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 0.672 m. At one point on this circle, the ball has an angular acceleration of 63.8 rad/s2 and an angular speed of 12.8 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

Answer 1

Answer:

a) a= 118.15 m/s²

b)θ=20.80°

Explanation:

Given that

r= 0.672 m

Angular acceleration ,α = 63.8 rad/s²

Angular speed ,ω = 12.8 rad/s

Tangential acceleration at

at = α .r

at ==63.8 x 0.672

at = 42.87 m/s²

Centripetal acceleration ac

ac= ω² r

ac =12.8² x 0.672

ac= 110.1 m/s²

So the total acceleration a

$a=\sqrt{at^2+ac^2}\ m/s^2$

$a=\sqrt{42.87^2+110.1^2}\ m/s^2$

a= 118.15 m/s²

Angle θ

$\tan\theta=\dfrac{at}{ac}$

$\tan\theta=\dfrac{42.87}{110.1}$

θ=20.80°