Elastic collision momentum transfer

[j1979p]

Consider a collision between a large and small body (golf club/hockey stick/baseball bat and a ball for example). How is it possible to calculate how much the momentum of parts of the body removed from the collision contact point have an effect on the momentum of the smaller body before it accelerates away?

Many people think that since the contact duration is so small (as little as half a millisecond in some cases), that only the part of mass in the vicinity of the collision matters (i.e. the effective mass).

However, how do you calculate this? Is it something to do with the speed of propogation of elastic stress waves?

 

[K^2]

Elastic collision conserves energy. When a club/bat hits a ball, it looses some of its momentum and some of its angular momentum. Both are carried away by the linear motion of the ball. (Ball can also have angular momentum, but it can usually be neglected at this stage.)

So you have 3 unknowns after collision: linear momentum of the ball, linear momentum of the bat, angular momentum of the bat; and 3 equations: conservation of momentum, conservation of angular momentum, and conservation of energy. If you write out the before/after condition on all 3 equations, and then solve them as a system for your 3 unknowns, you will find the correct momentum transfer.

Notice that momentum transfer technically has 3 degrees of freedom, but in a brief collision, momentum is only transferred in the direction that is normal to the surface of the ball at the impact point. This is what let's you reduce the problem to 3 unknowns instead of 9. (3 DOF for two linear motions and for one rotation.)

This approach does neglect a few factors. When a baseball bat hits baseball, for example, the bat vibrates quite violently. Typically, you can model that as energy loss in a collision, id est, assume it's not perfectly elastic, which is usually true. But even assuming perfectly elastic collision, you'll be surprised how accurate you can get, especially for something like the golf ball.

If any of this is still hazy, feel free to make up an example problem, and somebody here can probably walk you through solving it. Make sure you provide initial velocities for the ball and whatever you swing at it, as well as masses of the ball, and either the actual value or a way to calculate moment of inertia for the bat/club.

 

[j1979p]

This approach does neglect a few factors. When a baseball bat hits baseball, for example, the bat vibrates quite violently. Typically, you can model that as energy loss in a collision, id est, assume it's not perfectly elastic, which is usually true. But even assuming perfectly elastic collision, you'll be surprised how accurate you can get, especially for something like the golf ball.

True, but the vibrations occur long after the ball has left the bat and have no effect on the ball. I am more interested in the momentum/energy that is transferred to the ball.

If any of this is still hazy, feel free to make up an example problem, and somebody here can probably walk you through solving it. Make sure you provide initial velocities for the ball and whatever you swing at it, as well as masses of the ball, and either the actual value or a way to calculate moment of inertia for the bat/club.

OK if you need numbers, take:

clubhead mass: 200g
shaft mass: 120g
clubhead material: titanium
shaft material: steel
initial ball velocity: 0
ball mass: 45g
clubhead speed at impact: 45 m/s
coefficient of restitution between club and ball = 0.8

Now, the usual answer is to ignore shaft and say momentum (of clubhead) before = 9000 gm/s etc etc. However, we must calculate the angular momentum of the shaft too (probably a differential equation) and then try to find out how much of this angular momentum is transferred in the contact time of only 0.5 milliseconds!
The latter is the part of the problem I am most interested in. And I think more variables need to be brought in for this (e.g. stress wave velocity in steel)

 

[K^2]

I'll need length of the shaft and initial pivot point as well. And can I assume that impact happens in direction perpendicular to the arm of rotation?

Vibrations do impact collision, though. It has to do with the fact that the club isn't a true rigid body.

 

[j1979p]

Take length of shaft as 1m for simplicity. Pivot point is at the top (l = 0) of this shaft again for simplicity.

Yes, impact is perpendicular to the arm of rotation.

I understand vibrations impact slightly how the ball comes off the clubface but not very much since the majority of the vibrational energy is dissipated after the ball is gone.

 

[K^2]

Alright.

Center of mass.

$$R = \frac{M_{shaft}\frac{L}{2} + M_{head}L}{M_{shaft} + M_{head}}$$

Moment of inertia about CoM.

$$I_{head} = M_{head}(L-R)^2$$

$$I_{shaft} = \frac{M_{shaft}L^2}{12} + M_{shaft}\left(R-\frac{L}{2}\right)^2$$

$$I = I_{head} + I_{shaft}$$

v - CoM velocity of club.
ω - angular velocity of club.
u - CoM velocity of ball.
M = M_head+M_shaft - mass of club.
m - mass of ball.

$$v_i = v_{head}\frac{R}{L}$$

$$\omega_i = \frac{v_{head}}{L}$$

Conservation equations.

Energy.

$$\frac{M v_i^2}{2} + \frac{I \omega_i^2}{2} = \frac{M v_f^2}{2} + \frac{I \omega_f^2}{2} + \frac{m u_f^2}{2}$$

Momentum.

$$M v_i = M v_f + m u_f$$

Angular momentum.

$$I \omega_i = I \omega_f + m u_f L$$

Rewriting for v_f and ω_f.

$$v_f = \frac{M v_i - m u_f}{M}$$

$$\omega_f = \frac{I \omega_i - m L u_f}{I}$$

And going back to energy conservation with substitutions.

$$\frac{M v_i^2}{2} + \frac{I \omega_i^2}{2} = \frac{1}{2}M \left(\frac{M v_i - m u_f}{M}\right)^2 + \frac{1}{2}I \left(\frac{I \omega_i - m L u_f}{I}\right)^2 + \frac{m u_f^2}{2}$$

Collecting terms.

$$\frac{1}{2}\left( \frac{m^2}{M} + \frac{m^2 L^2}{I} + m \right)u_f^2 - (v_i m + \omega_i m L)u_f = 0$$

Note the constant term cancellation. This is a very good thing, since one of the two solutions is automatically u_f=0. Initial condition must satisfy the same conservation equations as final. If not, we messed up somewhere.

Excluding u_f=0 solution, final form.

$$\frac{1}{2}\left( \frac{m^2}{M} + \frac{m^2 L^2}{I} + m \right)u_f = v_i m + \omega_i m L$$

Numbers time.

(I just realized that I completely forgot to put in coefficient of restitution. So this collision just became perfectly elastic. The coefficient of restitution can be put into energy conservation equation, and it makes a royal mess. The constant term doesn't go away, so you have to solve a quadratic equation and get two solutions for u_f. Only one of these will make sense, though. I'm pretty sure the wrong solution will be imaginary, but I don't want to bother with actually going through it.)

R = 0.8125m
I_head = 0.00703125 kg m²
I_shaft = 0.02171875 kg m²
I = 0.02875 kg m²
v_i = 36.5625 m/s
ω_i = 45 s^-1

Drum roll, please.

u_f = 91.12 m/s.

Seems reasonable, and dimensions work out, so that's always nice. Feel free to try this with restitution coefficient in place. It shouldn't change the answer too much.

 

[j1979p]

Thanks K^2, you put a lot of work into that problem but it seems a bit familiar to me.

It seems you have simplified the problem (as many do) by not taking into account the fact that not all of the energy or momentum transfer that you have calculated to have been transferred to the ball below will actually have been transferred.

I'm sure the clubhead contribution you have calculated is pretty accurate but in the case of the shaft, some of the angular momentum of the shaft simply cannot have had time in the 0.5 milliseconds to have transferred anything to the ball! The reason is because the shaft is so long.

I would love to know how much of the shaft might have contribued and what determines this?

It is a part of collision physics that is so important (not just in sport) because most colliding objects are not perfectly rigid and so only the "effective" mass close to the point of collision should be used in momentum calculations. Yet despite its apparent importance, I cannot find anything on the topic on the internet!