quiz16_solution - Therefore a=g mR^2/(I+mR^2). Alternate...

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04/02/08 Physics 201 Quiz 16 An object of mass M is hung by a massless rope that is wound around a frictionless pulley whose moment of inertia is I and radius is R. Find the tension in the rope and the acceleration of the mass. Solution:  a=alpha R T R=I alpha=I a/R mg-T=ma=m T R^2/I.  Therefore T=mg/(1+mR^2/I)
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Unformatted text preview: Therefore a=g mR^2/(I+mR^2). Alternate solution using cons of Energy I omega^2/2+m v^2/2=(I/(R^2)+m)v^2/2=m g h compare to v^2=2 a h to get a=g mR^2/(I+mR^2). I really don't understand why this method doesn't seem to be taught....
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This note was uploaded on 03/06/2010 for the course PHYSICS 201 taught by Professor Walker during the Spring '08 term at Wisconsin.

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