Rotational Motion relationship between 2 disks

[gills]

Homework Statement

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?

Homework Equations

T = tau
w= omega
R = Radius
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

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.

 

[Doc Al]

what's conserved?

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

 

[gills]

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

hmmmm...i'm guessing momentum?

 

[Doc Al]

Make that angular momentum.

 

[gills]

Make that angular momentum.

ok, talk to me Doc!

 

[Doc Al]

Set up an equation for conservation of angular momentum.

 

[gills]

Set up an equation for conservation of angular momentum.

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

 

[gills]

Set up an equation for conservation of angular momentum.

I1w1 = I2w2

 

[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?

 

[Doc Al]

Wonderful.

 

[gills]

Wonderful.

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

the answer is 14.9 rad/s = 142rpm

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

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

 

[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?

 

[Doc Al]

You got it!

 

[gills]

You got it!

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?

 

[Doc Al]

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

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.)

 

[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.

 

[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