Kinetics\Biomechanics of throwing a basketball (Elbowjoint)

Question

I'm trying to make a simulation of a person throwing a basketball. My goal is to have at least a shoulder and elbow joint which I can assign with various torques over time to create a motion. When reaching a certain angle (between lower and upper arm/ in the elbow joint) the ball will release and will follow a trajectory, after which I want to tweak the torques to get the trajectory such that the ball will go in the net.

I don't think the projectile motion will give me that much trouble, but I am struggling very much with the translation of a Force(created by tricep) or torque on elbow joint to the angular velocity of the ball(just before release). Which is why I try to reach out to you guys.

For now I reduced the problem to only the lower arm turning around the elbow joint axis. I also left out the force of gravity in the drawing to keep it somewhat clear.

I tried to apply several examples, such as a bullet colliding with a beam that rotates around a axis but in that case variables such as speed of the bullet are given and make it hard for me to translate to 'my' situation.

I'm hoping someone can help me find the angular velocity of the ball in situation 2 (after a angular rotation of 30° and 30 N.m torque on the elbow joint).

Answer 1

First of all if you want to throw the ball after thirty degree of rotation of arm then according to me you should focus on the tangential velocity on that instant at which you have to release the ball.

I'm going to assume that the torque on the Catapult is going to remain constant all the time because you haven't told that if the arm is attached to the spring which is going to provide the torque to the arm.

So for that as you said torque is 30N.m so according to the relation:-

$τ=Iα$

Where tau is the torque on the arm, I is the moment of inertia of the rod and, alpha will be the angular acceleration.

For the calculation of moment of inertia I'm going to add the mass of the ball and diametrically add the diameter of the ball as well.

$I= \frac{1}{3} (M+m) (L+r)^2$

Taking M as the mass of rod and m as the mass of ball.

So it will come out to be.

$I = 0.16Kgm^2$

So by putting all this in the equation of torque we get:-

$30 = 0.16×α$

From here we get angular acceleration as

$α = 187.5$

Putting this angular acceleration in the equation of circular motion:-

$ω^2-u^2=2αθ$

Initial angular velocity will be zero hence final angular velocity will come out to be after putting theta as pi/6 :-

$ω = 14$

So here you got the angular velocity of the ball at the instant at which the arm is making 30 degree which you were asking for.

And hence you can find tangential velocity of the ball with which the ball will leave the Catapult.

That will come out to be:-

$V = ωr$

$V = 6.72 m/s$

This might happen that there may be some calculation mistakes so please go through the answer carefully and I hope you liked it!