Announcements 1/25/09

•  Quiz Tomorrow – Chapter 3 focus •  Driscoll 1019 (same room as last week)

•  Homework/recitation interplay….. –  (shhh!!! - answers to odd problems and

checkpoints are in the back)

Announcements 1/27/09

•  Refresh Cache on your browser when visiting course website to see the latest – Coming soon - lecture notes – Coming soon - Quiz sample problems

•  Comments about Questions During Class •  Comments about Quiz….

– As they relate to homework, reading, etc.

Topics: •  Position displacement, average and

instantaneous velocity and acceleration

•  Projectile motion •  Uniform circular motion •  Relative motion

Chapter 4: Motion in two and three dimensions

x

y

z

•  Vectors are needed to describe the 2-D or 3-D motion

•  Position vector: r = x i + y j + z k for example:

r = ( 3 m ) i + ( 2 m ) j + ( 4 m ) k

x

y

z

•  Displacement (change in position): from r1 to r2: ∆ r = r2 - r1 ∆ r = (x2 - x1) i + ( y2 - y1 ) j + ( z2 – z1 ) k

x

y

z

∆ r

r2 r1

0

1) What is the displacement between t = 1s and t = 3s?

2)  What is the velocity at t = 3 s ?

3) What is the acceleration at t = 3 s ?

Example

r3 = 32i – ( (2)(3) +1) j = 9i – 7 j

r1 = 12i – ( (2)(1) +1) j = 1i – 3 j

r = 1t2 i – ( 2 t +1) j (r in meters and t in seconds)

∆ r = 8 i + ( -4) j

v = dr/dt = 2t i – 2 j = 6 i – 2 j t=3s

a = dv/dt = 2 i + 0 j = 2 i t=3s

Velocity Vector •  Average velocity between t1 to t2

•  Instantaneous velocity

v points along the tangent of the path at that position

Acceleration vector •  Average acceleration between t1 to t2

•  Instantaneous acceleration

r = x i +  y j + z kv = vxi   + vy j  + vzka = axi +  ay j  + azk

All 3-D motion can be broken into 3 1-D motions along the 3 axes.

Three 1-D motions

•  Initial velocity: vo = vox i + voy j vox = vo cos Øo, voy = vo sin Øo

Projectile Motion in 2-D

x

y

vo

0 Øo

voy

vox

Acceleration Components

Consider gravity, g. Then, ax = az = 0; ay = -g

Projectile motion in 2-D

ay = -g, which then changes vy magnitude and direction.

ax = 0, which means vx is constant

Projectile motion in 2-D

•  The horizontal motion and the vertical motion are independent of each other

•  Horizontal motion: Motion with constant velocity vx = v0x = v0cosØ0

x – x0 = (v0cosØ0) t •  Vertical motion:

Motion of free-falling object vy = v0y + ayt = (v0 sinØ0) - g t y - y0 = v0y t + ½ a t2 = (v0 sinØ0) t - ½ g t2

assume the upward direction is positive

Projectile motion analyzed assume x0 = 0 and y0 = 0, x = ( v0cosø0 ) t (1) y = ( v0 sinø0 ) t - ½ g t2 (2) The equation of the path

From (1): t = x / (v0 cosø0) plug into (2): y = ( tanØ0 ) x - [g/2(v0cosØ0)2] x2

y = a x - b x2 Eqn of a Parabola!

x – x0 = ( v0cosø0 ) t (1) y – y0 = ( v0 sinø0 ) t - ½ g t2 (2) The horizontal range R

R = x - x0 when y - y0 = 0 From (2): 0 = ( v0 sinø0 ) t - ½ g t2 t = 2 ( v0 sinq0 ) /g or t = 0 R = ( v0cosø0 ) 2 ( v0 sinø0 ) /g = (v0

2/g) sin2ø0

Uses trig identity: sin 2ø = 2 sin ø cos ø sin 2ø is a max when the argument (2ø is 90˚)

Thus, when ø0 = 450, R = v02/g maximum

Check Point 5-5: A fly ball is hit to the outfield. During its flight(ignore the effect of the air). What happens to its

(a)  horizontal components of velocity? (b) vertical components of velocity? What are the (c)horizontal and (d)vertical

components of its acceleration during its ascent and its descent, and at the topmost point of its flight?

Projectile motion: 1)  Select a coordinate system 2)  Resolve the initial v vector into x and y

components 3)  Treat the horizontal motion and the vertical

motion independently 4)  Analyze the horizontal motion of the projectile

as a particle under constant velocity 5)  Analyze the vertical motion of the projectile as a

particle under constant acceleration ( a = -g )

Two identical balls are released at the same time and at the same height. One ball is dropped while the other is launched horizontally. Which ball will hit the floor first? Neglect air resistance.

1.  both hit at the same time 2.  launched (yellow ball) 3.  dropped (red) 4.  more information needed 5.  none of the above

Two identical balls are released at the same time and at the same height. One ball is dropped while the other is launched horizontally. Which ball will hit the floor first? Neglect air resistance.

1) both hit at the same time 2) launched (yellow ball) 3) dropped (red ball) 4) more information needed 0) none of the above.

Chapter 4: Relative Motion (1-D)

Velocity of a particle (P) depends on the reference frame of whoever is measuring the velocity

xPA = xPB + xBA

vPA = vPB + vBA

If the reference frame is moving at constant velocity (vBA is constant): aPA = aPB

Check point 4.7 : The table gives velocities (km/h) for Barbara and car P for three situations. For each what is the missing value and how is the distance between Barbara and car P changing.

vBA vPA vPB

(a) +50 +50

(b) +30 +40

(c) +60 -20

The table gives velocities (km/h) for Barbara and car P for three situations. For each what is the missing value and how is the distance between Barbara and car P changing?

vBA vPA vPB

(a)  +50 +50

(c)  +30 +40

(c) +60 -20

0

+70

+80

vPA = vPB + vBA

Relative motion in two dimensions

rPA = rPB + rBA

vPA = vPB + vBA

If vBA is constant: aPA = aPB

Sample problem: A motor boat can travel at 15 km/h in still water. A river flows at 9 km/h east. A boater wishes to cross from the south bank to a point directly opposite on the north bank. At what angle must the boat be headed?

1.  due north 2.  45o west of north 3.  53o east of north 4.  37o west of north 5.  none of the above

vr = 9 km/h

direction of vfinal = north

vboat = 15 km/h, but in which direction?

Position A

Position B

vr = 9 km/h

vfinal

Sample problem: A motor boat travels at a speed of 15 km/h in still water. A river flows at 9 km/h east. At what angle must the boat be steered (headed) to cross from the south bank to a point directly opposite on the north bank?

vboat = 15 km/h Ø

1) due north 2) 45o west of north 3) 53o east of north 4) 37o west of north 0) none of the above.

Sin Ø = 9/15 = 3/5 Ø = 37o

vr = 9 km/h

vfinal

Sample problem: How long does it take to cross if the river is 48 m wide?

vboat = 15 km/h Ø

sin Ø = 9/15 = 3/5 Ø = 37o

vfinal = 152 - 92 = 12.0 km/h

y = yo + vfinal t

t = (y - yo)/vfinal = 48m/12km/h = 4.0 x 10-3 h = 14.4 s

Chapter 4: Uniform Circular Motion

x

y

z

r

Ø arc length: s = r Ø

One-dimensional motion bent into an arc!

“Something goes in a circle at a constant velocity”

CERN high energy particle

physics experiments ( CERN =Conseil Européen pour la

Recherche Nucléaire)

2 km

David v. Goliath

or

Uniform circular motion – some facts

Period of revolution (time to complete one revolution) T = 2πr/v

Centripetal (center seeking )acceleration magnitude: a = v2 /r direction: radially inward

v and a: constant magnitude but vary continuously in direction

Chapter 4: Circular Motion

x

y

Ø ds = rdØ

differential arc length ds = rdθ

1 2 v1

v2 r1 r2

w is angular frequency

speed, v, is constant radius magnitude, r, is constant

Analysis of Uniform Circular Motion In Cartesian Coordinates