Physics 207, Lecture 15, Oct. 22 Goals: Chapter 11 Chapter 12 Employ conservative and non-conservative forces Use the concept of power (i.e., energy per time) Chapter 12 Extend the particle model to rigid-bodies Understand the equilibrium of an extended object. Understand rotation about a fixed axis. Employ “conservation of angular momentum” concept Assignment: HW7 due Oct. 29 For Monday: Read Chapter 12, Sections 7, 8 & 11 do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia” 1

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 Start Finish F F q = 0° Dx Work has units of energy and is a scalar!

Example: Hooke’s Law Spring (xi equilibrium) 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 (vi) on frictionless surface as shown below and xi is the equilibrium position of the spring? x vi m ti F spring compressed spring at an equilibrium position V=0 t

They are opposite, and equal (spring is conservative) Work signs x vi m ti F spring compressed spring at an equilibrium position V=0 t 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)

Conservative Forces & Potential Energy 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. This can be written as: ò W = F ·dr = - U ri rf Uf Ui U = Uf - Ui = - W = - F • dr ò ri rf

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

Exercise Work Done by Gravity 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. How much total work is done by gravity on the cart when it reaches its original position? (g = 10 m/s2) h = 1 m sin 30° = 0.5 m 1 meter 30° (A) 5 J (B) 10 J (C) 20 J (D) 0 J

Non-conservative Forces : If the work done does not depend on the path taken, the force involved is said to be conservative. If the work done does depend on the path taken, the force involved is said to be non-conservative. An example of a non-conservative force is friction: Pushing a box across the floor, the amount of work that is done by friction depends on the path taken. and work done is proportional to the length of the path !

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

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 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)

Power is the rate at which work is done. Work & Power: Power is the rate at which work is done. Average Power: Instantaneous Power: Units (SI) are Watts (W): 1 W = 1 J / 1s Example: A person of mass 80.0 kg walks up to 3rd floor (12.0m). If he/she climbs in 20.0 sec what is the average power used. Pavg = F h / t = mgh / t = 80.0 x 9.80 x 12.0 / 20.0 W P = 470. 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

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 Dynamics: A child’s toy, a physics playground or a student’s nightmare A merry-go-round is spinning and we run and jump on it. What does it do? We are standing on the rim and our “friends” spin it faster. What happens to us? We are standing on the rim a walk towards the center. Does anything change?

Overview (with comparison to 1-D kinematics) Angular Linear distance R from the rotation And for a point at a axis: x = R v =  R a =  R Here the a refers to tangential acceleration

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

System of Particles (Distributed Mass): 1 K= ½ m v2 = ½ m (w r)2 2 K= ½ m (2v)2 = ½ m (w 2r)2 w The rotation axis matters too! Twice the radius, four times the kinetic energy

System of Particles: Center of Mass (CM) If an object is not held then it will rotate about the center of mass. Center of mass: Where the system is balanced ! Building a mobile is an exercise in finding centers of mass. m1 m2 + m1 m2 + mobile

System of Particles: Center of Mass How do we describe the “position” of a system made up of many parts ? Define the Center of Mass (average position): For a collection of N individual pointlike particles whose masses and positions we know: RCM m2 m1 r2 r1 y x (In this case, N = 2)

Sample calculation: Consider the following mass distribution: XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters XCM = 12 meters YCM = 6 meters (24,0) (0,0) (12,12) m 2m RCM = (12,6)

System of Particles: Center of Mass For a continuous solid, convert sums to an integral. dm r y where dm is an infinitesimal mass element. x

Rotational Dynamics: What makes it spin? A force applied at a distance from the rotation axis gives a torque a FTangential F TOT = |r| |FTang| ≡ |r| |F| sin f Frandial r Only the tangential component of the force matters. With torque the position of the force matters Torque is the rotational equivalent of force Torque has units of kg m2/s2 = (kg m/s2) m = N m A constant torque gives constant angular acceleration iff the mass distribution and the axis of rotation remain constant.

Physics 207, Lecture 15, Oct. 22 Assignment: HW7 due Oct. 29 For Wednesday: Read Chapter 12, Sections 7, 8 & 11 do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia” 1