Bullet Hit Mechanics: Force, Momentum, and Rotation

In summary, the conversation discusses a project on simulating a bullet hitting a glass plate and the resulting movement of shards. The placement of the bullet determines the axis of rotation and there is no force keeping the other side of the shard in place. The further the bullet hits from the shard's center of mass, the more energy is transferred into rotation instead of translation.
  • #1
zoltaaaan
2
0
Well hello there! :D

I'm doing a project on graphical simulation of a bullet hitting a glass plate, and I have some describing the way the shards fly off upon impact.

Note that in this calculation I am not yet taking the force of gravity or any forces other than the force of the impact of the bullet.

To be more precise, say my shard is a triangle-based prism as shown in the attached figure. If the bullet hit that shard at the center of its mass, perpendicularly to the plane of the base of the prism, the prism would - given that it doesn't break - receive some momentum and fly off straight along the axis of the initial trajectory of the bullet, without any rotation.

However, should the bullet hit anywhere beside the axis through the mass center of the shard, for example at one of its vertices, as shown on the figure, beside the straight translation, some rotation would also occur.

My question is: where is the axis of this rotation, and what forces (probably inertial) keep the other side of the shard in place.

I hope I made myself clear enough, if not, please let me know and I will try to elaborate further.

Thank you.
 

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  • #2
It's not clear to me what you're asking. The impulse delivered by the bullet produces a translational acceleration of the center of mass and a rotational acceleration about the center of mass. The instantaneous axis of rotation at any moment can be determined by adding the two motions together. There's no force "keeping the other side in place".
 
  • #3
Well you kind of answered my question... :D I had a hunch that it would be rotating around its center of mass, but I had nothing to back that thought up, at least now I have your opinion :D

Could you please also comment on this thought:
I'm guessing that the further the bullet hits from the shard's center of mass, the more energy would be transferred into rotation, and less to translation of the shard. Am I correct?
 

FAQ: Bullet Hit Mechanics: Force, Momentum, and Rotation

1. What factors contribute to the force of a bullet hit?

The force of a bullet hit is primarily determined by the mass and velocity of the bullet. Other factors that can affect the force include the angle of impact, the type and hardness of the object being hit, and the type of bullet being used.

2. How does momentum play a role in bullet hits?

Momentum is a measure of an object's mass and velocity combined. In the case of a bullet hit, the momentum of the bullet is transferred to the object it hits, causing it to move or deform. The greater the momentum of the bullet, the greater the impact force will be.

3. Can rotation occur during a bullet hit?

Yes, rotation can occur during a bullet hit depending on the angle of impact and the shape of the bullet. If the bullet strikes an object at an angle, it may cause the object to rotate as well. Additionally, some bullets are designed to have a rotational motion as they travel through the air, which can also affect the rotation during impact.

4. How does the type of surface being hit affect the bullet's force and momentum?

The type of surface being hit can greatly affect the force and momentum of a bullet. A softer surface, such as flesh, will absorb more of the bullet's energy, resulting in a lower force and momentum upon impact. A harder surface, such as metal or concrete, will not absorb as much energy, resulting in a greater force and momentum upon impact.

5. Is there a way to calculate the force and momentum of a bullet hit?

Yes, the force and momentum of a bullet hit can be calculated using the principles of physics. The mass, velocity, and angle of impact can all be taken into account to determine the force and momentum of the bullet as well as the resulting force on the object being hit.

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