Question

You have a lab apparatus with two disks on an axle that are able to rotate freely. The disks are not initally in contact with each other and are both starting from rest. You apply a constant torque to the lower disk (disk a) to produce an angular acceleration of 10.0 rad/s^2 then remove the torque. Both disks hae the same moment of inertia, which is 1.25 kg m^2. What is the angular velocity of disk A at the end of the 5.00 s. What was the angular displacement of disk A during this time?

Answer 1

Even if you do not know the answer but know some concept of it maybe we could attempt it together. It was a problem on my test that I received partial credit on and my final is coming up so I would like to make sure I know how to complete this problem incase one similar pops up. If posting a picture of the work that I completed and the teachers notes would help someone it would be greatly appreciated. I would just like to have worked on it a little bit before I bring it to my teacher to ask her for help.

Answer 2

If you can just post a picture of the problem what would be helpful

Answer 3

Sure thing, give me a second. I will also post a photo of the work that I did do. So glad someone replied.

Answer 4

Question two obviously.

Answer 5

ok lets see your work please

Answer 6

never mind...Do you know the basic kinematic equations for circular motion assuming a constant angular acceleration?

Answer 7

yes. I also know torque equations. Like I posted before, I wasn't sure how to add the two together. Or how to use them together. I obviously had the gist of the problem correct because I received 15 out of 20. That doesn't mean that I understood it, and that's what I need to do.

Answer 8

ok so for part a)
alpha is constant = 10 rad/s^2
t = 5 seconds
so what equation can I use to find omega (angular velocity) at the end of 5s?

Answer 9

correct which was 50.0 rad/s

Answer 10

w=w0+at

Answer 11

ok for b) what equation can be used to solve for the angular dispalcement during the same time?

Answer 12

displacement...

Answer 13

w=change in theta/change in time

Answer 14

250 rad

Answer 15

please tell me if I'm wrong, obviously

Answer 16

no....
theta = theta (initial) + omega(initial)*t + 1/2*alpha*t^2
Does that look familiar?

Answer 17

no.

Answer 18

oh wait

Answer 19

yeah.

Answer 20

w=change in theta/change in time if for omega average not for omega at an instant

In part a) you solved for omega after time t=5s , which is omega at that instant in time

Answer 21

theta = theta (initial) + omega(initial)*t + 1/2*alpha*t^2

So you can use this to solve for b....what is
theta(initial) = ?
and omega (initial) = ?
these come from reading the problem carefully.

Answer 22

0

Answer 23

yes so the equation reduces to:
theta = 1/2*alpha*t^2 and you know alpha and t...solve for theta please

Answer 24

125

Answer 25

wati

Answer 26

oh yeah right

Answer 27

ok...you should put in the units...125 what?

Answer 28

rad/s

Answer 29

no those are the units for angular velocity (part a)...try again

Answer 30

rad

Answer 31

Great!

Answer 32

sorry I thought we were correcting my part A. was confused for a minute

Answer 33

ok for part c, do you know what principal to apply to solve it?

Answer 34

Iw=I0W0 ?

Answer 35

or L=iw where Li=Lf?

Answer 36

Close...yes we use conservation of angular momentum..

Answer 37

ooohhh

Answer 38

Now, the first disc a rotates and an omega value which you already solved for.
You now bring it into contact with disc b which is initially at rest.
Now the two discs begin to rotate together.
Do you know how to describe this mathematically?

Answer 39

inertia?

Answer 40

Ia*omegaa +Ib*omegab=( Ia + Ib)0*omega(together)

Answer 41

sorry misunderstood

Answer 42

The left hand side is the initial total angular momentum and the right side is the final total angular momentum ....agreed?

Answer 43

sure. obviously. I need to study that part more.

Answer 44

ok...now what does omegab = ? from the problem description?

Answer 45

finally would be the 10 rad/s2

Answer 46

final*

Answer 47

wait

Answer 48

you are confusing alpha with omega...
omegab is the initial angular velocity of disc b but disc b starts from rest ...therefore omegab = ?

Answer 49

0

Answer 50

actually, where I am confused is I call it W and A; which I shouldn't i'm aware.

Answer 51

the angular velocity is the greek letter omega
the angular acceleration is the grek letter alpha

yes...so rearranging the equation above with omegab = 0 gives:

omega(together) = Ia*omegaa/(Ia + Ib)

Answer 52

Does that help?

Answer 53

when you say together is that another way to say final?

Answer 54

yes...

Answer 55

okay I thought so, so when you gave me the equation for part b. I was looking through my list of kinematics for circular motion and I do not see any one where it has theta = theta (initial) ect.

Answer 56

you should look them up and note that they only apply when alpha is constant.(not a function of time)

If alpha is not constant but depends on time, then you need calculus to solve those more difficult problems.

Answer 57

the closet one out of the four that the teacher presented is thetha=omeg a t+ (1/2)at^2

Answer 58

he/she has just set omega(initial) = 0.

Answer 59

interesting. did you calculus to get to that equation out of the 4 equations?

Answer 60

use*

Answer 61

okay hold on

Answer 62

The four equations are derived from calculus but all you need to do at this time is memorize them.

Answer 63

Okay. Thank you. I will rework this with the equations you have given me. I'm still confused on how and why she set omega initial to zero but I will ask her about that.

Answer 64

your welcome. goodnight