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Relationship between Linear and Rotational Variables

  1. Mar 17, 2008 #1
    My main confusion is in the proof my professor showed us just before break. He came up with a relationship of (angular acceleration)=radius*(linear acceleration) which doesnt make sense, we are assuming a 90 degree angle, so from r x (angular acceleration) = (linear acceleration) wouldnt we get r*aa=la --> aa=la/r.

    I am in need of this to make the conversion in a problem where I can estimate the velocity of an action (and derive the velocity from the acceleration). But have to start with the basic Fnet=Iaa, and the only force is the torque.

    Any clarification would be great. THANK YOU!
  2. jcsd
  3. Mar 17, 2008 #2


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    You should use radians instead of degrees. Radian is a bit unusual in physics, in that it's a pseudo-unit. When converting from rotational movement to linear movement, "radian" can be simply dropped, and when converting from linear to rotational movement, "radian" can be added.

    Force x radius = torque, so force = torque / radius.
  4. Mar 18, 2008 #3


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    linear acceleration = radius x angular acceleration

    Hi killercatfish! :smile:

    Angular acceleration is 1/time^2.

    Linear acceleration is length/time^2.

    Radius*(linear acceleration) is length^2/time^2.

    It should be linear acceleration (tangential, with fixed radius) = radius*angular acceleration. :smile:

    Your professor must be wrong :frown: (or sadly misunderstood! :cry:).
  5. Mar 18, 2008 #4

    Doc Al

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    No it doesn't. Sounds like you (or he) have it backwards. How did he "prove" this?
    That's the correct relationship.
  6. Mar 19, 2008 #5
    Here is how I derived angular acceleration to linear acceleration:

    > Radius*theta = ArcLength;
    > d(Radius*theta)/dt = ds/dt;
    > Radius*omega = V;
    > d(Radius*omega)/dt = dV/dt;
    > Radius*alpha = a;
    > alpha = a/Radius;

    here is a link for a clearer image:

    And this is what I came up with for the impulse:

    > tau = Radius*`sinθ`*Force;
    > Fnet = I*alpha;
    > Fnet = I*a/Radius;
    > Fnet = tau;
    > MI := (1/2)*mass*(R^2+R[o]^2);
    > Radius*Force = MI*alpha;
    > Force = MI*`ΔV`/(`Δt`*Radius);
    > Force*`Δt` = MI*`ΔV`/Radius;

    here is a link for clearer image:

    Could someone A, let me know if this is correct, and B, help me to understand how this will give me the radius which is the point of breaking?

  7. Mar 19, 2008 #6


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    Breaking what? You haven't set out the question … :confused:
  8. Mar 19, 2008 #7
    HAHA! Oh man, thank you for pointing out the key flaw.. :)

    This is to describe the breaking of a piece of toilet paper off the roll. I have read the other post on the site, but it wasnt heavily equation laden.

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