Relationship between Linear and Rotational Variables

[killercatfish]

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 doesn't make sense, we are assuming a 90 degree angle, so from r x (angular acceleration) = (linear acceleration) wouldn't 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!

 

[rcgldr]

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]

linear acceleration = radius x angular acceleration

Hi killercatfish!

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.

Your professor must be wrong (or sadly misunderstood! ).

 

[Doc Al]

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 doesn't 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) wouldn't we get r*aa=la --> aa=la/r.

That's the correct relationship.

 

[killercatfish]

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"

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"

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!

 

[tiny-tim]

… how this will give me the radius which is the point of breaking?

Breaking what? You haven't set out the question …

 

[killercatfish]

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!