KE of off-center impacts in rotating and translating rigid bodies

In summary, the conversation discusses the energy available in off-center impacts of thrown objects. The focus is on understanding the kinetic energy transferred in the impact, particularly on the "advancing" side versus the "receding" side. The concept of completely inelastic collisions is mentioned, with the reminder that in such cases, there is no linear kinetic energy after the collision. The importance of considering both linear and angular momentum in calculating the energy transferred is emphasized. The participants acknowledge the complexity of the problem and the potential for a simpler solution that may have been overlooked.
  • #1
dpettig
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I'm trying to understand basic principles of ancient thrown weaponry. Let's say we have something like a bar with a known inertia tensor that is thrown from one end such that it is both rotating and translating. If it strikes something along either side of its center of mass (an off-center impact), how can we understand the energy available in the impact? For example if it strikes with its long axis roughly perpendicular to the vector of translation (see sketch), on one side (the "advancing" side relative to the combined rotation and translation vectors) the kinetic energy in the impact will be much higher than on the "receding" side. But even here not all the KE available in the body will be transferred to the impacted object (which for the sake of simplicity can be inelastic). I hope it's as simple as including the distance (r) from CM somewhere in the equation? Please forgive the terrible sketch.
 

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  • #2
dpettig said:
... (which for the sake of simplicity can be inelastic). ...
Working in the rest frame of the common center of mass: For completely inelastic collisions (objects stick together), there is no linear KE after the collision. The remaining rotational KE follows from the conserved angular momentum.
 
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  • #3
My apologies, I forgot the difference between elastic and inelastic collisions. This would be of the latter form, but all I care about is the KE available in the impact, so the specifics of the collision can be simplified as much as possible.
 
  • #4
A.T. said:
Working in the rest frame of the common center of mass: For completely inelastic collisions (objects stick together), there is no linear KE after the collision. The remaining rotational KE follows from the conserved angular momentum.
I think that this over-idealizes the situation.

[I've cast the problem as two men rather than a man and a weapon. It is easier to refer to body parts and is pleasantly symmetric]

In a completely inelastic collision of one man's hand on another man's hand, the hands will come to relative rest, but not necessarily the men. One can calculate the impulse that is required to deliver enough combined linear and angular velocity change on each man so that the two hands each come to relative rest. One can then calculate the resulting change in linear and rotational energy for each man.
 
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  • #5
The change in momentum from the impact would be split between linear and angular. You would have to find the final linear and angular velocity to get the appropriate amount of linear and rotational energy. Good luck finding that. 🤷‍♂️
 
  • #6
Thanks osilmag. Right, this is challenging, hence why I've read everything including a book on off-center impacts and haven't found a simple solution. But I am not a physicist by training, so I am suspicious that I've already come across the solution and didn't recognize it.
 

Related to KE of off-center impacts in rotating and translating rigid bodies

1. What is kinetic energy (KE)?

Kinetic energy is the energy possessed by an object due to its motion. It is a scalar quantity and is dependent on the mass and velocity of the object.

2. How is KE related to off-center impacts in rotating and translating rigid bodies?

When a rigid body is rotating and translating, it has both linear and angular motion. During an off-center impact, both types of motion contribute to the overall KE of the body.

3. How does the location of impact affect the KE of a rotating and translating rigid body?

The location of impact can affect the KE of a rotating and translating rigid body because it determines the direction and magnitude of the linear and angular velocities at the point of impact. A direct impact at the center of mass will result in a higher KE compared to an off-center impact.

4. What is the formula for calculating KE in a rotating and translating rigid body?

The formula for calculating KE in a rotating and translating rigid body is KE = 1/2 * m * v^2 + 1/2 * I * ω^2, where m is the mass of the body, v is the linear velocity, I is the moment of inertia, and ω is the angular velocity.

5. How can the KE of off-center impacts in rotating and translating rigid bodies be applied in real-world scenarios?

The understanding of KE in off-center impacts in rotating and translating rigid bodies is important in various industries such as sports, engineering, and transportation. It can help in designing more efficient and safer equipment, analyzing the effects of collisions, and predicting the behavior of objects in motion.

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