week 02 - velocity.pdf

7/27/2019 Week 02 - Velocity.pdf

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Vector - quantity with both magnitude (size) and dir

Scalar - quantity with magnitude only

Vectors:

Displacement

Velocity

Acceleration

Momentum

Force

Scalars:

Distance

Speed

Time

Mass

Energy


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Position (x) where a particle/object is located Position is represented in a coordinate system

as a point relative to the point of reference

Example:

(3)

(4,7)

(5,6,-9)

(10,30O,10O)

(40,30

O

,-10)

Find the cylindrical coordinates of the pointwhose Cartesian coordinates are (1, 2, 3)


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Find the Cartesian coordinates of the pointwhose cylindrical coordinates are (2, /4, 3)

Distance (d) how far you have traveled, regardless of direction You drive the path, and your odometer goes up by 8 Km Walking path from Physics building to Faculty building

Displacement (x) where you are in relation to where you started(point of origin) Displacement is the shorter directed distance from start to stop (green arrow). When you walk around a circle and coming back to your initial position where

you start to walk, your displacement is 0, no matter what is your path.

start

stop


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Speed is a scalar (how fast something ismoving regardless of its direction).Ex: v= 20 mph

Speed is the magnitude of velocity.

Velocity is a combination of speed anddirection. Ex: v = 20 mph at 15 south ofwest

The symbol for speed is v. The symbol for velocity is type written in

bold: v or hand written with an arrow: v


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During your 8 mi. trip, which took 15 min.,

your speedometer displays yourinstantaneous speed, which variesthroughout the trip.

Your average speed is 32 mi/hr. Your average velocity is 32 mi/hr in a SE

direction. At any point in time, your velocity vector

points tangent to your path. The faster you go, the longer your velocity

vector.

constant, rightward (+) velocity of +10 m/s

a rightward (+), changing velocity - that is, a car that is moving

rightward but speeding up oraccelerating


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constant, rightward (+) velocity of +10 m/s

a rightward (+), changing velocity - that is, a car that is moving

rightward but speeding up oraccelerating

Position in Cartesian coordinates system can bewritten in a form of

The velocity ifs found by differentiating a positionvector:

The acceleration is found by differentiating theposition vector twice

zyxr zyx

zyxv

zyxrr

v

zyx vvv

zyxdt

d

zyxrr

r

va

2

2

zyxdt

d

dt

dt

dd

dt

d


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r

r

As the coordinate r changes from time t1 to time t2= t1 + t, the unitvector changes by:

r

r

r rr

Recall: arc length = r

In this case, q = D and r = 1 r

r t We can rewrite , hence or, after taking the limit as t

approaches zero,

Thus, our first derivative is , so the components ofvare

t

r

dt

d

rrrv rrrdt

d

dt

dr

rrvrvr ;

Now we need to take another derivative to get

This is going to involve a time derivative of the unit vector, for which we usemuch the same procedure as before.

Since the unit vector is perpendicular to the unit vector, we

have the same geometry as before.

The change is now in the direction, and its length is again

, so finally we have:

term-by-term derivative to get

t r

dt

d

rvrra rrdt

d

dt

d

dt

d

dtd

r

r

dt

drrr

dt

drrrr

dt

d

rrra

r

r

r

rr


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Starting from the expression on the previous slide:

and our expressions for the unit vector derivatives:

we have finally:

Admittedly this looks a little complicated, so lets look at some special cases: r = constant (i.e. ball on a string):

Here, is the centripetal acceleration and is any

angular acceleration which may impose in swinging the stone.

r

dt

d

r

dt

d

r

rra

2

2

2

rrrr

rrrrr

rra 22

rrrr

rvrar /22 ra

dt

drrr

dt

drrrr

dt

d

rrra

Cars traveling along a clover-leafinterchange experience anacceleration due to a change in speedas well as due to a change in directionof the velocity.


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There are some special cases of motion to consider.

2) The particle moves along a curve at constant speed.at = v = 0 => a = an = v

2/r.

The normal component represents the time rate of change in thedirection of the velocity.

1) The particle moves along a straight line.r => an = v

2/r = 0 => a = at = v.

The tangential component represents the time rate of change inthe magnitude of the velocity.

3) The tangential component of acceleration is constant, at = (at)c.

In this case,

s = so + vot + (1/2)(at)ct2

v = vo + (at)ct

v2 = (vo)2 + 2(at)c(s so)

As before, so and vo are the initial position and velocity of the

particle at t = 0.


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week 02 - velocity.pdf

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