# String wrapped around a disk

## Homework Statement

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.

## Homework Equations

Krot = F*d
KE = 1/2 m v^2[/B]
K= 1/2 Iw^2

## 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[/B]

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ehild
Homework Helper

## Homework Statement

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
Correct.

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.

## Homework Equations

Krot = F*d
KE = 1/2 m v^2[/B]
K= 1/2 Iw^2

## 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[/B]
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:
So are you saying it is 4.42 + .5(m)(1.2687)^2 ?

ehild
Homework Helper
So are you saying it is 4.42 + .5(m)(1.2687)^2 ?
No.

SammyS
Staff Emeritus
Homework Helper
Gold Member
So are you saying it is 4.42 + .5(m)(1.2687)^2 ?
@ Westin,

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

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

Krot would be Work - Translational KE ?

SammyS
Staff Emeritus
Homework Helper
Gold Member
Krot would be Work - Translational KE ?
Yes,

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

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

I have one try left

SammyS
Staff Emeritus