University Physics with Modern Physics with Mastering Physics (11th Edition)

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10.66: The accelerations of blocks A and B will have the same magnitude a . Since the cord does not slip, the angular acceleration of the pulley will be . R a = α Denoting the tensions in the cord as , and B A T T the equations of motion are , 2 a R I T T a m g m T a m T g m B A B B B A A A = - = - = - where the last equation is obtained by dividing α τ I = by R and substituting for α in terms of a . Adding the three equations eliminates both tensions, with the result that
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Unformatted text preview: 2 / R I m m m m g a B A B A + +-= Then, . / R I R m R m m m g R a B A B A + +-= = The tensions are then found from . 2 ) ( 2 ) ( 2 2 2 2 R I m m R I m m m g a g m T R I m m R I m m m g a g m T B A B A B B B B A A B A A A + + + = + = + + + =-= As a check, it can be shown that . ) ( I R T T B A =-...
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