Question

a compact disc spins at 2.3 revolutions per second. an ant is walking on the CD and finds that it just begins to slide off the CD when it reaches a point 3.1cm from the CD center

Answer 1

@roadjester can you help us both on this please

Answer 2

so what are you looking for?

Answer 3

Is that the whole question?

Answer 4

@m806texas
@vsandrasue

Answer 5

did it cut off when it reaches a point of 3.1cm from the CD's center.

Answer 6

HUH?

Answer 7

a compact disc spins at 2.3 revolutions per second. an ant is walking on the CD and finds that it just begins to slide off the CD when it reaches a point 3.1cm from the CD's center.

Answer 8

can u help me find the answer?

Answer 9

@vsandrasue you have to type the question .....like .... whats the coefficient of friction or whats acceleration

Answer 10

I need the WHOLE question. EVERYTHING that was given to you.

Answer 11

oh the question is What is the coefficient of friction between the ant and the CD?

Answer 12

so is the movement of the ant perpendicular to the CD?

Answer 13

it doesnt say. was i used this formula Efx= max=mv2/r to find Us

Answer 14

Ummm, what does that notation stand for?

Answer 15

maybe im not understanding which formula to use to find?

Answer 16

You're on a spinning disk so this problem requires angular movement.

Answer 17

so there is no formula just an answer?

Answer 18

A compact disc spins at 2.3 revolutions per second. An ant is walking on the CD and finds that it just begins to slide off the CD when it reaches a point 3.1 cm from the CD's center.
(a) What is the coefficient of friction between the ant and the CD?

Incorrect: Your answer is incorrect.

(b) Is this the coefficient of static friction or kinetic friction?
static friction
kinetic friction
Correct: Your answer is correct.

Answer 19

i sent you the copy off my homework

Answer 20

\(\huge \vec F=m\vec a_c=m{v^2\over r}={m(r\omega)^2\over r}={\dfrac {mr^2\omega ^2 } r=mr\omega^2}\)

Answer 21

it says on the empty box i have to fill in( to enter a number)

Answer 22

And what is the "Incorrect: Your answer is incorrect." and "Correct: Your answer is correct."?

Answer 23

do i have to find the omega first?

Answer 24

there is to questions the second answer is correct. the first question was wrong which was the equation i feel i did wrong.

Answer 25

omega is given. it is your angular velocity which is currently in revolutions per second You need to convert it to radians per second.

Answer 26

2\(\pi\) radians=1revolution

Answer 27

ok i did that and the answer i got was .241

Answer 28

first answer was 144.513/60 this is my radius per/sec and got the answer of .241

Answer 29

0.241 rad/s?

Answer 30

you have 2.3 rev/s. to convert it to rad/s, you multiply by 2pi

Answer 31

this what i did i put 2.3*2pie divide by 60 seconds.

Answer 32

why did you divide by 60s?

Answer 33

i thought i needed to to change to radius/per second. i dont know what i did can you help me find the answer.

Answer 34

not radius per second; radian per second

Answer 35

radian is another unit to represent degrees

Answer 36

for a circle, it is 360 degrees
a circle is also 2\(\pi\) radians

Answer 37

ok radian. can u help me get the right answer please. this what i did
(2.3rev/sec)(2pie/1rev)(1min/60sec) which equaled (144.513/60) = .241 was that wrong?

Answer 38

why do you have a time conversion factor?
2.3 rev/s(2\(\pi\) rad/1rev)=14.45 rad/s

Answer 39

rad/s is the unit for omega

Answer 40

so what did i do wrong

Answer 41

don't divide by 60, keep the units at rad/s and that is your unit for omega

Answer 42

oh ok so its just 144.513 right

Answer 43

no, 14.4513

Answer 44

decimal is in the wrong place

Answer 45

your right and then use the mvsquared/r to get my Us

Answer 46

what is Us?

Answer 47

mue

Answer 48

u know that weird looking u the mue

Answer 49

ok use this
\(\large\mu\)

Answer 50

sorry it pronounced mu

Answer 51

backslash, open parentheses, backslash, mu, backslash, close parentheses

Answer 52

yes, do i need to solve for that to use the radius

Answer 53

I'm going to be honest, this problem is whack
Not sure if this will work but I'm thinking
\(\huge F_k=\mu_kn=\mu_kmg=mr\omega^2\), from there \(\huge \mu_k={\dfrac {r\omega^2}g}\)

Answer 54

Not sure if that will work since this is three-dimensional and angular, not translational

Answer 55

I get .6606189

Answer 56

0.6606189

Answer 57

ok thats what i got let me put in the answer and see if thats correct

Answer 58

that was correct thank you.

Answer 59

shouldn't thank me
thank @m806texas
I think she said you two were classmates.