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 0.681 m. At one point on this circle, the ball has an angular acceleration of 67.7 rad/s^2 and an angular speed of 18.6 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). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

$a_{c}=\omega^2\times r$

Put the value into the formula

$a_{c}=(18.6)^2\times0.681$

$a_{c}=235.5\ m/s^2$

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

$a_{t}=0.681\times67.7$

$a_{t}=46.104\ m/s^2$

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

$a=\sqrt{a_{c}^2+a_{t}^2}$

Put the value into the formula

$a=\sqrt{(235.5)^2+(46.104)^2}$

$a=239.97\ m/s^2$

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

$\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{46.104}{235.5})$

$\theta=11.0^{\circ}$

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°