Please, Read Question Below!

Question

A ball swings counterclockwise in a vertical circle at the end of a rope 1.29 m long. When the ball is 36.1° past the lowest point on its way up, its total acceleration is ( −17.1 î + 23.4ĵ) m/s². For that instant, do the following.

-Determine the magnitude of its radial acceleration.

: _____ m/s²

Determine the velocity of the ball.

magnitude: ______ m/s

direction: ________° counterclockwise from the +î direction

Answer 1

The total acceleration vector magnitude is given as a_T = { -17.1i + 23.4j } m/s², 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/s², and the radial acceleration is a_r = 23.3 j m/s². The total acceleration is:

a_{T = √} a_t² + a_r² , and a_T = √ (-17.1)²+(23.4)² = 29.0 m/s² is the resultant vector of a_t and a_r

Therefore:

a_t = -17.1 m/s² , and a_r = 23.4 m/s² and a_r = V_t²/r , V² = a_r.r = 23.4 m/s² . 1.29 m = 30.19 m²/s² , and Vt = √30.19 m²/s² = 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.