Question

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so that the ball in her hand moves in a circle. In one instance, the radius of the circle is 0.670 m. At one point on this circle, the ball has an angular acceleration of 64.0 rad/s2 and an angular speed of 16.0 rad/s

(a) Find the magnitude of the total acceleration (centripetal plus tangential) of the ball. m/s2
(b) Determine the angle of the total acceleration relative to the radial direction.

Answer 1

Part (a) the magnitude of the total acceleration (centripetal plus tangential) is 176.799 m/s²

Part (b) the angle of the total acceleration relative to the radial direction is 44.13⁰

Step-by-step explanation:

Given;

angular acceleration α = 64.0 rad/s²

angular speed ω = 16.0 rad/s

radius of the circle r = 0.670 m

Part (a) the magnitude of the total acceleration (centripetal plus tangential)

centripetal acceleration a = ω²r = (16)²*0.67 = 171.52 m/s²

tangential acceleration α = αr = 64*0.67 = 42.88 m/s²

the magnitude of the total acceleration = √(a² + α²)

= √(171.52² + 42.88²)

= 176.799 m/s²

Part (b) the angle of the total acceleration relative to the radial direction.

Let this angle be θ

tan θ = (radial or centripetal acceleration)/(total acceleration)

tan θ = 171.52 /176.799

tan θ = 0.9701

θ = tan⁻¹ (0.9701)

θ = 44.13⁰