lecture 5

Physics 207: Lecture 8, Pg 1

Lecture 5 Goals: (finish Ch. 4)Goals: (finish Ch. 4)

Address 2D motion in systems with acceleration (includes linear, projectile and circular motion)

Discern differing reference frames and understand how they relate to particle motion


Trajectory with constant acceleration along the vertical

What do the velocity and acceleration

vectors look like? x

t = 0

4y

x vs y

Position

Special Case:

ax=0 & ay= -g

vx(t=0) = v0 vy(t=0)

vy(t) = – g t

x(t) = x0 v0 t

y(t) = y0 - ½ g t2


Trajectory with constant Trajectory with constant acceleration along the vertical acceleration along the vertical


vectors look like?

Velocity vector is always tangent to the curve!

x

t = 0

4y

x vs yVelocity


Trajectory with constant acceleration along the vertical


vectors look like?

Velocity vector is always tangent to the curve!

Acceleration may or may not be!

x

t = 0

4y

x vs yAcceleration

PositionVelocity


Concept Check

E1. You drop a ball from rest, how much of the acceleration from gravity goes to changing its speed?

A. All of it

B. Most of it

C. Some of it

D. None of it

E2. A hockey puck slides off the edge of a horizontal table, just at the instant it leaves the table, how much of the acceleration from gravity goes to changing its speed?

A. All of it

B. Most of it

C. Some of it

D. None of it


Example ProblemExample Problem

Example Problem: If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel be for it hit the level ground (if g= -10 m/s2 & no air resistance)?


Example Problem

If the knife is thrown horizontally at 10 m/s second and the knife starts out at 1.25 m from the ground, then how far does the knife travel before it hits the level ground

(assume g= -10 m/s2 & no air resistance)? at t=0 the knife is at (0.0 m, 1.0 m) with vy=0 after t the kinfe is at (x m, 0.0 m) x = x0 + vx t and y = y0 – ½ g t2

So x = 10 m/s t and

0.0 m = 1.25 m – ½ g t2 2.5/10 s2 = t2 t = 0.50 sx = 10 m/s 0.50 s = 5 .0 m


Exercise 3Relative Trajectories: Monkey and Hunter

A. go over the monkey.

B. hit the monkey.

C. go under the monkey.

All free objects, if acted on by gravity, accelerate similarly.

A hunter sees a monkey in a tree, aims his gun at the monkey and fires. At the same instant the monkey lets go.

Does the bullet …


Schematic of the problem

xB(t) = d = v0 cos t yB(t) = hf = v0 sin t – ½ g t2

xM(t) = d yM(t) = h – ½ g t2

Does yM(t) = yB(t) = hf?

Does anyone want to change their answer ?

(x0,y0) = (0 ,0)

(vx,vy) = (v0 cos , v0 sin

Bullet v0

g

(x,y) = (d,h)

hf

What happens if g=0 ?

How does introducing g change things?

Monkey


Relative motion and frames of reference

Reference frame S is stationary Reference frame S’ is moving at vo

This also means that S moves at – vo relative to S’ Define time t = 0 as that time when the origins coincide


Relative motion and frames of reference

Reference frame S is stationary, A is stationary Reference frame S’ is moving at vo

A moves at – vo relative to an observer in S’ Define time t = 0 as that time (O & O’ origins coincide)

r’= r-v0t

O,O’

t=0

-v0

r

A -v0 t

O’

r

t later

A


Relative Velocity

The positions, r and r’, as seen from the two reference frames are related through the velocity, vo, where vo is velocity of the r’ reference frame relative to r r’ = r – vo t

The derivative of the position equation will give the velocity equation v’ = v – vo

These are called the Galilean transformation equations Reference frames that move with “constant velocity” (i.e., at

constant speed in a straight line) are defined to be inertial reference frames (IRF); anyone in an IRF sees the same acceleration of a particle moving along a trajectory. a’ = a (dvo / dt = 0)


Central concept for problem solving: “Central concept for problem solving: “xx” and “” and “yy” ” components of motion treated independently.components of motion treated independently.

Example: Man on cart tosses a ball straight up in the air. You can view the trajectory from two reference frames:

Reference frame

on the ground.

Reference frame

on the moving cart.

y(t) motion governed by 1) a = -g y

2) vy = v0y – g t3) y = y0 + v0y – g t2/2

x motion: x = vxt

Net motion: R = x(t) i + y(t) j (vector)


Exercise, Relative Motion

You are swimming across a 50. m wide river in which the current moves at 1.0 m/s with respect to the shore. Your swimming speed is 2.0 m/s with respect to the water.

You swim perfectly perpendicular to the current,how fast do you appear to be moving to an observer on shore?

2 m/s

1 m/s50mj m/s 2i m/s 1v

m/s 2.2 m/s 0.20.1 |v| 22

Home exercise: What is your apparent speed if you swim across?


Case 2: Uniform Circular MotionCircular motion has been around a long time


Circular Motion (UCM) is common so specialized terms

Angular position (CCW + CW -)



Angular position (CCW + CW -) Radius is r

r



Angular position (CCW + CW -) Radius is r Arc traversed s = r (if starting at 0 deg.)

r

s



Angular position (CCW + CW -) Radius is r Arc traversed s = r (if starting at 0 deg.) Tangential “velocity” vT = ds/dt

rvT

s



Angular position (CCW + CW -) Radius is r Arc traversed s = r Tangential “velocity” vT = ds/dt Angular velocity, ≡ d/dt (CCW + CW -) vT = ds/dt = r d/dt = r

r

vT

s


Uniform Circular Motion (UCM) has only radial acceleration

UCM changes only in the direction of v

1. Particle doesn’t speed up or slow down!

2. Velocity is always tangential, acceleration perpendicular !

path radial aa

0Ta

a

v

v

|v||v|

v

vv

Tvv

rrdt

d

dt

da TTTTTr /v /vvvv

v 2


Again

Centripetal Acceleration

ar = vT2/r = 2r

Circular motion involves continuous radial acceleration

vT

r

ar

Uniform circular motion involves only changes in the direction of the velocity vector

Acceleration is perpendicular to the trajectory at any point, acceleration is only in the radial direction.


Angular motion, signsAngular motion, signs

If

angular displacement

velocity

accelerations

are counter clockwise then sign is positive. If clockwise then negative


RecapRecap

Assignment: HW2, (Chapters 2,3 & some 4)

due Wednesday

Read through Chapter 6, Sections 1-4

lecture 5

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