Linear and angular momentum problem: Ball hitting a rod

[PeroK]

I am getting confused.

Assuming that the parameters I selected are OK and that my equation for angular momentum is correct then..

IbOb = IrOr (I am using O for Omega)
I calculated that...
Ib = Mb*Db^2 = (.1)(.4)^2 = 0.016
Ob = Vb/Db = 10/.4 = 25
Ir = (1/12)MrL^2 = (1/12)(1)(1)=.08333
so
Or = IbOb/Ir = (0.016)(25)/(0.08333) = 4.8

Recall at this time I am using the parameters I established, i.e.
Mb = .1 , Vb = 10, Mr = 1, and Db = .4

$KE_i = \frac 1 2 M_b v_b^2 = 5J$

$KE_f = \frac 1 2 I_r w^2 = 9.6J$

Your methods are unsound!

You need to do this properly or not at all.

 

[barryj]

I am trying to do this properly. I am not understanding the basic concept here. Let's forget about my initial statement that the ball is stationary after the collision, OK?

Now, are my equations correct, if not then they must be fixed.

I agree that "KEi=12Mbv2b=5J "
I don't see where you get "KEf=12Irw2=9.6J"

However, I am not considering KE at this point, just conservation of linear and angular momentum.

I think my parameters are realistic so I should be able to calculate the Or after the collision with the equations, if they are correct.Yes?

 

[PeroK]

I am trying to do this properly. I am not understanding the basic concept here. Let's forget about my initial statement that the ball is stationary after the collision, OK?

Now, are my equations correct, if not then they must be fixed.

I agree that "KEi=12Mbv2b=5J "
I don't see where you get "KEf=12Irw2=9.6J"

However, I am not considering KE at this point, just conservation of linear and angular momentum.

I think my parameters are realistic so I should be able to calculate the Or after the collision with the equations, if they are correct.Yes?

I'm not sure what you've understood and what you haven't. We've told you that you can't just plug in any old masses here and get a consistent solution.

Your parameters are not valid for this problem, because they don't support the ball having zero velocity after the collision. A ball that light would rebound. I must have told you that several times already.

I don't understand why you are plugging numbers in rather than solving the problem properly - using the algebraic variables. It's not that hard, surely.

A few posts back you seemed to be getting close to having all the equations, but you've drifted off into a numerical approach that leads nowhere.

Your lack of Latex makes things hard to read as well, it must be said.

 

[jbriggs444]

IbOb = IrOr (I am using O for Omega)

Let us attack just this equation for the moment. It asserts that $I_b$ (the moment of inertia of the ball just prior to impact) times $\omega_b$ (the initial angular velocity of the ball just prior to impact) is equal to $I_r$ (the moment of inertia of the rod) times $\omega_r$ (the angular velocity of the rod just after impact).$$I_b \omega_b = I_r \omega_r$$
That would follow from conservation of angular momentum but only on the assumption that the ball stops dead after the impact. I thought that we were solving three simultaneous equations and discovering whether the ball would or would not stop based on the results obtained.

I am not entirely comfortable describing the angular momentum of the ball as $I_b \omega_b$ since it is not an object rotating rigidly about the reference axis. Instead, I would use $\vec{r_b} \times \vec{p_b}$ where $\vec{r_b}$ is the displacement of the ball from the reference axis, and $\vec{p_b}$ is its momentum and $\times$ denotes the vector cross product.

The concern with using $I_b$ to denote the moment of inertia of the ball is that it is not constant.

 

[barryj]

Ah Ha, I see some light. As briggs said, my angular momentum equation is valid only if the ball stops and I keep saying let's not have this condition and this makes my equation wrong. I must learn LaTex also!

 

[jbriggs444]

Ah Ha, I see some light. As briggs said, my angular momentum equation is valid only if the ball stops and I keep saying let's not have this condition and this makes my equation wrong. I must learn LaTex also!

You can also use the clicky thing that looks like $\sqrt{x}$ above the editting panel. That will give you to access to symbols such as "ω".

I cannot find superscripts and subscripts there. But you can insert those using bbcode tags such as ω[sub]b[/sub]. Which renders as ω_b

 

[PeroK]

Ah Ha, I see some light. As briggs said, my angular momentum equation is valid only if the ball stops and I keep saying let's not have this condition and this makes my equation wrong. I must learn LaTex also!

You need to decide whether the ball has a known velocity of zero after the collision; or a to-be-determined velocity.

Having a zero velocity simplifies the equation, but puts constraints on the variables.

It's your problem, so which is it?

 

[jbriggs444]

Having a zero velocity simplifies the equation, but puts constraints on the variables.

If I understand correctly, the constraint on the variables can be seen as arising from the idea that the collision should not result in a net increase in kinetic energy. One can erase that constraint by placing a tiny blasting cap of exactly the right size on the rod at the point of collision.

 

[barryj]

Lets start out with zero just so I can get something that is correct.
If the ball stops then the linear KE of the rod + angular KE of the rod must be 5.
This would give a relationship between ω_r and V_r yes?

 

[jbriggs444]

Lets start out with zero just so I can get something that is correct.
If the ball stops then the linear KE of the rod + angular KE of the rod must be 5.
This would give a relationship between ω_r and V_r yes?

If the ball stops and you still insist on conserving kinetic energy, that gives you too many equations. They will likely be inconsistent. The obvious suggestion is to drop the requirement that kinetic energy be conserved and go with conservation of linear and angular momentum only.

 

[barryj]

If we don't insist on conserving KE then we can eliminate equation 1, yes and only use the conservation of linear momentuim and conservation of angular momentum equations, yes?

 

[PeroK]

If the ball stops and you still insist on conserving kinetic energy, that gives you too many equations. They will likely be inconsistent. The obvious suggestion is to drop the requirement that kinetic energy be conserved and go with conservation of linear and angular momentum only.

The problem as originally stated was quite nice I thought. You just have to assume that the ball hit the rod at just the right spot.

 

[jbriggs444]

If we don't insist on conserving KE then we can eliminate equation 1, yes and only use the conservation of linear momentuim and conservation of angular momentum equations, yes?

Yes. The angular momentum equation I'd suggested above would be$$\vec{r_b} \times \vec{p_b} = I_r\omega_r$$ Can you produce for us the linear momentum equation in a similar format before proceeding to solve? [It is easier to follow algebra than numeric computations]

 

[barryj]

m_bv_b = m_rv_r

 

[barryj]

Does the "thumbs up" mean I am correct? If so then we can solve for v_r given the other inputs, yes?

 

[barryj]

Is the angular equation above the same as
m_bv_bd = I_rω_r where d is the distance between the axis of rotation and the impact point?

 

[PeroK]

Is the angular equation above the same as
m_bv_bd = I_rω_r where d is the distance between the axis of rotation and the impact point?

Yes. This assumes, of course, that the ball is at rest after impact.

 

[barryj]

Can I make the statement that the ball comes to rest after impact? If so then the equation in post 42 and 52 are correct, yes? and we can solve for the final v_r and ω_r.

 

[PeroK]

Can I make the statement that the ball comes to rest after impact? If so then the equation in post 42 and 52 are correct, yes? and we can solve for the final v_r and ω_r.

You can assume that is the case. That gives you a specific problem to solve.

Or, you can assume it does not, and that gives you a more general problem to solve.

I had assumed that the source of this question would have made that clear?

 

[barryj]

Perok. Since I set initial parameters for the problem, i.e. elastic collision, masses, velocity of the ball, length of rod, distance of impact from the COM, then I do not think I can assume that the ball will be stationary after impact. Isn't this correct? So if it is not stationary, then many if not all of my equations are incorrect. yes?

 

[PeroK]

Perok. Since I set initial parameters for the problem, i.e. elastic collision, masses, velocity of the ball, length of rod, distance of impact from the COM, then I do not think I can assume that the ball will be stationary after impact. Isn't this correct? So if it is not stationary, then many if not all of my equations are incorrect. yes?

If this is some problem you made up and you want to assume specific numbers for the variables, then you cannot assume the final velocity of the ball is zero.

 

[barryj]

If we assume the ball is stationary after the collision, then as I think you pointed out earlier, then there must be some relationships between the masses of the objects such that KE is also conserved, assuming an elastic collision. Is this correct?

 

[PeroK]

If we assume the ball is stationary after the collision, then as I think you pointed out earlier, then there must be some relationships between the masses of the objects such that KE is also conserved, assuming an elastic collision. Is this correct?

I seem to recall saying that a number of times.

 

[barryj]

I have a thick skull.

 

[barryj]

I have not seen a problem of this type in any of my physics books. All of the problems I see tend to have the rod fixed to a pivot point and this makes the problem much easier. Can this problem even be solved at all?

 

[jbriggs444]

I have not seen a problem of this type in any of my physics books. All of the problems I see tend to have the rod fixed to a pivot point and this makes the problem much easier. Can this problem even be solved at all?

Yes. Once you have settled on which variant you are interested in, it can be solved.

I think that three variants are currently on the table:

1. Ball ends at rest. Rod is not anchored. Use conservation of linear and angular momentum to determine final linear velocity and angular velocity of rod.

2. Elastic collision. Rod is not anchored. Use conservation of linear and angular momentum and of kinetic energy to determine final linear velocity and angular velocity of rod and final velocity of ball.

3. Ball ends at rest and elastic collision. Rod is not anchored. Use conservation of linear and angular momentum and of kinetic energy to determine final linear velocity and angular velocity of rod and one additional parameter -- such as the point of impact.

 

[barryj]

Lets start with case #1 above. Given my initial parameters, i.e. Mass of ball = 0.1 kg, mass of rod = 1 Kg,velocity of ball = 10 m/sec, impact 0.4 m from rod center, can we assume the ball will be at rest after the collision. I don't think so.

#2 and #3 might be possible. Which one is the easiest to solve? Let's do that one?

 

[barryj]

I solved #3 with d = 0.866