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Torque Problem, acceration of mass. Help!

 
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Nov9-10, 02:06 PM   #18
 

Torque Problem, acceration of mass. Help!


I'm sorry, I feel stupid but I don't really understand. I know that Cx is the x-component of the center of mass, my issue is why Ax-Cx=k-rθ...Because lets say the center of mass of the cylinder doesn't move, then k-rθ is how much closer the mass is to the cylinder, in other words then Ax-Cx=k-rθ..but lets say the cylinder rotates that same amount, but this time Cx also moves, then the equation would predict the same result, but obviously it's not the same because now the cylinder is closer to the mass...

Sorry, but what is my constraint equation?

I have never had this much trouble understanding a problem

Thanks for your help by the way.
Nov9-10, 02:24 PM   #19
 
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Quote by mmmboh View Post
… lets say the cylinder rotates that same amount, but this time Cx also moves, then the equation would predict the same result, but obviously it's not the same because now the cylinder is closer to the mass...
no, because the mass has also moved, in this case the same amount as the cylinder (if the rotation is the same)
Sorry, but what is my constraint equation?
Ax - Cx = k - rθ …

a constraint equation is a geometrical equation rather than a force equation … like a body being constrained to move along a particular surface, or in this case the free length of string being constrained to be connected to the position and rotation of the bodies
Nov9-10, 02:33 PM   #20
 
Ok but, I used that constraint equation already to find α to plug it into my torque equation...

The three equations are T+t=ma
(T-t)r=Iα=0.5mr2α
Ax - Cx = k - rθ, so differentiating this twice I can solve for α.

So I get T+t=macylinder, and T-t=0.5m(acylinder-amass), and then I'm stuck.
Nov9-10, 02:44 PM   #21
 
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ah, but you also have t = -mamass
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acceleration, cylinder, string, torque
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