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A wheel of diameter 30.0 cm starts from rest and rotates with a constant angular acceleration of

`3.50ra(d)/(s^(2))`

. At the instant the wheel has completed its second revolution, compute the radial acceleration of a point on the rim in two ways. Part A Using the relationship

`a_(rad)=\omega ^(2)r`

. Express your answer in meters per second squared.

`a_(rad)=|,|(m)/(s^(2))`

Part B From the relationship

`a_(rad)=(v^(2))/(r)`

. Express your answer in meters per second squared.

`a_(rad )=,(m)/(s^(2))`