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Honors Physics 1Class 08 Fall 2013

Review of quiz 3

Integrating equation of motion

Work-Energy Theorem

Potential Energy Intro

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Integrating the equation of motion

If we know the force as a function of time, then it is in principle straightforward to solve the motion of a particle. Frequently though, we only know the force as a function of position (spring force, gravitational force...)We will now develop some tools for addressing such situations.

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Integrating the equation of motion

2 2

1 1

( )

First we address one dimensional motion.

( )

Integrate both sides with respect to x:

( )

The integral on the right is straighforward, since ( ) is known.

x x

x x

dvm F rdt

dvm F xdt

dvm dx F x dxdt

F x

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2 2 22 2 2 2

2 111 1

2 2

We use a a change of variables for the left side.

1 1 1 1

2 2 2 2

and using indefinite upper limit on the integral:

1 1( ') '

2 2

Since

i

t t t

tt t

x

ix

dxdx dt vdt

dt

dv dmv dt m v dt mv mv mvdt dt

mv mv F x dx

we can integrate again to find ( ).dx

v x tdt

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Example 1A mass m is thrown vertically upward with initial speed v0. How high does it rise?

(Assume g is constant and neglect friction.)

0

2 20 0

20 0

1 1( ') '

2 2

1At the peak v=0, so:

2which we could have found by Newton's law alone.

Note that the solution does not reference time at all.

x

x

F mg

mv mv F x dx mg x x

v g x x

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Example

A mass m is shot vertically upward from the surface of the earth at initial speed v0. Assuming that the only force is gravity (GmM/r2),

a)Find the height to which the mass rises.

b)Find the escape velocity.

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Example 2

2

2

2 20

20

20 2

' 1 1

'

1 1( )

To find the maximum height, we set ( ) 0.

1 1

and for escape velocity, we set and ( ) 0.

22 using

e

r

eR

e

e

ee e

GMmF

r

dr

r Rr

v r v GMr R

v r

v GMr R

r v

GM GMv gR g

R R

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Equation of motion in three dimensions

Take a small step r along the trajectory

(which we don't know yet.)

and using

and we can use a vector identity to convert this to:

1

2

dvF r m

dt

dvF r m r r v t

dtdv

F r m v tdt

dv dvdt dt

2

2

2

1

2 2 2

1 1

2 2

so now we have: 2

Now divide the trajectory up into many steps:

2

1 1( )

2 2 2

N

j j j jj

rb tb

b ara ta

d dv dvv v v v v

dt dt dt

m dF r v t

dt

m dF r r v t

dt

m dF r dr v dt mv mv

dt

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Escape velocity: General case

What if the projectile is not shot straight up?

2

2 2

2

2

ˆ ˆ

We don't know the trajectory, but we can still solve the problem.

ˆˆ

so

so we find out that the escape velocity does not

depend on details of trajectory.

GMm RF r mg r

r r

dr drr rd

RF dr mg dr

r

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The Work-Energy Theorem(in one dimension)

21We call the quantity the kinetic energy K.

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We call the force integral the work done moving the

particle from point a to b, .

Our relation now takes the form: .

The SI unit of wo

b

a

ab

ab b a

mv

Fdx

W

W K K

2

2

rk is the Joule:

kg m1J=1

s

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So what?

The work-energy theorem says that:

which suggests that we need to know everything about

the path (motion) before we start a calculation.

But wait - there are some useful special cases

b

ab b aa

W F dr K K

.

Conservative forces - where the work does not depend on path.

Constrained motion where the constraining force does no work.

- e.g. - roller coaster, pendulum