Calculating Angular Acceleration and Required Force for Forearm Throw

[Amerk2]

Hi,

I'm new here, and in dire need of help. This problem has been stumping me for several hours now, and I don't seem to be getting any closer to understanding or solving it.

The premise: Assume that a 1 kg ball is thrown solely by the action of the forearm, which rotates about the elbow joint under the action of the triceps muscle. The ball is accelerated uniformly from rest to 8.5 m/s in .35 seconds, at which point it is released.

A diagram is shown of a elbow, presumably in the 90 degree position.

A.) Find the angular acceleration.

B.) The force required of the triceps muscle.

I've got A, and I know it's correct at 78.341 rad/s^2.

Now B is supposed to be 670 N according to the answer key, but I'm not getting it.

My approach so far:

I(moment of inertia) * angular acceleration = torque
R(perpendicular - this is the length of the lever it seems) * Force = Torque

So,

I*angular a = RF

The real values I used were:

I = .118523 ( I used I = 1/3(ml^2) for that, the equation for the I of a long uniform rod with axis at one end). I used .31 m for its length, and 3.7 kg for its weight.

Angular acceleration was the one I supplied above for A, and R was .025 m (for the right side of the equation).

I'm not sure where I'm going wrong, but I know I'm missing the effect of the ball held in the hand. It weighs 1 kg, but I don't think I'm supposed to factor it into the moment of inertia, as no radius is given.

Any thoughts?

 

[Shooting Star]

Where's the diagram?

 

[Doc Al]

I'm not sure where I'm going wrong, but I know I'm missing the effect of the ball held in the hand. It weighs 1 kg, but I don't think I'm supposed to factor it into the moment of inertia, as no radius is given.

Treat the ball as a point mass. Its distance from the axis will be the length of the forearm.