Physics 121 Newtonian Mechanics Instructor: Karine Chesnel Feb 26 th, 2009 Review for Exam 2 Class website:

Mid-term exam 2 Fri Feb 27 through Tuesday Mar 3 At the testing center : 8 am – 9 pm Closed Book and closed Notes Only bring:- Pen / pencil - Calculator - Math reference sheet - dictionary (international) - your CID No time limit (typically 3 hours)

Midterm exam 2 Review: ch 5 – ch 8 Ch. 6 Newton’s laws applications Circular Motion Drag forces and viscosity Friction Fictitious forces Ch. 8 Conservation of Energy Mechanical energy Conservation of energy Ch. 7 Work and energy Work Kinetic energy Potential energy Work- kinetic energy theorem Ch. 5 The Laws of Motion Newton’s first law Newton’s second law Newton’s third law

Summary of the Laws of Motions Third Law: Action and reaction If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1. First Law: Principle of Inertia In a inertial frame, an isolated system remains at constant velocity or at rest Second Law: Forces and motion In an inertial frame the acceleration of a system is equal to the sum of all external forces divided by the system mass Ch.5 Laws of motion2/26/09 F1F1 F2F2

Review of basic forces The weight Object of mass m FgFg Normal reaction When two objects are in contact - N N The reaction exerted by the support on the object is NORMAL to the surface Force of tension The spring force tends to bring the object back to rest F Spring force F = - k x î 0 x T The tension exerted by a rope on the object is ALONG the direction of the rope Ch.5 Laws of motion2/26/09

` Forces of Friction Two regimes mgmg R F f If one applies a force F at which point the system starts to move? If F is smaller than a maximum value f max then the system does not move If F is larger than the maximum value f max the system starts to move and the friction is constant Static regimeKinetic regime Static regime Dynamic regime Ch.5 Friction2/26/09

Friction Summary of two regimes Static regime Kinetic regime In the static regime, the magnitude of the friction is equal to the force pushing the object When the system is on the verge to move Static coefficient of friction Once the system is moving Kinetic coefficient of friction Ch.5 Friction2/26/09

Resistive forces Low speed regime V F The equation of the motion is given by and V f = mg/b is the terminal speed Where  = m/b the time constant The motion starts at t = 0 with no initial speed The speed increases to reach the limit V f When t =  the speed value is  V = (1-1/e) V f ~ 0.63 V f V(t) VfVf t  (1-1/e)V f 0 Ch.6 Special applications of Newton’s law2/26/09

Ch.5&6 Laws of motion2/26/09 Define a frame of work that suits with the situation: either Cartesian coordinates (x, y) or polar coordinates ( ,  ) List all the forces applied on the system, for example: - the weight mg - the normal reaction of a support N - a force of tension T - a force of friction f … etc 2. List the forces 3. Apply Newton’s law General method To solve a given problem: 1. Define system Define the object you will consider and identify its mass m m 4. Define a frame and project Project the Newton’s law along each axis separately. Be careful with the SIGN!!

Newton’s second law Pitfalls to avoid This is an ABSOLUTE equation (vectors). Projection along specific axis The projection is not an absolute equation: the sign depends on your choice of axis orientation. Be CONSISTENT with your choice of axis! Vectorial equation Example m T mg Axis choice m T mg z Choice 1 m T mg z Choice 2 Ch.5 Laws of motion2/26/09

Ch.5&6 Laws of motion2/26/09 Newton’s second law Pitfalls to avoid This is a VECTORIAL equation What if forces are in different directions? Be careful: do not mix forces in different directions!! Examples m T mg  Take into account the direction, possibly by using inclination angles (  ). Project Newton’s law along each axis separately  H mg R 

Tangential and radial acceleration General case V1V1 V2V2 V3V3 a a a V is tangential to the trajectory Tangential acceleration The sign tells if the particle speeds up or slows down a t = dV/dt Centripetal acceleration The centripetal acceleration is toward the center of curvature a c = R  2 = V 2 /R Ch.6 Motion2/26/09

Problem (Attwood machine) m1m1 m2m2 Two objects of different mass are suspended at each end of a string with a frictionless pulley Will the system move? If so, in which direction and with what acceleration? Let’s apply Newton’s Law on each object: Object 1: m 1 a 1 = T 1 + m 1 g Object 2: m 2 a 2 = T 2 + m 2 g z Let’s project these equations along z axis Knowing that T 1 = T 2 and that a 2 = - a 1 we get m 1 a 1 = T 1 - m 1 g m 2 a 2 = T 2 - m 2 g So T 1 = m 1 a 1 + m 1 g T 2 = m 2 a 2 + m 2 g m 1 a 1 + m 1 g = -m 2 a 1 + m 2 g (m 2 - m1) g = (m 2 + m1) a 1 Ch.5&6 Newton’s law2/26/09 T1T1 T2T2 m1gm1g m2gm2g

m1m1 m2m2 Two objects of different mass are suspended at each end of a string with a frictionless pulley Will the system move? If so, in which direction and with what acceleration? z We have T 1 = T 2 and a 2 = - a 1 T1T1 T2T2 m1gm1g m2gm2g If m 2 > m 1 : then a 1 > 0 the red sphere moves down and green cube moves up If m 2 < m 1 : then a 1 < 0 the red sphere moves up and green cube moves down Problem (Attwood machine) Ch.5 Laws of motion2/26/09

Work of a force A particle moves under the action of a force F from initial point A to final point B dr F A B The total work done by the force F on the particle from point A to point B is Ch.7 Work and energy2/26/09 F A B dr If at any time along the path, the force F is perpendicular to the displacement, then: F dr A B  If the force is constant and working along a straight line

Conservative force A force is conservative when: its work does not depend on the path. The force conserves the energy Examples of conservative forces: - Gravity - Elastic force - Gravitational field - Electric force - Magnetic force - any constant force Path independence A The work done by a conservative force on a closed path is zero For conservative forces, we can express the work in terms of potential energy Ch.7 Work and energy2/26/09

Gravity potential energy We can express the work of the weight as a variation of a potential function Ep mgmg B H A Ch.7 Work and energy2/26/09

Elastic potential energy L0L0 x 0 F x F A B We can express the work of a spring force as a variation of the elastic potential Ep Ch.7 Work and energy2/26/09

Work and kinetic energy Defining the kinetic energy Using Newton’s second law Work- Kinetic energy theorem Ch.7 Work and energy2/26/09

Mechanical energy We define the mechanical energy E mech as the sum of kinetic and potential energies Ch.7 Work and energy2/26/09

Closed System with conservative forces only F cons There are no non-conservative forces working The mechanical energy is constant The mechanical energy is conserved between initial and final points Ch.7 Work and energy2/26/09

Read Textbook: Chapter 9 Next Class Homework assignment: Today Feb 26 th 7pm Problems 8: 5-7 Good luck on You exam!! Tuesday March 3 rd