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Honors Physics 1Class 07 Fall 2013

Center of mass – multivariable integration

Impulse – Integration of force

Integrating equations of motion

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Calculating the Center of Mass

1 1 1; ;com x com y com zx xdM y ydM z zdM

m m m

The center of mass is found by integration

over the volume.

This approach is used for calculating the average value of a quantity for any distribution.

qdVq q

dV

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Center of Mass Integration

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We calculate the center of mass for a uniform triangular plate with edges at

0, 0, 2

1 1 1( , ) ( , ) ( , ) ( , )

4Since the density is constant, = we can

x y and y a x

R rdm x y r x y dxdy x y r x y dxdyM M M

M M

A a

/2 2 /2

0 0 0

/22 3 3 3 3

20

pull it out of the integral.

ˆ ˆ

2

3 2 42

2 3 24 24 24 6

(which equals 2/3 of a/2.)

a a x a

x

a

R xi yj dxdyM

R x dxdy x dydx x a x dxM M M

ax x a a a a

M M a

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Computing Center of mass for a Shaped Object

1) Use symmetry to reduce the amount of direct calculation.

2) Set up the multivariable integral relation for calculating the average value along the axis of interest.

3) Do the integration in one of the variables (e.g.- y), putting the limitations on the shape (in terms of the other variables) into the limits of the integral.

4) Repeat for as many directions as needed until you get to the last one, where the upper and lower bounds are then the global max and min of the shape.

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A Math Interlude: Differentials

When a function of some variable is "continuous and smooth"*

then we can use the first term in the Taylor series to

evaluate the change in for a small change in :

is called a different

f x

f x

dfdf dx

dxdf

ial and specifically uses the linear

approximation.

This idea is very useful for converting between variables.

*continuous= no jumps**; smooth=no jumps in the derivative.

** no discontinuities.

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Changing the variable of integration

2

2

2

1/2 1/2

1/2 1/2

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Consider the integral:

A useful substitution is:

1so and then

2The integral then can be rewritten:

1 1 1

2 2 2

with

b bb a

a a

bx

a

u ubu ux u u

a u u

a

xe dx

x u

dxx u dx du u du

du

xe dx u e u du e du e e

u a

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Impulse

0 (0)

0

The relation between force and momentum for a system of

particles is: .

So and therefore ' ( ) (0)

The integral ' is called the impulse and is equal to

P tt

P

t

dPF

dt

Fdt dP Fdt dP P t P

J Fdt

the

change in momentum over the time the force is applied.

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Impulse example 1 (KK131)

0.2 kg ball bounces from floor in time 10-3 s. 8 m/s. Find average force.

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Propulsion example

Is it possible to build a jetpack using downward firing machine guns? (Yes!)

We need to know: – mass of a machine gun, (AK 47 = 4.8 kg)– mass of a bullet, (0.008 kg)– rate at which bullets can be fired. (10/s)– speed of exiting bullet wrt gun (715 m/s)

http://what-if.xkcd.com/21/

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Propulsion example: AK47 jet pack

one bullet: 0

0.008 715 10 57 thrust

Weight of AK47=47N

Average Thrust-Weight=10N

Average child weighs 200N, so 20 guns firing simultaneously

should lift her up.

b b

bb b

F t p

F t m v

N kg m bF F t m v R N

t b s s

My drawingRandall Munroe’s drawing

Only 3 seconds of liftWe need more ammunition and ammunition has mass.

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But what if the mass changes with time?When bullets leave the gun, they leave at velocity relative to the gun.

In time t we eject mass m.

Let's take the remaining mass of the gun at any time to be M.

P(t+ t)=M v+ mu

u

dP dv dmM u

dt dt dt

0

0

If there is no external force on the gun, then

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To find the gun velocity:

''

1 1' ln

'

f

i

f

i

vt

f iv

Mtf

iM

dv dMM udt dt

dv dMu

dt M dt

dvdt dv v v

dt

MdMu dt udM u

M dt M M

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Filling in some details on rocket discussion

View system from lab frame.

Mass at time t is M+ m

v(t)=v

v(t+ t)=v+ v

Initial momentum: ( )

Final momentum: ( ) ( )

P t M m v

P t t M v v m v v u

P M v mu