# I Center of percussion - baseball bat theorem

1. Jul 24, 2016

### sgh1324

look figure (b)

suppose that baseball deliver F through horizontal motion.
imagine that the O point of the system is same line of F (+x is F direction)

then before percussion, the angular momentum of the system is "0" because r and v of baseball are same direction (L = r x mv = 0)

so after percussion, the ball gets y direction velocity because if bat has angular momentum, the other angular momentum has minus sign and same size so that make 0 angular momentum

that means center of mass of baseball bat has y direction velocity cause of conservation of linar momentum so that make y direction linar momentum also be 0 before and after percussion

so if the baseball hits bat like figure (b), the bat never move just x direction like figure (b)

sth problem in my inference??

2. Jul 24, 2016

### jbriggs444

In the case of an off-center impact, the ball starts with non-zero angular momentum in the reference frame where the center of the bat is at the origin. There need not be (and can not be) any motion in the y direction as the result of a collision with force along the x axis.

3. Jul 25, 2016

### sgh1324

center of the bat is at the origin → i know that makes the ball starts with non-zero angular momentum

i mean that if i set the origin point it same line with ball's movement, that makes the ball and bat get zero angular momentum - and after the collision, angular momentum of totaol system is not anymore zero

4. Jul 25, 2016

### jbriggs444

What makes you say that the angular momentum becomes non-zero after the collision? It is not true.

Conservation of angular momentum applies in the absence of external torques. If total angular momentum is zero before the collision, it must be zero after.

The drawing shows a dark dot in the center of the bat. I have understood that as a simple marker for the center of mass of the bat. If that were a pivot fixed to an axle then angular momentum could result from the collision. That is because such a pivot can transmit an external torque.

5. Jul 26, 2016

### sgh1324

fig (b) has no pivot - it's a free laminar motion

i know no external force means conservation of angular momentum

and i understood fig (b) must be same angular momentum

but it's also true that my set of origin point(same line with F) makes zero angular momentum before percussion because r and v have 0 angle, and after collision non-zero angular momentum because bat the will spin and ball move -x direction

i just wanna know problem of my inference

6. Jul 26, 2016

### jbriggs444

What makes you say that the spinning bat will have non-zero angular momentum? It is not just spinning. It is moving rightward as well.

7. Jul 26, 2016

### sgh1324

physics world of mine

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8. Jul 26, 2016

### jbriggs444

What makes you think that the angular momentum of the spinning bat is non-zero?

9. Jul 26, 2016

### sgh1324

if spinning bat has two mass top and bottom of it - for easy calculation . that angular momentum i think is non-zero

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10. Jul 26, 2016

### jbriggs444

In that diagram, where is your reference axis? Recall that you had originally placed that axis at the point of impact of the ball. Not at one end of the bat. Not at the midpoint of the bat.

Angular momentum is only conserved if you are consistent in your choice of reference axis and do not move it around, willy-nilly.

11. Jul 26, 2016

I have previously done this calculation. What you need to do is find the "location" on the bat where the angular momentum of the bat, (swinging at the pivot point which is your hands), computed with that "location" as the origin, is zero. (For a uniform density I believe this location is 3/5 the length of the bat from your hands). The final angular momentum of the bat after it hits the ball is assumed zero. (Hypothetically, you can consider the ball brings the bat to a stop.) Since you have no change in angular momentum, [it started at zero angular momentum, and even if the bat is swinging backwards after impact (with your hands as the pivot point), it still has zero angular momentum when computed from that "location" as the origin], the net torque must be zero when computed from the center of percussion point (the "location") that you just found. The ball hits this point (the origin), so the torque from the ball is zero (i.e. the length of the moment arm is zero). Meanwhile the only other possible force is from your hands. But since there is no change in angular momentum and no other torques, your hands do not supply any torque and thereby feel zero force when the ball hits the bat. (Incidentally, this is the same problem as a swinging door=where should you locate the doorstop so that the hinges don't get stressed when the door hits the stop?)

Last edited: Jul 26, 2016
12. Jul 27, 2016

### sgh1324

OK PLZ UNDERSTAND MY QUESTION

Problem - if ball collasped with stick hangs two heavy masses and moved like test.png, After collision angular momentum must be zero cause no external forces there. But collision makes stick get spin, at origin point of that system feel non zero angular momentum

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13. Jul 27, 2016

### jbriggs444

I insist that the spinning bat has zero total angular momentum about the reference axis that you have originally chosen.

The fact that it has non-zero angular momentum about a different axis is completely and totally irrelevant.

The angular momentum of a system can change if you change the position of your reference axis. The angular momentum of a system can change if you choose to set your reference axis in motion. If you want to compute the angular momentum of the bat it is not good enough to mutter something about $R_a V_a sin\ a$. You need to define what those terms mean and relate them to the reference axis.

Last edited: Jul 27, 2016
14. Jul 27, 2016

### sgh1324

what makes you think that i calculated at different axes, before and after are same axes, they are just fixed simple axes. i calculate angular momentum about the reference axis after collision

15. Jul 27, 2016

### jbriggs444

The reference axis that you claimed you chose was one at rest at the position of impact of ball on bat. The reference axis you ended up using is apparently one at rest at the position of the center of mass of the bat just prior to impact.

Your insistence on invoking trig functions is also erroneous, but we need not go there. [The angular momentum of the spinning bat is constant -- it does not vary as the bat rotates]

Edit:

Looking more slowly through your most recent drawing, I see that this time the reference axis has not moved. That's good.

You show the ball rebounding at a speed of -v purely along the x axis. That may or may not be correct, depending on the mass of the bat and the coefficient of restitution. But since that motion is aligned with the axis of rotation, it is irrelevant.

You show one mass (a) representing the top end of the bat rotating counter-clockwise with a velocity of Va on a moment arm labelled Ra so that the velocity makes an angle "a" with the moment arm. That gives you an angular momentum term of $m_a V_a\ sin\ a$

You show another mass (b) representing the bottom end of the bat also rotating counter-clockwise with a velocity of Vb. on a moment arm labelled Rb so that the velocity makes an angle b with the moment arm. That gives you an angular momentum term of $m_b V_b\ sin\ b$

You claim that this sum is non-zero, that it is equal to "r x mv" and that it is equal to the angular momentum of the bat.

I agree that it is equal to the angular momentum of the bat.
I do not agree that it is equal to "r x mv" -- [actually it's r x 2mv but r=0, so that comes out OK]
I do not agree that is is non-zero.

Looking at a partially-rotated bat is not a particularly effective view. It adds unnecessary complexity to the problem. Better to look at the vertical bat immediately following the collision. Let us assume as is implicit in your most recent drawing that the ball collides exactly at the bottom end of the bat.

In that position, Va will be zero, Ra will be pointing straight up. sin a will be undefined but that is irrelevant. The angular momentum of mass a is zero.

Vb will be non-zero, Rb will be zero. sin b will be undefined but that is irrelevant. The angular momentum of mass b is zero.'

So the angular momentum of the bat is zero. By conservation of momentum, this angular momentum does not vary over time.

Edit again...

Now looking back at your depiction of the partially rotated bat, we see that you have drawn Va so that it points at an angle clockwise from the moment arm Ra. This is an important hazard when reasoning from drawings -- how do you know that the correct angle will not be counter-clockwise from the moment arm? In fact, it will be counter-clockwise as the top end of the bat begins to trace out a cycloid path. [I am fond of an apparently rigorous proof that all triangles are isoceles that proceeds based on a drawing with the same sort of flaw].

Last edited: Jul 27, 2016