# Torque Problem, acceration of mass. Help!

by mmmboh
Tags: acceleration, cylinder, string, torque
HW Helper
Thanks
P: 26,148
 Quote by mmmboh … 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
 P: 408 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.
 Sci Advisor HW Helper Thanks P: 26,148 ah, but you also have t = -mamass

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