Calculating angular frequency and velocity after a collision

In summary: The problem states that the bullets are approaching the disk "both with trajectories parallel to the x-axis and at distance h, -h from the y-axis respectively". This means that they are approaching from opposite sides of the disk, with one bullet at a distance h above the x-axis and the other at a distance h below the x-axis.
  • #1
Fibo112
149
3

Homework Statement



The Problem is the following: We have a uniform disk of radius r laying still with its center at the origin. Two bullets, with equal mass m and negligible size are approaching the disk, both with trajectories parallel to the x-axis and at distance h, -h from the y-axis respectively where h<r. The velocities are v1 and v2 whre v2>v1. Both bullets remain stuck in the disk immediately without penetrating it. The question is now to solve for v1 and v2 in dependence of the angular velocity and the velocity of the disk after the collision.

Homework Equations



mh(v1+v2)=(1/2 Mr^2 + (2mr^2))w

The Attempt at a Solution


I can solve the question without problems using this relation, which is also used in the solution. mh(v1+v2)=(1/2 Mr^2 + (2mr^2))w where w is the angular velocity. What bothers me is 1. why can we just assume that the disk will rotate about its center( at least I think this has to be assumed based on the moment of inertia we are using), 2. If the disk does rotate about its center it seems to me that it will have to continuously slow down unless the connecting line between the impact points of the bullets goes through the center.
 
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  • #2
Fibo112 said:

Homework Statement



The Problem is the following: We have a uniform disk of radius r laying still with its center at the origin. Two bullets, with equal mass m and negligible size are approaching the disk, both with trajectories parallel to the x-axis and at distance h, -h from the y-axis respectively where h<r. The velocities are v1 and v2 whre v2>v1. Both bullets remain stuck in the disk immediately without penetrating it. The question is now to solve for v1 and v2 in dependence of the angular velocity and the velocity of the disk after the collision.

Homework Equations



mh(v1+v2)=(1/2 Mr^2 + (2mr^2))w

The Attempt at a Solution


I can solve the question without problems using this relation, which is also used in the solution. mh(v1+v2)=(1/2 Mr^2 + (2mr^2))w where w is the angular velocity. What bothers me is 1. why can we just assume that the disk will rotate about its center( at least I think this has to be assumed based on the moment of inertia we are using), 2. If the disk does rotate about its center it seems to me that it will have to continuously slow down unless the connecting line between the impact points of the bullets goes through the center.
I think the solution requires the assumption: The bullets strike the disk simultaneously.

Also, I suspect there is a misprint in the problem statement. The problem statement says, "...both with trajectories parallel to the x-axis and at distance h, -h from the y axis ..." I think that is a mistake. I'm guessing h and -h should be defined as the distances from the x-axis, not the y-axis. In other words, I think the problem statement is saying that one bullet is traveling at a distance h above the x-axis, and the other is traveling in the opposite direction, and a distance h below the x-axis.
 
  • #3
collinsmark said:
I think the solution requires the assumption: The bullets strike the disk simultaneously.

Also, I suspect there is a misprint in the problem statement. The problem statement says, "...both with trajectories parallel to the x-axis and at distance h, -h from the y axis ..." I think that is a mistake. I'm guessing h and -h should be defined as the distances from the x-axis, not the y-axis. In other words, I think the problem statement is saying that one bullet is traveling at a distance h above the x-axis, and the other is traveling in the opposite direction, and a distance h below the x-axis.
Thanks for the reply. You are right, I meant distance h from the x axis. Could you ellaborate further on how the assumption of simultanious impact is sufficient.
 
  • #4
Fibo112 said:
Thanks for the reply. You are right, I meant distance h from the x axis. Could you ellaborate further on how the assumption of simultanious impact is sufficient.
Well, that means that the bullets strike the disk on opposite sides of the disk. That means that line that connects the impact locations goes through the disk's center.

Also, keep in mind the part of the problem statement that says, "Both bullets remain stuck in the disk immediately without penetrating it." I take that as meaning you should treat the bullets as point particles that remain on the surface of the disk; i.e., they don't penetrate deeper into the disk.

Given all that, what can you conclude about the system's (disk + bullets) final center of mass?
 
  • #5
collinsmark said:
Well, that means that the bullets strike the disk on opposite sides of the disk. That means that line that connects the impact locations goes through the disk's center.

Also, keep in mind the part of the problem statement that says, "Both bullets remain stuck in the disk immediately without penetrating it." I take that as meaning you should treat the bullets as point particles that remain on the surface of the disk; i.e., they don't penetrate deeper into the disk.

Given all that, what can you conclude about the system's (disk + bullets) final center of mass?
That it is the center of the disk. But wouldn't it also be conceivable that the disk ends up rotating about a different point?
 
  • #6
Fibo112 said:
That it is the center of the disk. But wouldn't it also be conceivable that the disk ends up rotating about a different point?
It might if its motion is restricted somehow, such as if the disk was pinned to a hinge or affixed to rails, etc.

But if the final system (disk + attached bullets combo) is allowed to freely move and freely rotate in space, without restriction, it will rotate around its center of mass.
 
  • #7
So are they coming in from opposite sides (+x and -x) or from the same side? I cannot find where it specifies that.
 
  • #8
would it be correct to say that the reason for this is that any rotation that is not about the center of mass requires addition forces to sustain it and since there are no forces acting after the collision it must rotate about its center of mass?
 
  • #9
scottdave said:
So are they coming in from opposite sides (+x and -x) or from the same side? I cannot find where it specifies that.
Yeah I forgot to mention that. Opposite sides
 
  • #10
Fibo112 said:
would it be correct to say that the reason for this is that any rotation that is not about the center of mass requires addition forces to sustain it and since there are no forces acting after the collision it must rotate about its center of mass?
That sounds reasonable to me. :smile:
 
  • #11
great, thanks for your help
 
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Related to Calculating angular frequency and velocity after a collision

1. How do you calculate angular frequency after a collision?

Angular frequency after a collision can be calculated by dividing the final angular velocity by the radius of the object. The formula is ω = v/r, where ω is the angular frequency, v is the final angular velocity, and r is the radius of the object.

2. What is the formula for calculating angular velocity after a collision?

The formula for calculating angular velocity after a collision is ω = (Iω₀ + mv₀r) / (I + mr²), where ω is the final angular velocity, I is the moment of inertia of the object, ω₀ is the initial angular velocity, m is the mass of the object, and r is the radius of the object.

3. Can angular velocity change after a collision?

Yes, angular velocity can change after a collision. It depends on the type of collision and the forces involved. In an elastic collision, the angular velocity will remain the same, while in an inelastic collision, the angular velocity may change due to the transfer of energy and momentum.

4. What factors can affect the calculation of angular frequency and velocity after a collision?

The factors that can affect the calculation of angular frequency and velocity after a collision include the mass, moment of inertia, and velocity of the objects involved in the collision, as well as the type of collision and any external forces acting on the objects.

5. How can the conservation of angular momentum be applied when calculating angular frequency and velocity after a collision?

The conservation of angular momentum states that the total angular momentum of a system will remain constant unless an external torque is applied. This principle can be applied when calculating angular frequency and velocity after a collision, as the total angular momentum of the system before and after the collision must be equal.

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