Find the angular speed of the cylinder

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SUMMARY

The discussion focuses on calculating the angular speed of a uniform cylinder with mass M and radius R, around which a flexible rope with mass m is wrapped. The cylinder starts with an initial angular speed ω0 and rotates without friction. After one complete revolution, the rope hangs vertically, and the final angular speed ωf can be expressed using the equation ωf = √{(mGR + 1/4MR^2ω0^2)/(1/2mv^2 + 1/4MR^2)}. The relationship between linear velocity v and angular speed ω is crucial for solving the problem.

PREREQUISITES
  • Understanding of rotational dynamics and conservation of energy principles.
  • Familiarity with the moment of inertia, specifically for a uniform cylinder.
  • Knowledge of the relationship between linear velocity and angular speed (ω = v/r).
  • Basic algebraic manipulation skills to solve equations involving multiple variables.
NEXT STEPS
  • Study the conservation of energy in rotational motion, focusing on potential and kinetic energy transformations.
  • Learn about the moment of inertia for different shapes, particularly the uniform cylinder.
  • Explore the relationship between linear and angular quantities in rotational systems.
  • Practice solving similar problems involving pulleys and rotating bodies to reinforce understanding.
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Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators seeking to clarify concepts related to angular motion and energy conservation.

PKay
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Homework Statement



Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R. The cylinder rotates without friction about a horizontal axle along the cylinder axis. One end of the rope is attached to the cylinder. The cylinder starts with angular speed ω0. After one revolution of the cylinder the rope has unwrapped and, at this instant, hangs vertically down, tangent to the cylinder.

Find the angular speed of the cylinder at this time. You can ignore the thickness of the rope. (Hint: Use Equation U=Mgycm.)
Express your answer in terms of the variables m, M, R, and appropriate constants.

Homework Equations



y = 1/2 M R^2
ω=v/r


The Attempt at a Solution



I know there's a post for the same problem. But I'm still confused. If I was supposed to ask that person on their thread or something, please tell me so I can delete this thread. I'm new to the forum. Anyways, this is what I tried:
mgh +1/2Iω02 = 1/2mv2 + 1/2Iwf2
mgR+1/2(1/2MR2ω02 = 1/2mv21/2(1/2MR2wf2
ωf = √{(mGR+1/4MR2ω02)/(1/2mv2+1/4MR2}

Apparently, according to masteringphysics, the answer does not depend on mv but now I'm a little stuck. I'm not sure what I should be substituting mv with. Thank you in advance!
 
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PKay said:

Homework Statement



Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R. The cylinder rotates without friction about a horizontal axle along the cylinder axis. One end of the rope is attached to the cylinder. The cylinder starts with angular speed ω0. After one revolution of the cylinder the rope has unwrapped and, at this instant, hangs vertically down, tangent to the cylinder.

Find the angular speed of the cylinder at this time. You can ignore the thickness of the rope. (Hint: Use Equation U=Mgycm.)
Express your answer in terms of the variables m, M, R, and appropriate constants.

Homework Equations



y = 1/2 M R^2
ω=v/r


The Attempt at a Solution



I know there's a post for the same problem. But I'm still confused. If I was supposed to ask that person on their thread or something, please tell me so I can delete this thread. I'm new to the forum. Anyways, this is what I tried:
mgh +1/2Iω02 = 1/2mv2 + 1/2Iwf2
mgR+1/2(1/2MR2ω02 = 1/2mv21/2(1/2MR2wf2
ωf = √{(mGR+1/4MR2ω02)/(1/2mv2+1/4MR2}

Apparently, according to masteringphysics, the answer does not depend on mv but now I'm a little stuck. I'm not sure what I should be substituting mv with. Thank you in advance!
Hello PKay. Welcome to PF !

If you are referring to this thread as being the previous post, then you are correct in starting this new thread. For one thing, that thread is rather old. Also, it has been closed.


There is a relationship between v and ω . That should enable you to eliminate one of them.
 
SammyS said:
Hello PKay. Welcome to PF !

If you are referring to this thread as being the previous post, then you are correct in starting this new thread. For one thing, that thread is rather old. Also, it has been closed.There is a relationship between v and ω . That should enable you to eliminate one of them.

Yes! I was referring to that thread. I didn't know it was closed. Glad I did it correctly.

I actually got it now. I actually still have a related question. C=2piR but why do we have to divide by 2 for the center of mass?
 
Last edited:

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