1. 1. The problem statement, all variables and given/known data

1. Consider a man throwing a baseball. Given the following, calculate the velocity of the ball as it leaves his hand:
1. The distance from his shoulder socket (humeral head) to the ball is 70 cm.
2. The distance from his humeral head to the points of insertion of the muscles moving his arm forward is 9 cm (this is a simplification, as the shoulder is a complex joint).
3. The velocity of the muscle shortening is 2.6 m/sec..

2. Relevant equations

Velocity = Distance / Time ????

3. The attempt at a solution

Here is my first attempt at this problem: 2.6 m/ sec = 260 cm/ sec.

70 - 9 = 61 cm

61 cm / 260 cm per second = 0.23 m /second?

Can anybody help me on how I should start this question? I would like to know how to start so I can begin to wrap my brain at it. Thank you. P.S. I don't know really know anything about velocity other than what I wrote above. Thank you!  2.

3. One way of finding that a formula is wrong is to notice that the units don't match. I don't know how you got this one: Originally Posted by searchme
61 cm / 260 cm per second = 0.23 m /second?
but it's certainly wrong. Dividing centimeters by centimeters per second will not yield a result in meters per second. Think of the arm as a rotating rigid body. Points on such a body move at velocities that are proportional to their distance from the axis of rotation.

Look at a door as it moves on its hinges. Parts that are close to the hinge move slowly. Parts closer to the lock move fast.  4. Thanks for the reply Leszek Luchowski. I converted 2.6 m / sec to 260 cm / sec.

There seems to be 3 points that make up what appears to be triangle:

First point: 70 cm
Second Point: 9 cm
Last Point: 61 cm

How do I calculate the velocity for one of these angles? Perhaps, if calculate all then I should add up the velocity for my answer.  5. The velocity at any point on a rigid body is the angular velocity multiplied by the radius. Both points on the arm have the same angular velocity, so
v1/r1=v2/r2
where v1 is the velocity where the muscle attaches (2.6 meters/sec), r1 is the radius at the point of attachment (9 cm), v2 is the velocity of the ball, and r2 is the radius of the arm at the ball (70 cm). We are solving for v2.  6. You were right Harold. Thank you!  Bookmarks
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