Question

A fast pitch softball player does a "windmill" pitch, moving her hand through a vertical circular arc to pitch a ball at 65 mph. The 0.19 kg ball is 50 cm from the pivot point at her shoulder.
- Just before the ball leaves her hand, what is its centripetal acceleration?
- At the lowest point of the circle the ball has reached its maximum speed. What is the magnitude of the force her hand exerts on the ball at this point?
- At the lowest point of the circle the ball has reached its maximum speed. What is the direction of the force her hand exerts on the ball at this point?

Answer 1

a) The centripetal acceleration is $1694 m/s^2$

b) The force on the ball is 321.9 N

c) The direction of the force is upward

Explanation:

a)

The centripetal acceleration of an object in circular motion is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circle

In this problem, we have:

r = 50 cm = 0.50 m (the radius of the circle is equal to the distance between the shoulder and the ball)

$v=65 mph \cdot \frac{1609 m/mi}{3600 s/h}=29.1 m/s$ is the speed

Substituting,

$a=\frac{29.1^2}{0.50}=1694 m/s^2$

b)

The force exerted by the hand on the ball is equal to the centripetal force, which is given by:

$F=ma$

where

m is the mass of the ball

a is the centripetal acceleration

In this problem, we have:

m = 0.19 kg

$a=1694 m/s^2$

Substituting, we find:

$F=(0.19)(1694)=321.9 N$

c)

In a uniform circular motion, the direction of the centripetal acceleration is always towards the centre of the circle.

As a result, since the  direction of the force is the same as the direction of the acceleration, the direction of the centripetal force is also towards the centre of the circle.

Therefore in this case, since the ball is at the lowest point in the circle, the  direction of the force is upward  (because it is towards the centre of the circular trajectory).

Learn more about centripetal force and acceleration:

brainly.com/question/2562955

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