Rotational Kinematics-YoYo Dropping from Ceiling

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SUMMARY

The discussion focuses on the analysis of a yo-yo's motion as it drops from a ceiling, specifically addressing the conversion of potential energy (PE) to kinetic energy (KE) in both translational and rotational forms. The correct final kinetic energy equation is established as KE = ½Mv² + ½Iω², where the moment of inertia I for a solid cylinder is I = (MR²)/2. The relationship between linear velocity (v) and angular velocity (ω) is clarified as v = ωR, leading to the derived velocity of the center of mass as v = sqrt(-4gh/3). The participants confirm the accuracy of the calculations and the simplification of the kinetic energy in terms of R, M, and v.

PREREQUISITES
  • Understanding of rotational dynamics and energy conservation principles.
  • Familiarity with the equations of motion, specifically v² = vi² + 2aX.
  • Knowledge of moment of inertia, particularly for a solid cylinder: I = (MR²)/2.
  • Basic grasp of angular velocity and its relationship to linear velocity in rolling motion.
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  • Study the principles of energy conservation in rotational systems.
  • Learn about the dynamics of rolling motion and the concept of rolling without slipping.
  • Explore the derivation of kinetic energy equations for various shapes and forms.
  • Investigate the applications of rotational kinematics in real-world scenarios, such as in machinery and sports.
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Students of physics, particularly those studying mechanics, educators teaching rotational dynamics, and anyone interested in the practical applications of energy conservation in rotational systems.

Darkalyan
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Homework Statement


Here's the problem in a JPG, with the image:

http://docs.google.com/Doc?id=d277r7r_50c9trfbdr

(the 25 points is b/c it's a practice test)

Homework Equations


The Attempt at a Solution



A)KE=0, PE=0 (defined to be at cylinder's initial position)B)PE=Mg(-h)C)Final KE=.5*M*(2gh)^2 (To get this, I used the v^2=vi^2 + 2aX equation, plugged it in for v^2 in KE=.5mv^2)However, I don't think my equations are right because if my KE was that, there would have to be extra energy put into the system for the yo-yo to rotate at all, thus there would be no rotational kinetic energy, which doesn't seem to jive with the original provlem. Where am I going wrong?
 
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Darkalyan said:
C)Final KE=.5*M*(2gh)^2 (To get this, I used the v^2=vi^2 + 2aX equation, plugged it in for v^2 in KE=.5mv^2)


However, I don't think my equations are right because if my KE was that, there would have to be extra energy put into the system for the yo-yo to rotate at all, thus there would be no rotational kinetic energy, which doesn't seem to jive with the original provlem. Where am I going wrong?

You are quite correct about you last eqn.

The final KE = ½Mv² + ½Iω²., where ω is related to v. This should take take of your problem.
 
Shooting Star said:
The final KE = ½Mv² + ½Iω²., where ω is related to v. This should take take of your problem.

Okay, so I know that I=(MR^2)/2, but you said that w is related to v. I don't see the relation?
 
Darkalyan said:
Okay, so I know that I=(MR^2)/2, but you said that w is related to v. I don't see the relation?

Without giving you a direct answer, think of the angular displacement of any point on the disk and its relationship to the distance traveled by the centre. What is the relationship between the v and ω of a wheel rolling without sliding? I am sure you can take it from here.
 
Hmm. Okay. So, if there's rolling w/out slipping, then v=wR. I then plugged that into the fact that -Mgh=½Mv² + ½Iω², and plugged into the w (that w=v/R). I did a bunch of algebra, and got that v=sqrt(-4gh/3), which works out to be a positive # b/c h is defined to be a negative distance fallen. Is that the right answer for the velocity of the center of mass?
 
Darkalyan said:
Hmm. Okay. So, if there's rolling w/out slipping, then v=wR. I then plugged that into the fact that -Mgh=½Mv² + ½Iω², and plugged into the w (that w=v/R). I did a bunch of algebra, and got that v=sqrt(-4gh/3), which works out to be a positive # b/c h is defined to be a negative distance fallen. Is that the right answer for the velocity of the center of mass?

Looks good, but I fail to see where the "lot" of algebra came in.

[-Mgh = ½Mv² + ½Iω² = (3/4)Mv², by putting ω=v/R , cannot be counted as a lot of algebra...:cool:]

However, looking at the original question, you are supposed to give the KE-rot in terms of R, M and v. That would just be (1/4)Mv².

I think that covers everything. Good job.
 
Shooting Star;1872431[-Mgh = ½Mv² + ½Iω² = (3/4)Mv² said:
I think that covers everything. Good job.

Sorry, I'm kind of lazy when it comes to simplification, so that's why I thought it was a 'lot' of algebra :). Thank you very, very much for your help.
 

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