# String wrapped around a disk

1. Apr 5, 2015

### Westin

1. The problem statement, all variables and given/known data
A string is wrapped around a disk of mass 2.1kg and radius 0.045m. Starting from rest, you pull the string with a constant force of 13N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance 0.13m, your hand has moved a distance 0.34m.

At this instant, what is the speed of the center of mass of the disk?

vcm= 1.2687 m/s

At this instant, how much rotational kinetic energy does the disk have relative to its center of mass?

Krot= ?

I provided the image for a visual aid.

2. Relevant equations

Krot = F*d
KE = 1/2 m v^2

K= 1/2 Iw^2
3. The attempt at a solution

Krot = F*d
Krot = (13)(.34m) = 4.42m
This is assuming all force from hand exerts to rotational KE, not sure what I am doing wrong.

My other attempt was trying to use energy conservation mgh=Iω^2/2

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2. Apr 5, 2015

### ehild

Correct.

Not the whole work will increase the rotational energy.

The work of the force is W=Fd. But that work makes the centre of the disk translate and it also rotates the disk abut its centre. The work is equal to the change of the kinetic energy, which is the sum of the translational and rotational KE-s.

Last edited: Apr 5, 2015
3. Apr 5, 2015

### Westin

So are you saying it is 4.42 + .5(m)(1.2687)^2 ?

4. Apr 5, 2015

No.

5. Apr 5, 2015

### SammyS

Staff Emeritus
@ Westin,

You should state what you mean by the word ' it ' .

Rotational K.E., Total energy, Work, ...

6. Apr 5, 2015

### Westin

Krot would be Work - Translational KE ?

7. Apr 5, 2015

### SammyS

Staff Emeritus
Yes,

Rotational Kinetic Energy is equal to the total work done minus the Translational Kinetic Energy .

8. Apr 5, 2015

### Westin

Krot = (.34*13) - (.5(2.1)(1.2687)^2) ?

I have one try left

9. Apr 5, 2015

### SammyS

Staff Emeritus
Isn't that the same as 4.42 + .5(m)(1.2687)^2 ?

That looks right, but if you tried it before, you might want to hold off.