UNIT 2 Ch 8 Physics

1 rev= -- degrees= --- radians

1 rev= 360 degrees= 2pi radians -(pi radians= 180 degrees)

problem: a particular bird's eye can just distinguish objects that subtend an angle no smaller than 3X10^-4 radians. how many degrees is this? How small an object can the bird just distinguish when flying at h= 100m?

1. (3x10^-4 degrees)(360 degrees/ 2pi radians) = 0.017 degrees 2. l= rtheta; for small angles l and chord length are approximately the same; l= r(theta)-> l= (100m)(3x10^-4 radians)= 3x10^-2 m= 3 cm

problem: for a child on the rotating carousel, determine that child's 1. tangential 2. centripetal and 3. total acceleration alpha= 0.060 rad/s^2 r= 2.5 v= 1.2 m/s

1. atan= r(alpha) (2.5)(0.06 radians/s^2)= 0.15 m/s^2 2. aR= v^2/r= 0.58 m/s^2 3. a= sqrt(atan^2 + aR^2)= 0.60 m/s^2

problem: a carousel is initially at rest. at t=0 it is given a constant angular acceleration alpha= 0.060 rad/s^2, which increases its angular velocity for 8 seconds. at t=8s, determine 1. the angular velocity and 2. the linear velocity of someone located 2.5 m from the center

1. avg alpha= delta O/ delta T -solve for Omega2= omega1 + alpha(deltat) -0.48 radians/s 2. v= r(omega)= 1.2 m/s

problem: is the lion faster than the horse? one child sits on a horse near the outer edge and one sits on a lion halfway out from the center. which child has greater linear v? which has the greater angular v?

1. child on outer edge travels a farther distance than the child on the inner part in the same time; child on outer edge has higher linear velocity 2. it is the same; object as a whole moves by 2 pi degrees in the same amount of time

kinetic energy of a cylinder

1/2(m)(1+[1/2mr^2/mr^2])v^2 =3/2(1/2mv^2) KE is 50% larger than if it was a mass moving along (1.5 times as much)

solid disk I=

1/2mr^2 -mass kinda all over

sphere

2/5m(r)^2

axis of rotation

An imaginary line around which rotation occurs

distance travelled by an object rolling without slipping

D travelled= R(delta theta) [similar to arc length]

reminder work=

F(distance cos theta)

if i have some object where max distance from origin of rotation is r, then biggest I is mr^2 but if there's any. mass inward then

I is smaller than mr^2

kinetic energy for an object rolling without slipping

Kinetic energy total= ½(m)(v^2) + ½(I)(omega^2) -has both rotation and translational motion

velocities of an object rolling without slipping

Vcm= R(omega) [which makes sense bc it is the same as the tangential velocity] Vtop: 2(Vcm)= 2(R)(omega) Vbottom: 0

each point of a rotating object has, at any moment,

a linear velocity v and linear acceleration a

acceleration of com of an object rolling without slipping

a(com)= R(alpha) [same at atan]

linear acceleration vs. angular acceleration

a-> alpha (delta omega/ delta t)

a centrifuge produces 15 g's (centripetal acceleration) when rotating 45 RPM; how big is the centrifuge

a=15(9.8)= 147m/s^2 45 rot/min *2pi/1 rot *1 min/60 seconds =4.71s^-1 ac=r(omega^2) r=ac/omega^2 147m/s^2/(4.71s^-1)^2 =6.65m so around 13.3 meters

problem: kids on a mary-go-round 1. which child experiences the largest angular acceleration

all the same -not impacted by radius -delta omega/delta t -> all have same omega and same alpha

when the object rotates at a constant speed, what acceleration is felt

alpha= 0 -> centripetal acceleration is felt -> force is pointed in

a centrifuge produces 15 g's when rotating 45 RPM; what is its angular acceleration

alpha= change in omega/ change in time 4.71^s-1-0s^-1/ 60 s= .078 s^-2

rolling: point of contact is

always stationary

rigid object

an object with a definite shape that doesn't change, so that the particles composing it stay in fixed positions relative to one another

All points in a rigid object rotate with the same

angular velocity *clockwise: +* *counterclockwise: -*

tangential acceleration is

antiparallel to the radius -centripetal is parallel **umm this seems wrong** maybe ac= parallel or anti? atan= perp idk

how to get a torque=0

apply force at the point of rotation d=0 so t=0 -motion is translational only

total magnitude of acceleration is vector sum of

atan and aR sqrt of [at^2 + ar^2] =r(sqrt[omega^4 + alpha^2])

a centrifuge produces 15 g's when rotating 45 RPM; if alpha is constant and omega speeds up uniformly, then the atan = what?

atan= r*alpha (6.65m)(0.078s^-2) =.5187 m/s^2 =.05g's

car is moving at 60 miles an hour means

axel is moving at 60 mph (center of tires) -bottom of tire is 0mph (static contact) -top 120mph

reminder

calculator mode!

average angular acceleration (alpha)

change in angular velocity/ time -(omega)/time -same for all points on an object

problem: a bike wheel rotates 4.50 revolutions. How many radians has it rotated?

conversion factors: 4.5 rev(2pi rad/ 1 rev)= 28.3 radians

angluar displacement

delta theta= theta2- theta 1 -change in angle

work energy theorem for rolling without slipping

deltaK 1/2m(1+I/mr^2)(v^2)initial- 1/2m(1+I/mr^2)(v^2)final= F(distancecostheta) -can use to find velocity

work energy theorem reminder

deltaKE= work

linear velocity is greater for points

farther from the axis

more friction,

faster it starts rotating until the 2 velocities come together

torque=

force x length (N*m) or (ft*lbs)

example: how many revs does it make in this time? given omega= 2000 rpm alpha= 20.0 s^-2

found: 10.5 s to reach full speed of 209.4s^-1 theta= theta0 + omega0(t) +1/2(alpha)(t^2) theta= 0 + 0 +1/2(20)(10.5^2)= 1,102.5 radians 1,102.5*(1 rev/2pi radians)= 175.4 revolutions

rotational part: starts off sliding, but friction provides a --- which gets the ball to rotate (roll) find the angular acceleration and angular velocity

friction acts at bottom on the floor T=(mewk*mg)(r) I=2/5(m)(r^2) alpha= T/I (mewk*mgr)/[2/5mr^2] alpha=5mewkg/2r use omega= omega0 + alphat

the centripetal acceleration is --- the farther out a point is

greater

what has the biggest moment of I

hoop evenly centered around rotation axis bc all mass is max distance from center

translational motion and rotational motion are governed

independently -translational-> force -rotational-> torque

if the floor had no friction

it would never start rotating and v0 wouldn't ever slow down

theta=

l/r -if l=r then theta is 1 radian -therefore, l= r(theta)

dynamics linear vs. rotational force and acceleration

linear F=ma so a=F/m -bigger the F, bigger the a -bigger the m, smaller the a rotational Torque is the force *Torque=r*F*sintheta moment of inertia is mass I= sum(m*r^2) angular acceleration= torque/ moment of intertia -bigger torque, bigger angular acceleration -bigger I, smaller angular acceleration

hoop/ring

m(r)^2 -all mass is on the outside

rotational mass is

moment of inertia (I)

rotational motion

motion around the c.o.m.

spherical mass

mr^2 -every bit of mass is r from the center

when u throw a bowling ball, is it rotating?

not until it hits the ground vi= vi omega=0 vi= slows down omega= rotating with omega what makes its v slow down? -friction can treat as two problems for linear and rotational motion

frequency

number of complete revolutions per second f=omega/2pi *omega= 2pif*

average angular velocity

omega; delta(theta)/ delta time -instaneous is the limit approaching zero -raidans/ s or just s^-1

v= v0 + at ->

omega= omega0 + alpha(t)

v^2= v0^2 + 2a(xf-xi) ->

omega^2 = omega0^2 + 2(alpha)(thetaf-theta0)

overall, velocity of rolling without slipping is dependent

on the moment of inertia bigger I, the slower it gets after pulling it a certain distance by a force the force changes the energy when it does work

T=1/f

period; s (s/rev)

linear velocity is --- to the radius

perpendicular

what does the linear velocity and tangential velocity (omega*r) tell us

plot v0=highest and goes down linearly vE (tangential)= goes up linearly what happens at their crossing point? cross: condition for rolling without slipping V=Vtan V=omega(r) after they cross the two become one and it rolls away at a constant velocity

example holding a 5 lb weight: holding it in hand at 90 distance from elbow to hand= 35 cm what is the torque of the weight acting on the elbow

r= 0.35 m F= 5lbs-> 22.3 N T=r*F(sin theta) Torque= (0.35)(22.3)(sin 90) =7.805

Torque=

rFsin(theta) -twisting force -r is distance from force to hinge point

angular acceleration (alpha)

rad/s^2 (1/s^2) -rate of change of angular velocity alpha= delta(omega)/delta(t)

aR is the (can also be aC)

radial or centripetal acceleration -directions is toward the center of the point's circular path

r is the

radius of the circle in which the particle is moving *different for different particles*

Hertz

revs/second

essentially, just add up all of the

rotational motions to get the sum of the whole I I and T can be used to find angular acceleration T=(I)(alpha) alpha=T/I

wheel diagram rolling without slipping

rotational: if omega goes clockwise: pt at top of wheel-> v=r(omega) pt at bottom of wheel-> v=-omega(r) translational: if whole thing is translating right with v=(omegar) total: top pt= 2(omega)(r) mid pt= (omega)(r) bottom pt= (0)

linear part: the bowling ball is --- and has --- friction how do we find v0

sliding, kinetic F=-Mew(k)*Fn ->-Mew(k)*mg a=F/m a=[-Mew(k)(mg)]/m a=-mew(k)*g v=v0+at v=v0-mewk*g(t)

Rolling without slipping depends on

static friction between the rolling object and the ground -static friction determines how fast a car/bike/etc can accelerate

in general, I=

sum(mr^2) -depends on the body though

theta for torque is

the angle between the radius and force

one radian is defined as

the angle subtended by an arc whose length is equal to the radius

rotational kinetic energy

the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity 1/2(I)(omega^2) ->still in Joules

rotational inertia

the resistance of an object to changes in its rotational motion

once something starts rolling without slipping

there's no longer kinetic friction, its just static friction

the definitions for angular velocity and angular acceleration are just like those for their linear counterparts except that

theta replaces linear displacement (x), omega replaces v, and alpha replaces a

x=x0 + v0(t) + 1/2(a)(t^2) ->

theta= theta0 + omega0(t) + 1/2(alpha)(t^2)

a centrifuge produces 15 g's when rotating 45 RPM; how many times can someone go around as it goes to its max speed

theta= theta0 + omega0(t) + 1/2(alpha)(t^2) theta= 0+0+ 1/2 (0.078s^-2)(60s^2)= 140 radians 140 rads* 1 rev/2pi rads= 22 revs

example holding a 5 lb weight: holding it in hand at 90 distance from elbow to hand= 35 cm what is F of the bicep that is 5 cm away

torque= r(Fsintheta) r=torque/F*sintheta F=(7.805 Nm)/(.05)= 156.1 N-> 35.1 lbs

the motion of a rigid object can be analyzed as the

translational motion of the object's center of mass, plus rotational motion about its center of mass

example: how long does it take to reach full speed given omega= 2000 rpm alpha= 20.0 s^-2

use omega= omega initial + alpha(t) t=(omega-omega0)/alpha omega: 2000rev/min*2pi rad/ 1 rev * 1min /60 secs= 209.4s^-1 t=(209.4-0)/(20)= 10.5 s

find tangential velocity of the rotational motion of the edge point

v(edge)= omega(r) vede= 5mewkg/2(t)

linear vs. angular velocity

v-> omega (delta theta/ delta t)

linear velocity

v=deltal/deltat OR v[r*delta(theta)]/deltat AND SINCE deltatheta/deltat= omega *v=r(omega)*

problem: it's popular to put oversize wheels on jeeps and trucks so they ride higher. when someone does this the speedometer reads: higher, lower, the same speedometer measures how fast tires rotate (measures revs/ min) (does not measure with respect to ground)

v=r(omega) how fast rotating= omega reading lower than you're actually going consider tire: 1 rev with radius r -> distance travelled l= 2(pi)r in one rev the bigger the tires, the more distance you cover the speedometer is calibrated for a certain tire radius so if you put bigger tires it still is calculating based on smaller tires but you're actually traveling further each rev

aR/aC=

v^2/r OR (aR= (r*omega)^2/r) aR=r(omega)^2

what does this tell us about the velocity of a regular dragging box not being rolled?

v^2= 2Fx/m which means *v=sqrt[2ax]* because F/m is a

if rolling without slipping object is initially at rest, what is the equation for finding final velocity using work energy theorem

v^2= [2F(distance)]/ [m(1+I/mr^2)] v= sqrt[2ax/(1+I/mr^2)]

work for rotational motion is

work= torque(delta theta) work= Force*r*(delta l/r) work= F(delta l) -> force times distance as expected

linear vs angular position

x-> theta

problem: centrifuge acceleration. a rotor is accelerated for 30s from rest to 20, 000 rpm (revolutions per minute). (a) what is its average alpha? (b) through how many revolutions has the centrifuge rotor turned during its acceleration period, assuming constant angular acceleration?

(a) w0= 0. wf= 2pi(f)= 2100 rad/s -> avg alpha= w-w0/ delta t = 70 rad/s^2 -> (70 rads/s)(1rev/2pi rad)= 11 rev/ sec (b) theta= w0(t)+ 1/2(alpha)(t)^2= 3.15x10^4 rad -> 3.15x10^4 rad/2pi rad/rev= 5.0x10^3 rev -> theta = [omega^2-omega0^2]/2(alpha)= 3.15 x 10^4 rad

the kinetic energy of a hoop is

*(1/2)(m)[1+mr^2/mr^2]v^2 =(2)(1/2mv^2) two times KE of what KE would be if it was just a typical mass moving along

angular position of a rotating object is

*theta* how far it has rotated -a point moves through a distance l measured along the circumference of its circular path [-l is r(theta)] [-C=2pir]

what's another way to write the equation for rolling without slipping?

*½(m)(v^2) + ½(I)([v/r]^2) *(1/2)[m+(I/r^2)]v^2* -remember to distribute *(1/2)(m)[1+I/mr^2]v^2

8.2 constant angular acceleration

----

problem: how many radians in our field of vision does the sun take up? given r of sun: 7 x10^5 km given distance to sun: 150 x 10^6

->distance is "radius" ->diameter is delta l (subtends) so 2(7x10^5km)/ 150 x10^6 km= 1/100 radians 0.01 radians

problem: kids on a mary-go-round 2. which child experiences the largest tangential (linear) acceleration

-kid hanging on the edge ->r(omega) -impacted by radius

problem: kids on a mary-go-round 3. which child experiences the largest centripetal acceleration

-kid hanging on the edge ->v^2/r ->r(omega)^2 -impacted by radius

tangential acceleration

-tangent to circular path atan= r(alpha) OR r(delta omega)/deltat OR deltav/deltat

same problem but holding at 45 degrees 1. torque 2. force on bicep

Torque= r(F)(sin theta) Torque= (0.35 m)(22.3N)(sin 45)= 5.52 Nm Force= Torque/rsintheta = 156.1N-> 35 lbs *not much changed*