phys 204 u1-6

Set up for questions

1. What am i finding?

2. Identify directions

3. Identify angle and quadrant

4. add ± signs to vectors

5. choose equations to use

reference angle

theta ref = tan^-1( |y/x|) (REF ANGLE ALWAYS +)

polar / standard angle in Q1

same as ref angle

polar / standard angle in Q2

(-,+) , 180 - ref angle

polar / standard angle in Q3

(-,-) , 180 + ref angle

polar / standard angle in Q4

(+,-) , 360 - ref angle

use polar / standard angle

when it asks for angle counterclockwise from +x axis

terminal speed

zero acc, net force = 0, velocity not 0

position time graph

slope = v avg = ~x / t

velocity time graph

slope = a avg = (~v / t)

Area under v vs t graph = displacement

(Area under v vs t graph / t ) = v avg

avg speed

total distance / total time

no acceleration

constant velocity

t

time eq w velocity

t = ~x / v

cartesian coordinates

x = cos() * r

y = sin()*r

cartesian coordinates for forces

Fx = cos() * F

Fy = sin() * F

East of North , North of West

angle goes from north to east , west to north

Cartesian unit vector expression

2D → ~r = ~xi + ~yj

3D → ~r = ~xi + ~yj + ~zk

magnitude of x n y components

|v| = sqrt( (vix² + vjy²) )

r

position vector ( r = xi + yj + zk)

~r

displacement vector (rf - ri) = (~r = ~xi + ~yj + ~zk)

Asks for “Magnitude of displacement” or “How far from origin”

displacement vector → |~r| = sqrt( ~x² + ~y² )

“direction” “orientation”

asking for angle

Magnitude andOrientation

asking for one magnitude and one angle

v avg

~r / t

v

dr / dt

a avg

~v / t

a

dv / dt

speed and or direction changes

causes acceleration

direction changes

causes acceleration

velocity vector form eq

v = sqrt( vx² + vy² )

in projectile motion v

vy changes vx constant

no matter where theta is measured from (above or belos +x axis)

vx is always v*cos() and vy is always v*sin()

“just clears”

barely passes over top

acceleration in projectile motion (x)

ax=0

equation of vxi in projectile motion

vxi or vxf = vi*cos() , (cz vx is constant)

equation of vyi in projectile motion

vyi = vi*sin()

equation of vyf in projectile motion

vyf = vyi sin() + (ay * t)

in projectile motion vyf and vyi

have the same magnitude but opposite direction

in projectile motion acceleration

is always downwards (but v can be up or down)

(~x) range eq projectile motion

~x = (vi² * sin(2 theta) ) / g

only works when yi=yf (object leaves from ground n lands on ground)

max range in projectile motion when yi=yf

45 deg

max height eq in projectile motion (how much higher the object travels from where it started NOT from ground)

hmax = (vi² * sin()²) / 2g

time to max height in projectile motion

tymax = ( vi * sin() ) / g

max total flight time when yi=yf

when object is launched straight up (vi is vertical) (90 deg)

direction of acceleration

same direction as Net force

action reaction forces

act on different objects or they cancel out

apparent weight

N (normal force) → what the scale reads

accelerating upwards

N = ma+mg

accelerating downwards

N = mg-ma

Uniform

means speed is constant (not velocity)

in circular motion v is

tangent to circle

in circular motion v and r

v and r are perpendicular

in circular motion r

from center outwards to the circle

in circular motion ac

always towards the center

in circular motion ac and r

ac and r pointing opposite directions

in circular motion ac and v

ac and v are perpendicular

radius in circular motion

the distance from the center of rotation to the object whose rotation ur analyzing

At the top of a vertical circle, when asked for the minimum speed to stay in contact or minimum speed to keep the string from becoming loose

ac = g, N=0, T=0 (tension)

At the top of a vertical circle, if the object is moving at a speed faster than the minimum speed required

ac>g → additional inward force needed N or T

A string / object exerts force on an object

only if its attached directly to the object, doesn’t count if it acts somewhere else in the overall system

in centripetal / radial acceleration

direction is changing

radial acceleration

changes the direction, ar = -v² / r

centripetal acceleration

always towards the center (inward), ac =|v² / r| (always +)

period eq

T = (2*pi*r) / v

angular speed eq

w = v/r or w= (2*pi)/T

linear velocity / tangential speed in circular motion eq

v = r*w

centripetal acceleration eq with angular speed

ac = r*w²

tangential acc / speed

speed is changing → at = dv/dt

no change in speed → at = 0

in uniform circular motion at

at = 0 because uniform = constant speed and no change in speed → at=0

at is only for

non uniform circular motion

total magnitude of acceleration eq in circular motion

a = sqrt( ac² + at²)

ac and ar are

ac = |ar|

speed and velocity @max n min height in free fall

max height: speed=0 , v=0

min speed: speed = max , v= max (downward)

speed and velocity @max n min height in projectile motion

max height: vx = constant (can be zero if launched at 90deg) , vy = 0, speed (y) = min , speed(x) = constant, a=g

min height: vx = same, vy = nonzero, speed (y) = max, a = same

speed and velocity @max n min height in vertical circular motion

max height: vx = nonzero, speed(x) = min, vy=0, speed(y)=0, total v = horizontal, total speed = min, ac = downwards

min height: vx = same, speed(x) = same, vy=same, speed(y)=0, total v = same, total speed = max, ac = upwards

speed and velocity in horizontal circular motion

vx = nonzero, speed(x) = constant, vy=0, speed(y)=0, total v = nonzero (tangent to circle), total speed = constant, ac = inwards, net force = toward center

centrifugal force

not same as centripetal force, fictitious force that pushes outwards, only used in a rotating frame

rotating frame

your pov is rotating too

Inertial frame

non accelerating, outside of rotation POV, reference frame where Newtons Laws apply normally.

Non Inertial frame

rotating/accelerating, inside rotation POV, ref frame where imaginary forces may b introduced (centrifugal force)

objects rotating together

objects have the same w

acceleration in atwood machine

a = (m2-m1) / (m2+m1)

% increase

( (v2-v1) / v1 ) * 100

unit of density

kg/m³

in → cm

1 = 2.45

ft → m

1 = 0.3

mi (miles) → km

1 = 1.6

nm (nanometer) → m

1 = 10^-9

area under acceleration time graph

area = ~v

in circular motion there always has to be a force towards the center

force(s) that acts as the centripetal force