Question

A CD has a mass of 13.9 g, an inner diameter of 1.5 cm, and an outer diameter of 11.9 cm. Suppose you toss it, causing it to spin at a rate of 4.5 revolutions per second.

If your fingers were in contact with the CD for 0.29 revolutions while it was acquiring its angular velocity and applied a constant torque to it, what was the magnitude of that torque?

Answer 1

COOL QUESTION MAN!!
i'm working on it

Answer 2

okey, previously question asked the moment of inertia of the CD which is 2.500×10-5 kg*m^2. i thought we might use Torque=( Moment of Inertia of Com) * ( angular momentum of com ) since we know the moment of inertia. But i failed to find the angular momentum.

Answer 3

also the answer is 5.483×10-3 N*m.

Answer 4

i think that the part of the question "If your fingers were in contact with the CD for 0.29 revolutions while it was acquiring its angular velocity and applied a constant torque to it" may have something in it

Answer 5

okey, previously question asked the moment of inertia of the CD which is 2.500×10-5 kg*m^2. i thought we might use Torque=( Moment of Inertia of Com) * ( angular acc of com ) since we know the moment of inertia. But i failed to find the angular acc. Angular acc i meant

Answer 6

yup

Answer 7

are you sure about the moment of inertia??

Answer 8

one thing that confuses me is that while the guy puts his finger on the CD does he slow down the spin rate?

Answer 9

no the fingets provide the torque to the disc
and i think that here, the equation for rotational kinetics would be applied
the final ang. vel.=4.5 , the initial = 0
and the ang disp.=0.29 so by the equation
(omegaf)^2-(omegai)^2=2(theta)(ang. accn)

Answer 10

and yes we will get the ang. accn from here and we will multiply it by the moment of inertia

Answer 11

is ang. vel. initially 0?

Answer 12

yes i think so as the sentence "in contact with the CD for 0.29 revolutions while it was ACQURING its angular velocity" says

Answer 13

according to my calc we dont get 5.483*10^-3

Answer 14

even i am getting 4.38 x 10^-2

Answer 15

i got moment of inertia = 0.001256

Answer 16

in si units

Answer 17

i used this and found the answer 2.5*10^-5

Answer 18

i think it should be ((R2)^2 - (R1)^2)
i mean according to me but i'm searching on it

Answer 19

i found the answer at last! You are THE BEST. thanks a lot. silly me forgot to convert rev to radians.

Answer 20

photo from university physics pearson 13. international edition. I usually trust a book if it is like 1500 pages :D

Answer 21

oh yeah man even i forgot to convert it
and thanks for calling me the best
and thanks for uploading such a good question
i mean thanks lot
and i am looking foward to buy this book as you use it

Answer 22

and yup here's your medal to respond to the question so actively

Answer 23

:D thanks man. I am gonna take the exam so i better be active. No one likes to repeat a year right?

Answer 24

yeah....hahaha

Answer 25

which class are you studying in???

Answer 26

The class is called physics 101 and im "trying" to be a civil engineer. Class gets significantly harder if you dont go to lecture :D

Answer 27

to become a civil eng. is a very great thing and the profession has many adventurous things to do and i know that you will surely be a "succesful" civil eng. and all the best for your future:)

Answer 28

Thanks ! Same goes for you, you are far more than just a 11th grader! Hope you do amazing in life!

Answer 29

and also i felt that you are the most sensible person , i've met on openstudy
and thanks a lot for your wishes

Answer 30

Well you helped me a lot and i appreciate your efforts! This is like volunteer work everyone needs to be very polite i think!

Answer 31

yes i think so

Answer 32

that everyone needs to be very polite

Answer 33

See you later! Thanks again for your company!

Answer 34

you're wecome
adios amigo hasta pronto(lol):)

Answer 35

and even thanks