Question on rotational dynamics

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The discussion revolves around deriving the equation α = mgr/(I + mr²) related to rotational dynamics in a lab setting involving a mass, pulley, and a rotating disk. Participants express confusion about the problem's requirements and the lack of clarity in the initial description of the experiment. Key concepts such as moment of inertia (I), mass (m), and radius (r) are highlighted as essential for understanding the derivation. A solution is provided that connects the string's acceleration to the disk's angular acceleration and outlines the tension in the rope. The conversation emphasizes the importance of clear communication when presenting experimental setups to avoid misunderstandings.
mathlete
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Here is the question (it's part of a lab):

"From your lectures in mechanics you should be able to derive:
\alpha = \frac{mgr}{I+mr^2}
This equation ignores the effect of the pulleys and the mass of the connecting string"

The lab was basically setting up a mass on a string over a pulley, connected to a rotation horizontal disc which would oscillate back and forth (if this information is needed).

Don't know where to start - I thought maybe conservation of energy, PEi + KEi = PEf + KEf, but I can't get anywhere.
 
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What is the question? Do they want you to derive that equation?
 
Sirus said:
What is the question? Do they want you to derive that equation?
Yes, that was the whole question. Don't know what to do..
 
Anyone have an idea?
 
Since you haven't bothered with the simple courtesy of defining your terms, nope.
 
arildno said:
Since you haven't bothered with the simple courtesy of defining your terms, nope.
If you don't know that I is moment of inertia and m is mass, you probably can't answer it anyway.
 
I of what?
m of what?
r of what?
 
For what it's worth, here's your solution:
The magnitude of yhe string's acceleration is related to the disk's angular acceleration by:
|a|=r\alpha
where the ang. acc. is considered greater than zero.
Hence, the tension in the rope is given by:
T=mg-mr\alpha
Hence, the momwnt-of momwntum wquation about the disk's center is:
mgr-mr^{2}\alpha=I\alpha

Next time, show a proper measure of respect towards your readers than some muddøed half-baked account of an experiment..
 
arildno said:
For what it's worth, here's your solution:
The magnitude of yhe string's acceleration is related to the disk's angular acceleration by:
|a|=r\alpha
where the ang. acc. is considered greater than zero.
Hence, the tension in the rope is given by:
T=mg-mr\alpha
Hence, the momwnt-of momwntum wquation about the disk's center is:
mgr-mr^{2}\alpha=I\alpha

Next time, show a proper measure of respect towards your readers than some muddøed half-baked account of an experiment..
I was just reading the problem straight from the book, verbatim. Thanks for the answer though :blushing:
 
  • #10
All right, just an advice:
Next time you relate an EXPERIMENT you've done, you need to remember that others can't read your mind. As you very sketchily described the experimental set-up, it was liable to several interpretations, not just one.

In particular, the role of the horizontal disk is most unsatisfactorily described!
It took some time to figure out the real experiment behind your words..:wink:
 
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