Question

A softball pitcher while winding up swings their arm at the shoulder in the anteroposterior plane in a circle. If at release, the ball (0.2 kg) is moving at 20 m/s, what is the minimum and maximum force exerted on the ball during rotation. Assume constant velocity around the circle and the arm is 0.75 m long. Remember which direction gravity is affecting the ball at the top of the circle and at the bottom. Маx = Min = 5.3333 N Max 108.62 N, Min 104.70 N Max 7.29 N, Min 3.37 N 15.1433 N, Min Max = -4.4767 N Мах — Min 106.6667 N

Answer 1

Jn R 0.7m the ba Maxii frce extrded mv2 R Mayx o. g.8 /08.6 N Minimum for exerted n the bell Frein my 2 6.2x (20) Fmio 104.76 thuy Tet angwer op tron

Answer 2

SOLUTION :

When the ball is at the top, maximum force is exerted since in this position 

Weight of the ball adds to the centripetal force. 

Hence,

 F max 

= m v^2 / r + mg 

= 0.2 * 20^2 / 0.75 + 0.2 * 9.8

= 108.63 N .

When the ball is at the bottom of circular motion, weight of the ball is opposite to centripetal force,

So, force exerted will be minimum.

Hence,

 F min 

= m v^2 / r - mg 

= 0.2 * 20^2 / 0.75 - 0.2 * 9.8

= 104.71 N .

Therefore, 2nd option F max = 108.62 N and F min = 104.70 N is the ANSWER.