Physics 207: Lecture 15, Pg 1 Lecture 15 Goals: Chapter 11 Chapter 11  Employ the dot product  Employ conservative and non-conservative forces  Use the concept of power (i.e., energy per time) Chapter 12 Chapter 12  Extend the particle model to rigid-bodies  Understand the equilibrium of an extended object.  Understand rigid object rotation about a fixed axis.  Employ “conservation of angular momentum” concept Assignment: l HW7 due March 10 th l For Thursday: Read Chapter 12, Sections 7-11 ” do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia”

Physics 207: Lecture 15, Pg 2 l Useful for finding parallel components A  î = A x î  î = 1 î  ĵ = 0 A  B = (A x )(B x ) + (A y )(B y ) + (A z )(B z ) l Calculation can be made in terms of components. Calculation also in terms of magnitudes and relative angles. Scalar Product (or Dot Product) A  B ≡ | A | | B | cos  You choose the way that works best for you! î A AxAx AyAy  ĵ

Physics 207: Lecture 15, Pg 3 Scalar Product (or Dot Product) Compare: A  B = (A x )(B x ) + (A y )(B y ) + (A z )(B z ) Redefine A  F (force), B   r (displacement) Notice: F   r = (F x )(  x) + (F y )(  z ) + (F z )(  z) So here F   r = W More generally a Force acting over a Distance does Work

Physics 207: Lecture 15, Pg 4 Work in terms of the dot product Ingredients: Force ( F ), displacement (  r ) Looks just like a Dot Product!   r displacement F Work, W, of a constant force F acts through a displacement  r : If the path is curved at each point and

Physics 207: Lecture 15, Pg 5 Remember that a real trajectory implies forces acting on an object l Only tangential forces yield work!  The distance over which F Tang is applied: Work a v path and time a a a = 0 Two possible options: Change in the magnitude of v Change in the direction of v a = 0 a a a a = + a radial a tang a = + l F radial F tang F = +

Physics 207: Lecture 15, Pg 6 Energy and Work Work, W, is the process of energy transfer in which a force component parallel to the path acts over a distance; individually it effects a change in energy of the “system”. 1.K or Kinetic Energy 2.U or Potential Energy (Conservative) and if there are losses (e.g., friction, non-conservative) 3. E Th Thermal Energy Positive W if energy transferred to a system

Physics 207: Lecture 15, Pg 7 A. U  K B. U  E Th C. K  U D. K  E Th E. There is no transformation because energy is conserved. A child slides down a playground slide at constant speed. The energy transformation is

Physics 207: Lecture 15, Pg 8 Exercise Work in the presence of friction and non-contact forces A. 2 B. 3 C. 4 D. 5 A box is pulled up a rough (  > 0) incline by a rope-pulley- weight arrangement as shown below.  How many forces (including non-contact ones) are doing work on the box ?  Of these which are positive and which are negative?  State the system (here, just the box)  Use a Free Body Diagram  Compare force and path v

Physics 207: Lecture 15, Pg 9 Work and Varying Forces (1D) l Consider a varying force F(x) FxFx x xx Area = F x  x F is increasing Here W = F ·  r becomes dW = F x dx F  = 0° Start Finish Work has units of energy and is a scalar! F xx

Physics 207: Lecture 15, Pg 10 How much will the spring compress (i.e.  x = x f - x i ) to bring the box to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (v i ) on frictionless surface as shown below with x i = x eq, the equilibrium position of the spring? xx vivi m titi F spring compressed spring at an equilibrium position V=0 t m Example: Hooke’s Law Spring (x i equilibrium)

Physics 207: Lecture 15, Pg 11 Work signs xx vivi m titi F spring compressed spring at an equilibrium position V=0 t m Notice that the spring force is opposite the displacement For the mass m, work is negative For the spring, work is positive They are opposite, and equal (spring is conservative)

Physics 207: Lecture 15, Pg 12 Conservative Forces & Potential Energy l For any conservative force F we can define a potential energy function U in the following way: The work done by a conservative force is equal and opposite to the change in the potential energy function. W = F ·dr ≡ -  U  riri rfrf UfUf UiUi

Physics 207: Lecture 15, Pg 13 Conservative Forces and Potential Energy l So we can also describe work and changes in potential energy (for conservative forces)  U = - W l Recalling (if 1D) W = F x  x l Combining these two,  U = - F x  x l Letting small quantities go to infinitesimals, dU = - F x dx l Or, F x = -dU / dx

Physics 207: Lecture 15, Pg 14 Exercise Work Done by Gravity l An frictionless track is at an angle of 30° with respect to the horizontal. A cart (mass 1 kg) is released from rest. It slides 1 meter downwards along the track bounces and then slides upwards to its original position. l How much total work is done by gravity on the cart when it reaches its original position? (g = 10 m/s 2 ) 1 meter 30° (A) 5 J(B) 10 J(C) 20 J(D) 0 J h = 1 m sin 30° = 0.5 m

Physics 207: Lecture 15, Pg 15 Home Exercise: Work & Friction Two blocks having mass m 1 and m 2 where m 1 > m 2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e.  > 0) which slows them down to a stop. l Which one will go farther before stopping? l Hint: How much work does friction do on each block ? (A) (B) (C) (A) m 1 (B) m 2 (C) They will go the same distance m1m1 m2m2 v2v2 v1v1

Physics 207: Lecture 15, Pg 16 Exercise: Work & Friction W = F d = -  N d = -  mg d =  K = 0 – ½ mv 2 -  m 1 g d 1 = -  m 2 g d 2  d 1 / d 2 = m 2 / m 1 (A) (B) (C) (A) m 1 (B) m 2 (C) They will go the same distance m1m1 m2m2 v2v2 v1v1

Physics 207: Lecture 15, Pg 17 Home Exercise Work/Energy for Non-Conservative Forces l The air track is once again at an angle of 30° with respect to horizontal. The cart (with mass 1 kg) is released 1 meter from the bottom and hits the bumper at a speed, v 1. This time the vacuum/ air generator breaks half-way through and the air stops. The cart only bounces up half as high as where it started. l How much work did friction do on the cart ?(g=10 m/s 2 ) 1 meter 30° (A) 2.5 J (B) 5 J (C) 10 J (D) –2.5 J (E) –5 J (F) –10 J

Physics 207: Lecture 15, Pg 18 Home Exercise Work/Energy for Non-Conservative Forces l How much work did friction do on the cart ? (g=10 m/s 2 ) W = F  x is not easy to do… l Work done is equal to the change in the energy of the system (U and/or K). E final - E initial and is < 0. (E = U+K) Use W = U final - U init = mg ( h f - h i ) = - mg sin 30° 0.5 m W = -2.5 N m = -2.5 J or (D) 1 meter 30° (A) 2.5 J (B) 5 J (C) 10 J (D) –2.5 J (E) –5 J (F) –10 J hfhf hihi

Physics 207: Lecture 15, Pg 19 A Non-Conservative Force Path 2 Path 1 Since path 2 distance >path 1 distance the puck will be traveling slower at the end of path 2. Work done by a non-conservative force irreversibly removes energy out of the “system”. Here W NC = E final - E initial < 0  and reflects E thermal

Physics 207: Lecture 15, Pg 20 Work & Power: l Two cars go up a hill, a Corvette and a ordinary Chevy Malibu. Both cars have the same mass. l Assuming identical friction, both engines do the same amount of work to get up the hill. l Are the cars essentially the same ? l NO. The Corvette can get up the hill quicker l It has a more powerful engine.

Physics 207: Lecture 15, Pg 21 Work & Power: l Power is the rate at which work is done. Instantaneous Power: Average Power: l A person, mass 80.0 kg, runs up 2 floors (8.0 m). If they climb it in 5.0 sec, what is the average power used? P avg = F h /  t = mgh /  t = 80.0 x 9.80 x 8.0 / 5.0 W l P = 1250 W Example: Units (SI) are Watts (W): 1 W = 1 J / 1s

Physics 207: Lecture 15, Pg 22 Work & Power: l Power is also, If force constant, W= F  x = F ( v 0  t + ½ a  t 2 ) and P = W /  t = F (v 0 + a  t)

Physics 207: Lecture 15, Pg 23 Exercise Work & Power A. Top B. Middle C. Bottom l Starting from rest, a car drives up a hill at constant acceleration and then quickly stops at the top. (Hint: What does constant acceleration imply?) l The instantaneous power delivered by the engine during this drive looks like which of the following, Z3 time Power time

Physics 207: Lecture 15, Pg 24 Chap. 12: Rotational Dynamics l Up until now rotation has been only in terms of circular motion with a c = v 2 / R and | a T | = d| v | / dt l Rotation is common in the world around us. l Many ideas developed for translational motion are transferable.

Physics 207: Lecture 15, Pg 25 Rotational Variables l Rotation about a fixed axis:  Consider a disk rotating about an axis through its center: l Recall : (Analogous to the linear case )  

Physics 207: Lecture 15, Pg 26 Rotational Variables... At a point a distance R away from the axis of rotation, the tangential motion:  x (arc) =  R  v T (tangential) =  R  a T =  R   R v =  R x 

Physics 207: Lecture 15, Pg 27 Overview (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a T =  R Here a T refers to tangential acceleration

Physics 207: Lecture 15, Pg 28 Lecture 15 Assignment: l HW7 due March 10 th l For Thursday: Read Chapter 12, Sections 7-11 ” Do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia”