# Homework Help: Inellastic collision question

1. Dec 11, 2009

### Samuelb88

1. The problem statement, all variables and given/known data
A small object with an initial velocity collides with and sticks to a long thing rod that is free to translate as well as rotate. Motion here is on a horizontal frictionless surface.

Find where the small object would collide with the rod to maximize the loss of energy in the collision and this loss of energy? Explain.

2. Relevant equations

3. The attempt at a solution
So since the rod which the bullet collides into is free to translate and rotate, thus, conservation of linear and angular momentum is conserved since the surface is frictionless.

I'm a little unsure of what equations should I use... Right now i'm using conservation of angular momentum, and the definition of the coefficient of restitution, setting it equal to zero...

So if d is the length of the thin rod, and x is the distance at which the bullet collides with the rod, I can find the angular velocity

$$l_0 + L_0 = L_f$$

$$xmv + 0 = I_0w$$

Therefore

$$w = \frac{xmv}{I_0}\right)$$

Then, I find the velocity using the COR

$$0 = \frac{xw - v_f}{v_0}\right)$$

Therefore

$$v_f = xw = x \frac{xmv}{I_0}\right)$$

Finally I set up the loss of K.E. function...

$$f(x) = \frac{mv_0^2}{2}\right) - ( \frac{(m+M)v_f^2}{2}\right) + \frac{I_0w^2}{2}\right) )$$

Substituting the values of v_f, and w into the function, i can differentiate to determine the maximum value of x in order to maximize a loss of K.E. in the system...

My question is am I setting this up right? after differentiating, and setting my derivative equal to zero, I keep getting a negative root... Should I be using different equations?

Any help would be greatly appreciated. :D

2. Dec 11, 2009

### denverdoc

It might be I'm missing something, but it seems at first blush you want KE as a function of X. In other words you need an explicit equation for the rotational energy in terms of x, and since Io can be computed in terms of the length of the rod and the impact point, some of the "variables" can be eliminated.. Unless you're planning on taking a set of partials and solving these simultaneously for minima, I'm not sure you have enough there. Have you looked at the two extremes--striking point blank center vs the end?

3. Dec 11, 2009

### Samuelb88

Well for my class, we're not calculating moments via integration but rather, given a table of moments.

The moment for a long thin rod, rotating about its center is I = 1/12 * ML^2, where L is the length of the rod.

Anywho, to answer your question, I can get explicitly define an angular velocity w, and a linear velocity v in terms of x, thus getting an explicit eq. of K.E. lost in terms of x. But I feel my equations are wrong.

Since the rod is free to translate and rotate, post-collision, the angular momentum is split into translation and angular motion, perhaps I should use...

$$l_0 = (m+M)v_f + I_0w$$

and the COR

$$0 = \frac{(L/2-x)w - v_f}{v_0}$$

where x is the distance from one end of the rod, thus, its distance from the center of the rod (whose length is L) is L/2 - x.

Thus using the two equations above, I can explicity define w, v, in terms of x.

So, substituting w, v, into my equation for the loss of K.E. of the system...

$$f(x) = K_0 - (\frac{(m+M)v_f^2}{2}\right) + \frac{I_0w^2}{2}\right) )$$

Another question I had is since the bullet sticks to the rod (inelastic collision), should I still be using the same moment, I_0 when calculating rotational K.E.? I deduced, since i'm treating the bullet as a point-mass object, it has no moment, thus, the when calculating rotational K.E., the rod is rotating about its center, therefore using its initial moment. Does this make sense?

4. Dec 11, 2009

### denverdoc

Sorry if that was unclear. But no not asking for integration of Io--just that the Io can be computed from the sum of 1/12... and a point mass at distance r from the cg. You may need to use parallel axis theorum or otherwise compensate for the fact that the axis of rotation of the composite bullet/rod will not be about .5L.

5. Dec 12, 2009

### Samuelb88

Hmm, yes good point. But i'm still confused. The numbers that I keep coming up with seem wrong...

Here's what's given:

Inelastic collision
Motion is over a horizontal frictionless surface.
v_0 = 42. m/s (velocity of bullet)
m = .10 kg (mass of bullet)
M = 1.9 kg (mass of rod)
d = .100 m (distance from one end of the rod which the bullet collides)
L = .960 m (length of rod)
Rod is free to translate + rotate.

Determine the K.E. lost in the collision.

Conservation of A.M. gives

$$l_0 = L_f$$

$$Rmv_0 = R(m+M)v_f + I_c_mw_f$$

where R = L/2 - d = .48 - .10 = .38

$$(.38)(.1)(42.) = (.38)(2)(v_f) + (\frac{ML^2}{12}\right) + (2)(.38)^2)w_f$$

$$1.596 = .76v_f + .43472w_f$$

And using the COR

$$e = 0 = \frac{Rw_f - v_f}{v_0}\right)$$

$$v_f = .38w_f$$

Thus, substituting the expression of v_f above into the conservation of AM equation,

$$1.596 = .76(.38w) + .43472w_f$$

Therefore

$$w_f = 2.21 rad/s$$, and $$v_f = .84 m/s$$

Those numbers seem ridiculously low since the bullet had an initial velocity of 42 m/s, despite having less mass than the rod. Does this make sense?

I'm very confused as to how to use conservation of angular momentum in this case since the (1) the collision is inelastic, and (2) the rod + bullet, post-collision; is free to translate as well as rotate. Also, (3) since linear momentum is conserved too, should I be using P_0 = P_f as well as l_0 = L_f ?

6. Dec 12, 2009

### denverdoc

Here is what I tried just to get a handle on the problem as the algebra was getting messier than I could deal with.

I considered a case where there is no translation--in this case, I pretend that the rod pivots about its center.

I consider that angular momentum is conserved: at the instant of collision
angular momentum of bullet= (v/r)*m*r^2=mvr as you have.

After collision, the Io=mr^2+(ML^2)/12 where L=2r
So we have .1(r^2)+(1.9*4(r^2)/12=r^2(.1+0.633), ie the contribution of the rod is about 6 times that of the bullet. So angular velocity is reduced by a factor approx 1/7.

In the case where the bullet strikes dead center, the velocity is reduced by a factor of 1/20.

Looking at the energy shows that much more KE is retained by the purely rotating system vs the perfectly translating system.

So my best guess is hitting the rod at the very end where translation will be minimal is the best case. But this is far from a very good answer. I'll ask one of the mentors to look at it.

7. Dec 12, 2009

### ideasrule

I haven't read every post in detail, but I'll give a few comments.

First, how is the ball hitting the stick? Is it coming in perpendicular to the stick, or at an angle? The solution gets a bit more complicated if it's coming in at an angle, so for this post I'll assume it isn't.

To the OP: I noticed that you set the COR to 0 in your solution. Don't do that; the lowest possible COR is almost certainly not 0. Just leave the COR out of the picture and express the loss of kinetic energy as a function of x, the distance from the impact point to the rod's center. Then you'll be able to tell which x causes the most energy loss.

To start off, final KE=0.5mv^2 + 0.5Iw^2 and delta-KE=0.5mv^2+0.5Iw^2-0.5*m_ball*v_i^2. So we need to find v, I, and w.

v is easy enough to find: just use the conservation of momentum. Initial momentum is m_ball*v_i, and it's equal to (m_stick+m_ball)Vf where Vf is the final speed of the center of mass of the stick-ball system.

To find w, you need to use the conservation of angular momentum. First, choose a pivot point about which you plan to calculate angular momentum. I suggest choosing the center of the rod, because that simplifies calculations. The initial angular momentum is just the angular momentum of the ball, which is mv_i*x (remember, x is the distance from the center to the ball's impact point). Final angular momentum has two components: angular momentum due to the linear movement of the ball-stick system's center of mass, and the angular momentum of the ball-stick system about its center of mass. The angular momentum due to translational motion is easy enough to find since you've already found v_f, so you just need to find the center of mass. For the angular momentum due to rotation, you'll have to calculate the moment of inertia about the center of mass (don't forget the moment of inertia of the attached ball!). The algebra gets quite complicated, so it's good to draw everything out, avoid taking shortcuts, label all distances/lengths, etc.

Good luck, and tell us if you've solved it!

8. Dec 13, 2009

### denverdoc

Likewise re good luck;

From the above post, there is a seeming paradox--that no matter where the rod is struck, its translational momentum will be the same. So whether it is spinning or not, the velocity of the center of the mass of the composite system is the same.

Yet the angular momenta also must be conserved. It seems like the same impulse is contributing momenta two different ways, each of which is conserved--like money for nothing, chicks are free. If so this is proof enough. Anything which contributes to rotation will increase the KE of the composite after collision. So the greater the rotaion the better. In most case, intuiton is a guiding light (quantum physics excepted). I'ze be lookin for a frozen pond with a buddy to hit a couple of yardsticks with shuffleboard disks.

9. Dec 13, 2009

### denverdoc

I worked on this problem some more this morning and came up with the following:

Let X + be the new center of mass relative to the center of the rod which is defined as x=0.

X=m(x)/(M+m) where M is the mass of the rod and m is the mass of the bullet.

Now to find the I about this axis, I use the parallel axis theorum: Neglecting the bullet so far

I = ML^2/12 + X^2(M)

Now adding the contribution from the bullet whose contribution will be m(x-X)^2

The quantity x-X = M(x)/(M+m).

This gives I=ML^2/12+ x^2(m*m*M+m*M*M)/(M+m)^2 which can be simplified to

I=ML^2/12+(x^2(mM/(M+m))

Using w*I= mv*x w=mvx/(ML^2/12 + 2mMx^2/(M+m)) Factoring out the M from the denominator gives: w=(mv/M)*x/(L^2/12+ x^2*2m/(M+m))

Solving for dW/dx and setting to 0 gives the curious result:

L^2(M+m)/(24m)=x^2 In this case worked out to be outside the boundary of the rod, but seems had the bullet been say equal to the rod, there is an optimal point to impart maximal spin.