Atangential = r alpha = .608 (65.3)
= 39.7 m/s^2
Ac = omega^2 r = (19.7)^2 (.608)
= 236 m/s^2
so mostly centripetal
A = sqrt (236^2 + 39.7^2)
=239.3 m/s^2 I get
angle to radius vector = close to 180 because mostly centripetal
tan angle above -x axis = 39.7/236
angle above -x axis = 9.55 deg
so angle to radius out vector = 180-9.55
= 170.45 degrees
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 0.608 m. At one point on this circle, the ball has an angular acceleration of 65.3 rad/s2 and an angular speed of 19.7 rad/s. (a) Find the magnitude of the total acceleration (centripetal plus tangential) of the ball. (b) Determine the angle of the total acceleration relative to the radial direction.
I already found that the total acceleration is = to 235.95 m/s^2
i am having issues with the angle. I have taken all trig functions with the total acceleration and 65.3 with no luck :(
1 answer
1 answer