial speed is 0, the above equation becomes Solving the equation for vf, we obtain Monday, Oct. 7, 2002 1 1 W mv2f mvi2 2 2 1 2 W mv f 2 2W 2 36 vf 3.5m / s m 6.0 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu 13 Work and Energy Involving Kinetic Friction • Some How do you think the work looks like if there is friction? – Why doesn’t static friction matter? Ffr M M vi vf d Friction force Ffr works on the object to slow down The work on the object by the friction Ffr is W fr F fr d cos180 F fr d KE Ffr d The final kinetic energy of an object with initial kinetic energy, friction force and other source of work is KE f KEi W Ffr d t=0, KEi Monday, Oct. 7, 2002 PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu Friction Engine work t=T, KEf 14 Example 7.8 A 6.0kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction mk=0.15 by a constant horizontal force of 12N. Find the speed of the block after it has moved 3.0m. Fk M F vi=0 M WF F d F d cosq 12 3.0 cos 0 36J vf d=3.0m Work done by friction Fk is Thus the total work is Work done by the force F is Wk Fk d m k mg d cosq 0.15 6.0 9.8 3.0 cos180 26J W WF Wk 36 26 10( J ) Using work-kinetic energy theorem and the fact that initial speed is 0, we obtain 1 W WF Wk mv2f 2 Monday, Oct. 7, 2002 Solving the equation for vf, we obtain PHYS 1443-003, Fall 2002 Dr. Jaehoon Yu vf 2W 2 10 1.8m / s m 6.0 15