A curved wire is in the shape of r = 3θ. If a bead, point A, moves along the curve at a constant rate of dθ/dt = 8 rad/s, what is the acceleration magnitude at A? Distance units are meters.

Acceleration Vectors of Bead

The total acceleration is a combination of both the r and θ accelerations.
     

The r terms can be found by taking derivatives,
     

The θ terms are given,
     

Combining equations and terms give,
     

The total acceleration is
     a = (a_r² + a_θ²)^0.5 = ( -201.1² + 384²)^0.5 =
     

v = 433.5 m/s²