v relate force to potential energy v - department of physicspage 1 physics 207 – lecture 16...

Physics 207 – Lecture 16

Physics 207: Lecture 16, Pg 1

Lecture 16

•• Chapter 11 (Work)Chapter 11 (Work)

v Employ conservative and non-conservative forces

v Relate force to potential energy

v Use the concept of power (i.e., energy per time)

Assignment:

l HW7 due Tuesday, Nov. 1st

l For Monday: Read Chapter 12, (skip angular momentum and explicit integration for center of mass, rotational inertia, etc.)

Exam 2 Exam 2 7:15 PM Thursday, Nov. 37:15 PM Thursday, Nov. 3thth

•• Chapter 12 Chapter 12

v Define rotational inertia

v Define rotational kinetic energy


Definition of Work, The basics

Ingredients: Force ( F ), displacement ( ∆∆∆∆ r )

θθθθ ∆∆∆∆ rr

displa

cem

ent

FFWork, W, of a constant force F

acts through a displacement ∆∆∆∆ r :

W = F ·∆∆∆∆ r (Work is a scalar)(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?



Units:

N-m or Joule Dyne-cm or erg

= 10-7 J

BTU = 1054 J

calorie= 4.184 J

foot-lb = 1.356 J

eV = 1.6x10-19 J

cgs Othermks

Force x Distance = Work

Newton x

[M][L] / [T]2

Meter = Joule

[L] [M][L]2 / [T]2


Net Work: 1-D Example (constant force)

l Net Work is F ∆∆∆∆x = 10 x 5 N m = 50 J= 10 x 5 N m = 50 J

l 1 N m 1 Joule (energy)

l Work reflects positive energy transfer

∆∆∆∆x

l A force F = 10 N pushes a box across a frictionless floorfor a distance ∆∆∆∆x = 5 m.

Fθ = 0°

Start Finish



Net Work: 1-D 2nd Example (constant force)

l Net Work is F ∆∆∆∆x = = --10 x 5 N m = 10 x 5 N m = --50 J50 J

l Work again reflects negative energy transfer

∆∆∆∆x

F

l A force F = 10 N is opposite the motion of a box across a frictionless floor for a distance ∆x = 5 m.

θ = 180°Start Finish


Work: “2-D” Example (constant force)

l (Net) Work is Fx ∆∆∆∆x = F cos(= F cos(-45°) ∆∆∆∆x = 50 x 0.71 Nm = 35 J= 50 x 0.71 Nm = 35 J

∆∆∆∆x

F

l An angled force, F = 10 N, pushes a box across a frictionless floor for a distance ∆x = 5 m and ∆y = 0 m

θ = -45°

Start Finish

Fx



ExerciseWork in the presence of friction and non-contact forces

A. 2

B. 3

C. 4

D. 5

l A box is pulled up a rough (µ > 0) incline by a rope-pulley-weight arrangement as shown below.

v How many forces (including non-contact ones) are doing work on the box ?

v Of these which are positive and which are negative?

v State the system (here, just the box)

v Use a Free Body Diagram

v Compare force and path v


Work and Varying Forces (1D)

l Consider a varying force F(x)

Fx

x∆x

Area = Fx ∆xF is increasing

Here W = F ·∆∆∆∆ rbecomes dW = F dx

∫=f

i

x

x

dxxFW )(



• 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 ?

∆x

vo

m

to

F

spring compressed

spring at an equilibrium position

V=0

t m

∫=f

i

x

x

dxxFW )(box

∫ −=f

i

x

x

dxkxW box

2

02

12

2

1 v mxk =∆

Example: Work Kinetic-Energy Theorem with variable force


Compare work with changes in potential energy

l Consider the ball moving up to height h

(from time 1 to time 2)

l How does this relate to the potential energy?

Work done by the Earth’s gravity on the ball)

W = F • ∆x = (-mg) -h = mgh

∆U = Uf – Ui = mg 0 - mg h = -mg h

∆U = -W

h

mg

mg



Conservative Forces & Potential Energy

l 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 ∫

ri

rf Uf

Ui


A Non-Conservative Force, Friction

l Looking down on an air-hockey table with no air flowing (µ > 0).

l 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 à and reflects Ethermal


A. U à K

B. U à ETh

C. K à U

D. Kà ETh

E. There is no transformation because

energy is conserved.

A child slides down a playground slide at

constant speed. The energy transformation

is



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 = Fx ∆x

l Combining these two,

∆U = - Fx ∆x

l Letting small quantities go to infinitesimals,

dU = - Fx dx

l Or,

Fx = -dU / dx


Equilibrium

l Example

v Spring: Fx = 0 => dU / dx = 0 for x=xeq

The spring is in equilibrium position

l In general: dU / dx = 0 à for ANY function establishes equilibrium

stable equilibrium unstable equilibrium

U U



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.


Work & Power:

l Power is the rate, J/s, at which work is done.

l Average Power is,

l Instantaneous Power is,

l If force constant, W= F ∆x = F (v0 ∆t + ½ a∆t2)

and P = W / ∆t = F (v0 + a∆t)

t

WP

∆=

dt

dWP =



Work & Power: Example

Average

Power:

l 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?

l Pavg = F h / ∆t = mgh / ∆t = 80.0 x 9.80 x 8.0 / 5.0 W

l P = 1250 W

Example:

1 W = 1 J / 1st

WP

∆=


Exercise Work & Power

A. Top

B. Middle

C. Bottom

l Starting from rest, a car drives up a hill at constant acceleration and then suddenly stops at the top.

l The instantaneous power delivered by the engine during this drive looks like which of the following,

Z3

time

Po

we

rP

ow

er

Po

we

r

time

time



Exercise Work & Power

l P = dW / dt and W = F d = (µ mg cos θ − mg sin θ) dand d = ½ a t2 (constant accelation)

So W = F ½ a t2 à P = F a t = F v

l (A)

l (B)

l (C)

Z3

time

Po

we

rP

ow

er

Po

we

r

time

time


Chap. 12: Rotational Dynamics

l Up until now rotation has been only in terms of circular motion with ac = v2 / R and | aT | = d| v | / dt

l Rotation is common in the world around us.

l Many ideas developed for translational motion are transferable.



Rotational motion has consequences

How does one describe rotation (magnitude and direction)?

Katrina


Rotational Variables

l Rotation about a fixed axis:

v Consider a disk rotating aboutan axis through its center:

l Recall :

(Analogous to the linear case )

ω

θ

/R (rad/s) 2

Tangentialv

Tdt

d === πθω

dt

dx=v



System of Particles (Distributed Mass):

l Until now, we have considered the behavior of very simple systems (one or two masses).

l But real objects have distributed mass !

l For example, consider a simple rotating disk and 2 equal mass m plugs at distances r and 2r.

l Compare the velocities and kinetic energies at these two points.

1 2ω


For these two particles

l Twice the radius, four times the kinetic energy

1 K= ½ m v2

2 K= ½ m (2v)2 = 4 ½ m v2

ω

2

2212

1212

242

21 )()(vvK rmrmmm ωω +=+=

22

2

2

121 ][K ωmrmr +=




1 K= ½ m v2

2 K= ½ m (2v)2 = 4 ½ m v2

ω

22

2

2

121 ][K ωmrmr +=

2

pointsmassAll

2

21 ][K ω∑= mr



ω

inertia) ofmoment theis (where

][K

2

21

Rot

2

pointsmassAll

2

21

IIK

mr

ω

ω

=

= ∑



Rotation & Kinetic Energy...

l The kinetic energy of a rotating system looks similar to that of a point particle:

Point ParticlePoint Particle Rotating SystemRotating System

∑=i

iirm

2I

2

21 Iω=K2

21 vmK =

v is “linear” velocity

m is the mass.

ω 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.∑

=≡

N

iii

rm1

2I

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:

mm

mm

L



Calculating Moment of Inertia...

l For a single object, I depends on the rotation axis!

l Example: I1 = 4 m R2 = 4 m (21/2 L / 2)2

L

I = 2mL2I2 = mL2

mm

mm

I1 = 2mL2


Lecture 16

Assignment:

l HW7 due Tuesday Nov. 1st

v relate force to potential energy v - department of physicspage 1 physics 207 – lecture 16...

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