Deriving angular acceleration for a mass-pulley-disc system

[mathlete]

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.

 

[Sirus]

What is the question? Do they want you to derive that equation?

 

[mathlete]

What is the question? Do they want you to derive that equation?

Yes, that was the whole question. Don't know what to do..

 

[mathlete]

Anyone have an idea?

 

[arildno]

Since you haven't bothered with the simple courtesy of defining your terms, nope.

 

[mathlete]

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.

 

[arildno]

I of what?
m of what?
r of what?

 

[arildno]

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..

 

[mathlete]

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

 

[arildno]

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..