# Homework Help: Rotational Motion relationship between 2 disks

1. Oct 26, 2007

### gills

1. The problem statement, all variables and given/known data
Two disks are mounted on a frictionless vertical shaft of neglible radius.

The lower disk, of mass 440g and radius 3.5cm, is rotating at 180rpm on the frictionless shaft of neglible radius. The upper disk, of mass 270g and radius 2.3cm, is initially not rotating. It drops freely down the shaft onto the lower disk, and frictional forces act to bring the two disks to a common rotational speed.

(a) What is that speed?
(b) What fraction of the initial kinetic energy is lost to friction?

2. Relevant equations

T = tau
w= omega
m1 = mass lower disk
m2 = mass upper disk
I = rotational inertia = (1/2)mR^2 (for disks)
upper disk = ud
lower disk = ld
alpha = angular acceleration
a(tan) = tangential linear acceleration
t=time

Ok, i will just pop out some equations:

T = I*alpha
w = w0 + alpha*t
a(tan) = alpha*R
K(rotational) = (1/2)Iw^2

3. The attempt at a solution

can we somehow use the K(rotational) equation to solve both?

well first i converted (inital omega of the lower disk) w0(ld) 180rpm = 18.8 rad/s
w0(ud) = 0

we know that wf(ud) = wf(ld) and we need to figure that out

the m has to be in kg so -->
m(ld) = 0.440kg m(ud) = 0.270kg

Kf - Ki = deltaK lost from frictional force?

(1/2)(m1 + m2)*wf^2 - (1/2)(m1)w0^2 = delta K lost?

Any help would be great.

2. Oct 26, 2007

### Staff: Mentor

what's conserved?

When the upper disk drops onto the lower disk, is anything conserved?

3. Oct 26, 2007

### gills

hmmmm...i'm guessing momentum?

4. Oct 26, 2007

### Staff: Mentor

Make that angular momentum.

5. Oct 26, 2007

### gills

ok, talk to me Doc!

6. Oct 26, 2007

### Staff: Mentor

Set up an equation for conservation of angular momentum.

7. Oct 26, 2007

### gills

I'm reading the chapter as we speak...give me a few moments...

8. Oct 26, 2007

### gills

I1w1 = I2w2

9. Oct 26, 2007

### gills

[(1/2)M1*R1^2]*w0 = wf[(1/2)M2*R2^2 +(1/2)M1*R1^2] -->

wf = [(1/2)M1*R1^2]*w0 / [(1/2)M2*R2^2 +(1/2)M1*R1^2]

how's that look?

10. Oct 26, 2007

### Staff: Mentor

Wonderful.

11. Oct 26, 2007

### gills

it certainly is...because it's correct!

I'm starting to love you Doc Thanks...AGAIN!

Now, i'm moving onto the next part...

12. Oct 26, 2007

### gills

Ok doc.

To determine energy lost to frictional force, energy is not conserved. Therefore-->

Ki - Kf = $$\Delta$$K ---> K lost

Of course we'll be using K for rotational motion which =

(1/2)Iw^2 -->therefore --->

Ki = (1/2)I1 * w1i^2 -->no angular velocity on I2 so = 0 -->
Kf = (1/2)wf^2 * [I1 + I2]

(1/2)I1 * w1i^2 - (1/2)wf^2 * [I1 + I2] = $$\Lambda$$K

Is this the right idea?

13. Oct 26, 2007

### Staff: Mentor

You got it!

14. Oct 26, 2007

### gills

yes, it's right. 20.6% change in K.

Doc, is it possible to solve this problem using a different method?

Such as Tau = I*alpha or using cirlcular motion equations?

15. Oct 26, 2007

### Staff: Mentor

This is by far the easiest way, since you don't have to know anything about the details of the forces between the two disks or the time it takes them to reach a common speed. (But you should be able to solve it by making up some generic assumptions about forces.)

16. Nov 3, 2007

### tinkertas

I am working on a very similar problem but am having trouble calculating the kinetic energy lost to friction. Is this just the change in kinetic energy? If so, my answer is off by 0.3% which is a lot for this particular problem.

17. Nov 3, 2007

### gills

follow the steps i went through and you shouldn't have a problem getting that number. Don't round the significant figures toward the end, and you should get that number.

K_i - K_f = K_lost