String wrapped around a disk

  • Thread starter Westin
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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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  • #2
ehild
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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.
 
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  • #3
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So are you saying it is 4.42 + .5(m)(1.2687)^2 ?
 
  • #4
ehild
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So are you saying it is 4.42 + .5(m)(1.2687)^2 ?
No.
 
  • #5
SammyS
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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, ...
 
  • #6
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Krot would be Work - Translational KE ?
 
  • #7
SammyS
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Krot would be Work - Translational KE ?
Yes,

Rotational Kinetic Energy is equal to the total work done minus the Translational Kinetic Energy .
 
  • #8
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Krot = (.34*13) - (.5(2.1)(1.2687)^2) ?

I have one try left
 
  • #9
SammyS
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Krot = (.34*13) - (.5(2.1)(1.2687)^2) ?

I have one try left
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.
 

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