Applying law of momentum conservation

In summary: The next step is to solve for v1 and v2 in terms of ω1 and v. Then you can substitute those expressions into the energy equation to find an equation that involves only ω1 and v. That equation is a quadratic equation for ω1.Nice. Your equations look correct to me. The next step is to solve for v1 and v2 in terms of ω1 and v. Then you can substitute those expressions into the energy equation to find an equation that involves only ω1 and v. That equation is a quadratic equation for ω1.In summary, we discussed a problem involving a rotating rod with two balls at its ends and a third ball with a different mass colliding with
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Homework Statement


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Suppose, a rod of negligible mass has two balls(A & B) at its two ends. Both the balls have same mass m. The rod is rotating around its center C with a uniform angular velocity ##\omega##. Another ball D (mass M) has a velocity v. At some time, A and D collide with each other. Now, how to apply the law of momentum conservation in this case? (The actual problem was different. But I need to understand how to apply the law of momentum conservation in such case.)[/B]

Homework Equations

The Attempt at a Solution


In this case, there are both linear and angular momentum. That confuses me.
Let the length of the rod be 2R.
And the final angular velocity of the rod is ##\omega _f## and final velocity of D is ##v_f##;
Now, I thought,
## m \omega R - m \omega R + Mv = m \omega _f R - m \omega _f R + Mv_f##
So, it becomes, ## v = v_f ##
But, it's not true.
 
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It is important to specify as part of the problem statement whether the rod is rotating about a fixed axis at C or whether it initially rotating freely in space (not attached to a fixed axis of rotation).
 
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Another thing to ask: do D and A have an elastic or inelastic collision?

Also, looking at:
[itex] mωR − mωR +M = mω_fR − mω_fR +Mv_f [/itex]
I believe mωR should be added to the other mωR because they spin in the same direction, i.e. they have the same value for ω
 
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TSny said:
It is important to specify as part of the problem statement whether the rod is rotating about a fixed axis at C or whether it initially rotating freely in space (not attached to a fixed axis of rotation).
Fixed axis.
Brian T said:
Another thing to ask: do D and A have an elastic or inelastic collision?
Elastic collision.
 
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If the axis is fixed, then the total linear momentum of the system consisting of the rod and ball D will not be conserved. Can you see why?
 
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TSny said:
If the axis is fixed, then the total linear momentum of the system consisting of the rod and ball D will not be conserved. Can you see why?
That's a very good point. There will be some force acting on the center C from outside. Then how to calculate the final angular velocity ?
 
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Besides linear momentum, what other type of momentum should you think about here?
 
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TSny said:
Besides linear momentum, what other type of momentum should you think about here?
Surely, angular momentum. But, what is the angular momentum of D ?
 
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Did you cover the definition of angular momentum of a point particle?
 
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TSny said:
Did you cover the definition of angular momentum of a point particle?
So, I think the equation will be,
## 2m \omega r^2 + Mvr = 2m \omega _f r^2 + Mv_f r##
And the equation for energy conservation,
##2\cdot \frac {1}{2} m r^2 \omega ^2+ \frac{1}{2} Mv^2 = 2\cdot \frac {1}{2} m r^2\omega _f ^2 + \frac{1}{2} Mv_f ^2 ##
 
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Looks good. (r = R)
 
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TSny said:
Looks good. (r = R)
In the actual problem, the picture was in the vertical plane. The ball D was falling down with the accelaration of ##9.8 ms^{-2}##.
So, in this case, I think, there will be a torque because of collision and another torque which is equal to ##MgR##.
Is it so?
 
  • #13
Generally for collision problems you assume that the collision forces are much stronger than the forces of gravity. So you assume that you can neglect the gravity force during the collision. This is sometimes referred to as the "impulse approximation". If you did want to include the effect of the weight of D during the collision, then you would need to know the time duration of the collision in order to determine the angular impulse caused by Mg.
 
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  • #14
TSny said:
Generally for collision problems you assume that the collision forces are much stronger than the forces of gravity. So you assume that you can neglect the gravity force during the collision. This is sometimes referred to as the "impulse approximation". If you did want to include the effect of the weight of D during the collision, then you would need to know the time duration of the collision in order to determine the angular impulse caused by Mg.
That's good!
Now, what would happen if the center was not fixed?
 
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Think about how many "unknowns" appear in the final state now that the entire rod can "recoil". How many conserved quantities are there?

I need to quit for the night. Will check back tomorrow.
 
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TSny said:
Think about how many "unknowns" appear in the final state now that the entire rod can "recoil". How many conserved quantities are there?

I need to quit for the night. Will check back tomorrow.
I think, after collision, the rod will have two type of motion: translational and rotational.
Let, the final translational velocity of the rod be ##v_1## and final velocity of ball D be ##v_2## ; and final angular velocity of rod be ##\omega_1## ;
So, applying the law of (linear) momentum conservation:
##0 + Mv = 2mv_1 + Mv_2## (assuming the rod as a point particle at its center of mass)
And angular momentum conservation:
##2m\omega r^2 + Mvr = 2m \omega _1 r^2 + Mv_2r##
Energy conservation:
##\frac{1}{2} \cdot 2mr^2 \omega ^2 + \frac {1}{2} Mv^2 = \frac{1}{2} \cdot 2mr^2 \omega _1^2 + \frac {1}{2} \cdot 2mv_1^2+ \frac {1}{2} Mv_2^2##
Three unknowns : ##\omega _1, v_1, v_2## ; three equations ;
Looks good?
 
  • #17
Nice. Your equations look correct to me.
 

What is the law of momentum conservation?

The law of momentum conservation states that the total momentum of a closed system remains constant, unless acted upon by an external force. This means that in a collision or interaction between objects, the total momentum before and after the interaction is equal.

How is the law of momentum conservation applied?

The law of momentum conservation can be applied by using the equation P1 = P2, where P1 is the total initial momentum and P2 is the total final momentum. This equation can be used to solve for unknown velocities or masses in a collision or interaction.

What types of systems does the law of momentum conservation apply to?

The law of momentum conservation applies to closed systems, which are systems where no external forces are acting on the objects involved. This includes collisions between objects, explosions, and other interactions.

What are the limitations of the law of momentum conservation?

The law of momentum conservation is based on the assumption that there are no external forces acting on the system. In reality, there may be small external forces present, which can affect the accuracy of the calculations. Additionally, the law of momentum conservation does not take into account energy losses due to factors such as friction.

Why is the law of momentum conservation important in science?

The law of momentum conservation is important in science because it helps us understand and predict the motion of objects in collisions and interactions. It is also a fundamental principle in physics and is used in a variety of real-world applications, such as in the design of car safety features and in space missions.

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