Relationship between Linear and Rotational Variables

Main Question or Discussion Point

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!
 

Answers and Replies

rcgldr
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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.
 
tiny-tim
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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:).
 
Doc Al
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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,
No it doesn't. Sounds like you (or he) have it backwards. How did he "prove" this?
we are assuming a 90 degree angle, so from r x (angular acceleration) = (linear acceleration) wouldnt we get r*aa=la --> aa=la/r.
That's the correct relationship.
 
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:
http://killercatfish.com/RandomIsh/images/Derivation.png" [Broken]

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:
http://killercatfish.com/RandomIsh/images/Formula.png" [Broken]

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?

Thanks!
 
Last edited by a moderator:
tiny-tim
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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.

Thanks!
 

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