A 2.72kg ball is attached to a .65m rod that weighs 1.24kg. The rod rotates around it’s axis 125 degrees. After rotating 125 degrees the ball is let go from the rod. How much torque is required to make the ball’s exiting velocity 18m/s.
2. Relevant equations
vf=vi + at
3. The attempt at a solution
First I solved for how much distance the ball is traveling in an arc before being released=1.43m
L = ((125)*2pi*.6558m)/360 = 1.43m.
Then I found it’s needed acceleration to reach 18m/s.
Then I solved for the amount of time it will take to reach 18m/s
vf=vi+at, 18=113*t, t=.1558s
Now this is where I got stuck.
F=m*a, so i’m assuming (2.72kg)*113.28m/s^2=308.12N
Is this correct? I think this may involve angular motion which I am inexperienced with. Ps, this is for a catapult idea/robot so any help is gratefully appreciated.