Linear and angular momentum problem: Ball hitting a rod

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  • #1
barryj
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Homework Statement:
Ball of mass mb and velocity vb hits rod of length L , Rod pivots about the center. What is the angular momentum aafter impact?
Relevant Equations:
I = 1/12 (mR^2)
I = mR^2
Homework Statement:: Ball of mass mb and velocity vb hits rod of length L , Rod pivots about the center. What is the angular momentum aafter impact?
Homework Equations:: I = 1/12 (mR^2)
I = mR^2

See the attached figure. I understand the concept of linear and angular momentum separately but I am confused as to how to use both at the same time. In the attached diagram ball B hits a rod that pivots in the center. In this case we have a classic case of conservation of angular momentum. However, if the rod is free to move, i.e. not connected at the center to the pivot, then the rod would move horizontally as well as rotating. I am having trouble finding an explanation as to how to add the linear momentum to the equation. Are they handled separately or can I have one equation for the angular part and another for the linear part?

I know that initially I would have a linear momentum of the ball being Mb X Vb. Would this be equal to the Mr X Vr where Vr is the linear velocity of the center of mass?
 

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  • #2
PeroK
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Homework Statement:: Ball of mass mb and velocity vb hits rod of length L , Rod pivots about the center. What is the angular momentum aafter impact?
Homework Equations:: I = 1/12 (mR^2)
I = mR^2

Homework Statement:: Ball of mass mb and velocity vb hits rod of length L , Rod pivots about the center. What is the angular momentum aafter impact?
Homework Equations:: I = 1/12 (mR^2)
I = mR^2

See the attached figure. I understand the concept of linear and angular momentum separately but I am confused as to how to use both at the same time. In the attached diagram ball B hits a rod that pivots in the center. In this case we have a classic case of conservation of angular momentum. However, if the rod is free to move, i.e. not connected at the center to the pivot, then the rod would move horizontally as well as rotating. I am having trouble finding an explanation as to how to add the linear momentum to the equation. Are they handled separately or can I have one equation for the angular part and another for the linear part?

I know that initially I would have a linear momentum of the ball being Mb X Vb. Would this be equal to the Mr X Vr where Vr is the linear velocity of the center of mass?

Both linear and angular momentum are conserved. Equations are equations and if they include common variables they can be combined in any way that is mathematically valid.

PS remember that angular momentum is defined relative to a given point. So, you are also free to choose the point about which you measure angular momentum of the system.
 
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  • #3
PeroK
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Homework Statement:: Ball of mass mb and velocity vb hits rod of length L , Rod pivots about the center. What is the angular momentum aafter impact?
Homework Equations:: I = 1/12 (mR^2)
I = mR^2

I don't understand your equation (1). Can you explain that? Why are those two quantities equal?
 
  • #4
haruspex
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@PeroK has answered your main question, but in solving the actual problem you seem to have forgotten about what happens to the ball afterwards. More information is needed, e.g. it sticks to the rod, or bounces elastically, or happens to come to rest, etc.
 
  • #5
barryj
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Referring to my attached sketch, if the ball strikes the free rod at the center of mass, then there would be no rotation only a linear translation then I could say that mb X vb = mr X vr. Would this be a good equation if the ball stikes the rod off center?

I just realized that the initial angular momentum is incorrect because the center of toation was taken from the end of the rod not the center.
 
  • #6
barryj
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Oh, I forgot. this is an elastic collision and the ball remains stationary after the collision.
 
  • #7
haruspex
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Oh, I forgot. this is an elastic collision and the ball remains stationary after the collision.
Ok. But if you take the same set up (rod length, masses) and just compare with and without pivot then the ball will not end up stationary in both. (I think.)
 
  • #8
barryj
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I think that whether or not the ball remains stationary might depend on the initial velocity. But my biggest confusion is how to work with both linear and angular momentum when the rod is free to rotate and translate.
 
  • #9
PeroK
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I think that whether or not the ball remains stationary might depend on the initial velocity. But my biggest confusion is how to work with both linear and angular momentum when the rod is free to rotate and translate.

It can't depend on the initial velocity. It could depend on ##D##. But, that's not really relevant.

You are okay with one equation at a time, but not with two? What do you do when both linear momentum and linear KE are conserved? Those are two different equations.

PS how do you find the total linear momentum of a rigid body?
 
  • #10
barryj
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My understanding is that in an elastic collision, both KE and Momentum are conserved. I don't see how this answers my basic question above. I start with a ball of mass mb moving at a velocity of vb that hits a free rod. The rod then starts rotating and translating. How are these two effects put together?
 
  • #11
haruspex
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I start with a ball of mass mb moving at a velocity of vb that hits a free rod. The rod then starts rotating and translating. How are these two effects put together?
Just write down the conservation equations independently, as you did.
 
  • #12
barryj
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Could I say...

KE of ball before impact = (linear)KE of the center of mass of the rod + the rotational KE of the rod?

You could also think of the ball as having rotational momentum, or could you, about what axis? so could you think of the ball as initially having some rotational KE? You can see why I am confused.
 
  • #13
PeroK
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Could I say...

KE of ball before impact = (linear)KE of the center of mass of the rod + the rotational KE of the rod?

That's correct.

You could also think of the ball as having rotational momentum, or could you, about what axis? so could you think of the ball as initially having some rotational KE? You can see why I am confused.

The ball has no rotational KE.
 
  • #14
jbriggs444
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Could I say...

KE of ball before impact = (linear)KE of the center of mass of the rod + the rotational KE of the rod?

You could also think of the ball as having rotational momentum, or could you, about what axis? so could you think of the ball as initially having some rotational KE? You can see why I am confused.
The ball has some initial KE. The equations do not care whether it is "rotational", "linear" or something else. The ball likely has some initial angular momentum. How much will depend on what axis of rotation you choose.

"Rotational kinetic energy" as such need not enter in.
 
  • #15
barryj
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OK, the ball has initial KE of 1/2 Mb*Vb^2
After impact there is both rotational and linear KE, yes/no?
So could I write one equation equating the initial KE of the ball = Rotational KE of the rod plus the Linear KE of the rod? something like..(1/2)Mb*Vb^2 = (1/2) Ib*wb^2 + (1/2)Mrod*Vrod^2
 
  • #16
PeroK
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OK, the ball has initial KE of 1/2 Mb*Vb^2
After impact there is both rotational and linear KE, yes/no?
So could I write one equation equating the initial KE of the ball = Rotational KE of the rod plus the Linear KE of the rod? something like..(1/2)Mb*Vb^2 = (1/2) Ib*wb^2 + (1/2)Mrod*Vrod^2

That's definitely your first equation. There are two more needed. One for linear momentum; and one for angular momentum about a point of your choice.
 
  • #17
barryj
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Then for linear momentum I would say that
Mball*Velocityball = Massrod * Velocityrod since mass1 is stationary after impact
take center of rod as axis of rotation then
mD^2 * Velocityball/D = (1/12)Massrod * D^2 *Omegarod

Maybe I should learn LaText or post a new set of equations.
 
  • #18
PeroK
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Then for linear momentum I would say that
Mball*Velocityball = Massrod * Velocityrod since mass1 is stationary after impact
take center of rod as axis of rotation then
mD^2 * Velocityball/D = (1/12)Massrod * D^2 *Omegarod

Maybe I should learn LaText or post a new set of equations.

The linear equation looks right.

The AM equation is not right. You seem to be taking the AM of the ball and the rod about different points: the ball relative to the top of the rod and the rod about its COM. Those are different points.

To be clear: you must specify the point about which you are measuring the AM of the system.
 
  • #19
barryj
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Yes, the AM of the ball around the COM should be m(D/2)^2 * Vball/(D/2)
So, I am going to have three equations to work with, yes?

I think I will take a break, get my thoughts together and get back in a bit.
 
  • #20
PeroK
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Yes, the AM of the ball around the COM should be m(D/2)^2 * Vball/(D/2)
So, I am going to have three equations to work with, yes?

I think I will take a break, get my thoughts together and get back in a bit.

I don't understand ##D/2##. It would be simpler to introduce, ##d = D - L/2##, which is the offset from the COM.

I also don't understand the square term. The AM of a point mass is simply ##mvd## in this context.
 
  • #21
barryj
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I of the ball is mR^2 AM = I*Omega

You are correct. I have the d confused. I need to take a break and redraw my diagram and get my equations correct. Don't forget me as I shall return
 
  • #22
barryj
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I have attached an updated diagram with equations.
I think these equations agree with what we have been discussing.
I will now have to manipulate them to get the results I want.
At this point, I have forgotten what I am looking for :-)
 

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  • #23
barryj
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Just for fun I assigned some values to see what I would get.
Mr = 1 kg, Mb = .1 kg, L = 1m, D = 0.4m
Vb = 10 m/sec

I calculagte Ib == mr^2 = (.1)(.4)^2 = 0.016
I calculatge Ir = (1/12)mL2 = (1/12)(1)(1)^2 = .0833

Using eq 2 I get Vr = 1
Using Eq 3 I get omega(r) = 4.8

If I put these into the KE equation, #1, I get 0.7 not close to 5 as I expected since (1/2)(.1)(10)^2 = 5.

Hmmmmmmm....
 
  • #24
haruspex
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I have attached an updated diagram with equations.
I think these equations agree with what we have been discussing.
I will now have to manipulate them to get the results I want.
At this point, I have forgotten what I am looking for :-)
A couple of errors in there, but later fixed up, so maybe transcription errors.
In the second KE equation you start with 1/12 when you mean 1/2.
In the first angular momentum equation you have the angular velocities squared.
 
  • #25
barryj
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You are totally correct on both accounts.
 
  • #27
barryj
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No it didn"t
 
  • #28
haruspex
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No it didn"t
You have chosen that it is an elastic collision and the ball comes to rest. That will only happen with the right relationship between the masses, but you have plugged in arbitrary masses.
Let the ball's mass be unknown. With the resulting movement of the rod, that gives you three unknowns in total. Conservation of energy, linear momentum and angular momentum gives you three equations. Solve.
 
  • #29
PeroK
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Just for fun I assigned some values to see what I would get.
Mr = 1 kg, Mb = .1 kg, L = 1m, D = 0.4m
Vb = 10 m/sec

I calculagte Ib == mr^2 = (.1)(.4)^2 = 0.016
I calculatge Ir = (1/12)mL2 = (1/12)(1)(1)^2 = .0833

Using eq 2 I get Vr = 1
Using Eq 3 I get omega(r) = 4.8

If I put these into the KE equation, #1, I get 0.7 not close to 5 as I expected since (1/2)(.1)(10)^2 = 5.

Hmmmmmmm....

It turns out that one constraint for a solution to this problem is that ##M_b \le M_r \le 4 M_b##.

Also, depending on the ratio ##M_r/M_b##, the ball must hit the rod at a specific offset ##d##. For example, if ##M_b = M_r##, then ##d = 0## and you have the (hopefully familiar) case of an equal mass elastic collision.

There can't be a solution for all variables. In fact, the ball can only stop when there is a strict relationship between ##M_b, M_r, d## and ##L##.

The one truly independent variable is the impact velocity ##v_b##.
 
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  • #30
barryj
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Given that my assumed variables will not result in a stationary ball after inmpact I assume we can still solve this problem for Vr and Or after the collision. I am using O for (omega)

Using the linear momentum equation I get (0.1)(10 = (a)(Vr) so Vr = 1

Using the angular momentum equation (0.16)(10/0.4) = (.0833)Or so Or = 4.8

If we calculate the KE after the equation it is (.5))(1)(1)^2 + (.5)(.0833)(4.8)^2 = 1.45J
The KE of the ball before the collision is (.5)(.1)(10)^2 = 5J

Clearly thee is still some KE in the ball after the collision so I assume that with these variables, the ball does not come to rest after the collision. Is this correct? If so, then we can assume that the equations are correct and that the initial assumption that the ball comes to rest is wrong. I am learning a lot from this exchange.
 
  • #31
barryj
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"It turns out that one constraint for a solution to this problem is that Mb≤Mr≤4Mb. "
I don't understand why this is so at this time but I will continue to study this problem.
 
  • #32
PeroK
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Given that my assumed variables will not result in a stationary ball after inmpact I assume we can still solve this problem for Vr and Or after the collision. I am using O for (omega)

Using the linear momentum equation I get (0.1)(10 = (a)(Vr) so Vr = 1

Using the angular momentum equation (0.16)(10/0.4) = (.0833)Or so Or = 4.8

If we calculate the KE after the equation it is (.5))(1)(1)^2 + (.5)(.0833)(4.8)^2 = 1.45J
The KE of the ball before the collision is (.5)(.1)(10)^2 = 5J

Clearly thee is still some KE in the ball after the collision so I assume that with these variables, the ball does not come to rest after the collision. Is this correct? If so, then we can assume that the equations are correct and that the initial assumption that the ball comes to rest is wrong. I am learning a lot from this exchange.
Not quite. The problem is that, given the assumptions, you must find the angular velocity of the rod. In solving the problem you may find constraints between the various quantities.

Like the one I found for the masses.

You have to work algebraically I'm afraid!
 
  • #33
barryj
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Well, given the initial masses and velocities it seems to me that I can use the conservation of angular momentum and conservation of linear momentum to fine Vr and Or. Isn't Or = 4.8?

I did some algebra and if the equations are correct why are my results of Vr and Or not correct?
 
  • #34
PeroK
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Well, given the initial masses and velocities it seems to me that I can use the conservation of angular momentum and conservation of linear momentum to fine Vr and Or. Isn't Or = 4.8?

I did some algebra and if the equations are correct why are my results of Vr and Or not correct?
What result? You're supposed to get am expression for ##\omega## in terms of the other variables.
 
  • #35
barryj
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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
 

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