ap physics test (Friction, Circular Motion, Drag Forces)

Fc =

(Fnet in circular motion)

(mv)^2/r

ac =

(acceleration in circular motion)

v^2/r

Ff > musFn

(friction is greater than max static friction)

incorrect, max static friction will always be the maximum amount of friction

when will something start moving

when its force overcomes the force of max static friction (muFn)

tension at the lowest point of a circular path

Fc = T - Fg

tension at the highest point of a circular path

Fc = T + Fg

when does something become “slack”?

tension = 0

Fc note

Fc (as net force) = the force next to Fc - the opposite force

ex: at the top of the path, Fc = Fg - Fn

(Fc is always towards center)

if both forces are on the same side as Fc, then Fc = Fg + Fn (upside down makes Fn down)

minimum speed in centripetal motion

smallest speed = smallest centripetal force (which is only gravity applying force)

(mv)^2/r = fg → (mv)^2/r = mg → (m cancels) - > v^2/r = g (idk if needs to be v = but. yeah)

if applied force is angled up, can Fn be equal to weight?

no, the vertical component of the applied force adds onto Fn, making it greater than weight.

(and horizontally, the horizontal component of Fx = Ff but in total, F > Ff)

if terminal speed is doubled,

it is accelerating up, which means that it has to slow down to reach terminal velocity (rather than speeding up, like usual)

its moving down, but accelerating up, so its slowing down

(on the free body diagram, in this case, Fd is bigger than Fg. Fd decreases until equal to Fg, then it hits terminal velocity)

do you ever actually reach terminal velocity?

NO. so the correct wording in an answer choice is “approaching vterminal” (not approaching 0 btw. cuz 0 is no velocity. and there Is velocity)

if you are given a static coefficient and asked to find the minimum acceleration needed so the block does not slide down (block is Next to what is moving)

a = g/mu (10/the coefficient of friction)

centripetal velocity & acceleration

velocity = tangential. goes Outward

acceleration = always to center

friction on an incline

Ff = mgsin(theta)

1 revolution (centripetal motion)

((2(pi)r)^2)/1

acceleration for 1 revolution a second fills this in for v in ac = v^2/r

what happens to radius in centripetal motion if velocity is doubled (2v)

ac = v^2/r

what happens is that r turns into 4r

(fill in test values! i find that easiest)

maxmium speed around a curve without slipping (max static friction)

Ffs = musFn → fill in, m cancels → musg = v^2/r

then fill in values

force of friction note

Ff = muFn, so friction stuff depends actually on what is vertical, not horizontal. (i assume if Fn was sideways, unlike how it usually is, it would depend on the horizontal. so like. cos in the answer instead of sin)

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ex: when filled in, if applied force is at an angle, f = mu(W - Fsin(theta))

(Ff = muFn, then Fn + Fy = Fg → Fn = Fg - Fy. fill in, Fn = W - Fsintheta. this turns into the equation above, rearranged for Ff)

to find acceleration in friction

use an ma = equation

like ma = F - Ff and whatnot (bigger - smaller)

tension in a pendulum

Fc = T - Fg

(T has to be bigger or else it would .. fall.)

what is required for something to accelerate down a ramp

Ff < mgsin(theta)

(Ff = mumgcos(theta). this has to be smaller than the other for it to accelerate)

static friction (not on an incline) + tension

not MAX static friction, Ff = whatever is on the other side. in this case, Ff = T. the string is holding up another block, so T = mg of the other block (i think??? it works for a question on the study guide)

moving in a straight line at a constant speed

net force = 0

velocity (in a straight line)

v = d/t

radius doubled

acceleration is half as great

constant speed while moving on 2 different lengths of track, what is the speed?

the same

when a car goes around a turn, which force provides centripetal force?

friction

direction of centripetal force in a circle

towards center

relationship between mass & terminal velocity

IDK ITS LINEAR OR EXPONENTIAL DEPENDING ON THE GRAPH U GET (some drag forces are squared some arent so. Maybe thats why)

why do objects reach terminal velocity

Because while an object is falling, Fd & Fg will balance out .