Roger N. answered • 06/01/21

. BE in Civil Engineering . Senior Structural/Civil Engineer

The total acceleration vector magnitude is given as a_{T} = { -17.1i + 23.4j } m/s^{2}, The components of the total acceleration are the tangential and radial accelerations such that the i component is the tangential acceleration, and the j component is the radial acceleration in a Cartesian Vector system

a_{t} = -17.1 i m/s^{2}, and the radial acceleration is a_{r} = 23.3 j m/s^{2}. The total acceleration is:

a_{T = √ }a_{t}^{2} + a_{r}^{2} , and a_{T }= √ (-17.1)^{2}+(23.4)^{2} = 29.0 m/s^{2} is the resultant vector of a_{t} and a_{r}

Therefore:

a_{t} = -17.1 m/s^{2} , and a_{r} = 23.4 m/s^{2} and a_{r} = V_{t}^{2}/r , V^{2} = a_{r}.r = 23.4 m/s^{2} . 1.29 m = 30.19 m^{2}/s^{2} , and Vt = √30.19 m^{2}/s^{2} = 5.5 m/s, where V_{t} is the tangential velocity

Since the i component vector is -ve, the tangential acceleration vector is negative to the left in the clockwise direction, while the radial vector is +ve and toward the center of rotation. This means that the positive +i axis is to the right pointing counterclockwise and the total acceleration vector must be in the 2nd quadrant where θ = tan^{-1}( 23.4/-17.1) = -53.8º rotating clockwise from the -i axis to the left. Its actual position is 180°- 53.8° = 126.2° rotating counterclockwise from the +i axis

P.S. the position of the ball at 36.1° from the bottom of the circle is not needed because the total acceleration vector is given and known. You just have to realize that the i and j components of the total acceleration vector are actually the tangential and radial acceleration of the ball.