Mechanics Problem

Discussion in 'Homework Help' started by lendo1, May 8, 2011.

  1. lendo1

    Thread Starter Active Member

    Apr 24, 2010
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    Say you have a pulley, with radius, R, and mass, M.

    This pulley is a disc with rotational inertia given by: I = (1/2)MR^2

    This pulley has a string wrapped around it with negligible friction, so that the pulley can be dropped, with the end of the string attached to a fixed point. The pulley, thus, appears to roll down the string.

    The pulley is dropped. What is the acceleration of the pulley in terms of the given variables and fundamental constants?
     
  2. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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  3. lendo1

    Thread Starter Active Member

    Apr 24, 2010
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    Very good Kermit, this is on Earth. So you have two forces acting on the pulley, the force of gravity (M*g), and the upward pull of the string. What is the acceleration?
     
  4. Georacer

    Moderator

    Nov 25, 2009
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    Is that your problem?

    [​IMG]

    Wouldn't it be nice if you had posted that image instead of me?

    Try this set of equation for starters:
    \left{ \begin{array}{c} \Sigma F = M \cdot \alpha \\<br />
\Sigma M = I \cdot \gamma \\<br />
\alpha = \gamma \cdot R \end{array} \right

    Where \alpha is the linear acceleration,
    \gamma is the rotational acceleration and
    M is the cylinder's torque.
     
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  5. lendo1

    Thread Starter Active Member

    Apr 24, 2010
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    That's it, thanks for the drawing! I tried using those equations (and others) and every time I have ended up setting torque to torque [(Rotational Inertia) * (Angular acceleration)] = (Tension * Radius), and my answers never seem to make any sense... too many variables cancel out; there is a property I am forgetting or an equation I am misusing, but I can't figure out where!
     
  6. Georacer

    Moderator

    Nov 25, 2009
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    Care to write the set of equations I gave you down and work with them a bit? The Homework Help section requires you to do some work too, as you might remember.

    Can you find an expression for the sum of the forces (as a function of the forces I noted on the sketch?)
    Can you find an expression for the sum of the torques, with a reference to the center of the cylinder?

    If you do that, using the proposed set of equations, you will end up with three equations and three unknowns.
     
  7. lendo1

    Thread Starter Active Member

    Apr 24, 2010
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    Let me show you what I came up with, please point me in the right direction :)

    My variables: T-tension, g-gravitational acc, a - linear acceleration, alpha - angular acceleration, R- radius, M - mass, I - Rotational Inertia

    Sum of forces:
    T - M*g = -M*a

    Sum of torques:
    TorqueNet = ?

    Isn't there only one torque acting on the disc? Torque = T * R ? Which has to be equal to (I * alpha)? That is the loop I keep finding myself in. I don't see how I can make a reference to the center of the cylinder using anything other than R, but that puts me back into this same situation where I*alpha = T*R, (which leaves me with two unknowns, tension AND linear acceleration).

    I ended up with a = (2T)/M
    Substituting that into my sum of forces equation gives me T = -Mg.... which can't be right
     
  8. Georacer

    Moderator

    Nov 25, 2009
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    Ok, another info income shot:
    \left{ \begin{array}{l}<br />
Weight-Tension=M \cdot \alpha \\<br />
Tension \cdot R=I \cdot \gamma\\<br />
\alpha=R \cdot \gamma \end{array}
     
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  9. lendo1

    Thread Starter Active Member

    Apr 24, 2010
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    Well I tried substituting into (Tension*R = I*/gamma) and I ended up with a = (2/3)*g. Definitely a believable result, but that means acceleration is independent of mass and radius... which I suppose makes sense. Is this right?

    Thanks!
     
  10. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    UPWARD pull of a string?

    You need to word your questions more carefully.

    Quote: "The pulley is dropped. What is the acceleration....?"

    You DID NOT give the pulley any definite rotation to start with, but simply defined a formula giving rotational inertia. Within the confines of your question my answer was totally appropriate----whether you find it helpful or not, would be up to you.

    Better questions get better answers.

    Kermit,
    Earth
     
  11. lendo1

    Thread Starter Active Member

    Apr 24, 2010
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    Yes, the UPWARD pull of the string. Relative to the disc, the force of tension from the string is UP, opposing the downward force of gravity. It was a tricky problem, Kermit, no worries.
     
  12. Georacer

    Moderator

    Nov 25, 2009
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    What Kermit is trying to teach you is to pick your words more carefully.
    The string isn't moved upwards and therefore the word "pull" isn't suitable. A better phrase would be "and the upwards tension from the string".

    My calculations agree with your answer.
     
  13. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    It is very easy to leave out pertinent details when one is trying to describe a problem. The problem being the asker is deeply immersed in the mechanics of the situation and unknowingly leaves out IMPORTANT details that someone not so involved with the problem NEEDS to be informed of. Go back and re-read your original post as if you don't know anything at all about the problem. You will see the details that tell others that the (yo-yo) is already spinning when it is dropped are not present. You simply ask what happens when you drop it, without saying anything about the starting conditions. You assumed that the rotational inertia formula was all that was needed, and for YOU it was, but the rest of us would have to be mind readers to gather this info from your original posting. Hence my rather 'snarky' answer. Your witty comebacks are all well and good, no offense taken, but as Georacer said, that was not the point I was trying to convey with the wiki link on earths gravity.
     
  14. lendo1

    Thread Starter Active Member

    Apr 24, 2010
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    Next time I ought to include a diagram, too. Well, thanks to both of you, I'll be back to exploit this forum next time I need help.
     
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