Employ conservative and non-conservative forces Lecture 16 Chapter 11 (Work) Employ conservative and non-conservative forces Relate force to potential energy Use the concept of power (i.e., energy per time) Chapter 12 Define rotational inertia Define rotational kinetic energy Assignment: HW7 due Tuesday, Nov. 1st For Monday: Read Chapter 12, (skip angular momentum and explicit integration for center of mass, rotational inertia, etc.) Exam 2 7:15 PM Thursday, Nov. 3th 1

Definition of Work, The basics Ingredients: Force ( F ), displacement (  r )   r displacement F Work, W, of a constant force F acts through a displacement  r : W = F · r (Work is a scalar) Work tells you something about what happened on the path! Did something do work on you? Did you do work on something?

Net Work: 1-D Example (constant force) A force F = 10 N pushes a box across a frictionless floor for a distance x = 5 m. Start Finish F q = 0° x Net Work is F x = 10 x 5 N m = 50 J 1 N m ≡ 1 Joule (energy) Work reflects positive energy transfer

Net Work: 1-D 2nd Example (constant force) A force F = 10 N is opposite the motion of a box across a frictionless floor for a distance x = 5 m. Start Finish q = 180° F x Net Work is F x = -10 x 5 N m = -50 J Work again reflects negative energy transfer

Work: “2-D” Example (constant force) An angled force, F = 10 N, pushes a box across a frictionless floor for a distance x = 5 m and y = 0 m Start Finish F Fx q = -45° x (Net) Work is Fx x = F cos(-45°) x = 50 x 0.71 Nm = 35 J

Exercise Work in the presence of friction and non-contact forces A box is pulled up a rough (m > 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 2 3 4 5

Work and Varying Forces (1D) Area = Fx Dx F is increasing Here W = F · r becomes dW = F dx Consider a varying force F(x) Fx x Dx

Example: Work Kinetic-Energy Theorem with variable force How much will the spring compress (i.e. x = xf - xi) to bring the box to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ? to vo m spring at an equilibrium position x F V=0 t m spring compressed

Compare work with changes in potential energy Consider the ball moving up to height h (from time 1 to time 2) How does this relate to the potential energy? Work done by the Earth’s gravity on the ball) W = F  Dx = (-mg) -h = mgh DU = Uf – Ui = mg 0 - mg h = -mg h DU = -W mg h mg

Conservative Forces & Potential Energy If a conservative force F we can define a potential energy function U: The work done by a conservative force is equal and opposite to the change in the potential energy function. (independent of path!) W = F ·dr = - U ò ri rf Uf Ui

A Non-Conservative Force, Friction Looking down on an air-hockey table with no air flowing (m > 0). Now compare two paths in which the puck starts out with the same speed (Ki path 1 = Ki path 2) . Path 2 Path 1

A Non-Conservative Force Path 2 Path 1 Since path2 distance >path1 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 WNC = Efinal - Einitial < 0  and reflects Ethermal

There is no transformation because energy is conserved. A child slides down a playground slide at constant speed. The energy transformation is U  K U  ETh K  U K ETh There is no transformation because energy is conserved. STT11.1 Answer: B

Conservative Forces and Potential Energy So we can also describe work and changes in potential energy (for conservative forces) DU = - W Recalling (if 1D) W = Fx Dx Combining these two, DU = - Fx Dx Letting small quantities go to infinitesimals, dU = - Fx dx Or, Fx = -dU / dx

Spring: Fx = 0 => dU / dx = 0 for x=xeq Equilibrium Example Spring: Fx = 0 => dU / dx = 0 for x=xeq The spring is in equilibrium position In general: dU / dx = 0  for ANY function establishes equilibrium U U stable equilibrium unstable equilibrium

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

Work & Power: Power is the rate, J/s, at which work is done. Average Power is, Instantaneous Power is, If force constant, W= F Dx = F (v0 Dt + ½ aDt2) and P = W / Dt = F (v0 + aDt)

Work & Power: Example Average Power: 1 W = 1 J / 1s Example: A person, mass 80.0 kg, runs up 2 floors (8.0 m) at constant speed. If they climb it in 5.0 sec, what is the average power used? Pavg = F h / Dt = mgh / Dt = 80.0 x 9.80 x 8.0 / 5.0 W P = 1250 W

Top Middle Bottom Exercise Work & Power Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top. The instantaneous power delivered by the engine during this drive looks like which of the following, Z3 time Power Top Middle Bottom

Exercise Work & Power P = dW / dt and W = F d = (m mg cos q - mg sin q) d and d = ½ a t2 (constant accelation) So W = F ½ a t2  P = F a t = F v (A) (B) (C) Power time Power Z3 time Power time

Chap. 12: Rotational Dynamics Up until now rotation has been only in terms of circular motion with ac = v2 / R and | aT | = d| v | / dt Rotation is common in the world around us. Many ideas developed for translational motion are transferable.

Rotational motion has consequences Katrina How does one describe rotation (magnitude and direction)?

Rotational Variables Rotation about a fixed axis: Consider a disk rotating about an axis through its center: Recall : (Analogous to the linear case )  

System of Particles (Distributed Mass): Until now, we have considered the behavior of very simple systems (one or two masses). But real objects have distributed mass ! For example, consider a simple rotating disk and 2 equal mass m plugs at distances r and 2r. Compare the velocities and kinetic energies at these two points. w 1 2

For these two particles 1 K= ½ m v2 Twice the radius, four times the kinetic energy w 2 K= ½ m (2v)2 = 4 ½ m v2

For these two particles 1 K= ½ m v2 w 2 K= ½ m (2v)2 = 4 ½ m v2

For these two particles

Rotation & Kinetic Energy... The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

Calculating Moment of Inertia where r is the distance from the mass to the axis of rotation. Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: m m L m m

Calculating Moment of Inertia... For a single object, I depends on the rotation axis! Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2 I1 = 2mL2 I2 = mL2 I = 2mL2 m m L m m

Lecture 16 Assignment: HW7 due Tuesday Nov. 1st 1