Rod, Clay, and an Inelastic Collision

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Homework Help Overview

The problem involves a thin rod and a clay ball undergoing an inelastic collision on a frictionless table. The rod is struck at a specific point, and the discussion centers around determining the translational and rotational motion of the rod after the collision using principles of conservation of momentum.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using conservation of angular momentum and linear momentum to analyze the motion post-collision. There are differing opinions on whether to calculate the translational velocity directly or to consider the rotational motion about the center of mass of the system.

Discussion Status

Some participants have offered guidance on how to approach the problem, suggesting methods for calculating both translational and rotational velocities. There is an ongoing exploration of the definitions and assumptions regarding the center of mass and the axes of rotation.

Contextual Notes

Participants are questioning the reference points for their calculations and the implications of the inelastic collision on the motion of the composite object. There is also a mention of the complexity involved in determining the center of mass after the collision.

e(ho0n3
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Problem: A thin rod of mass M and length L rests on a frictionless table and is struck at a point L/4 from its center of mass by a clay ball of mass m moving at a speed v (the velocity vector is perpendicular to the rod). The ball sticks to the rod. Determine the translational and rotational motion of the rod after the collision.

I can use conservation of angular momentum to determine the rotational velocity of the rod and the clay about the center of mass of the rod. Then, I'd figure I could use this angular speed to find the velocity of the clay and then use conservation of linear momentum to find the velocity of the center of mass of the rod. According to my calculations, the center of mass of the rod is moving in the same direction of the velocity vector v. Does this make sense?
 
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That seems like an awful lot of work to simply calculate the velocity of the center of mass after collision. Just use conservation of linear momentum throughout to calculate translational velocity. Use conservation of angular momentum to find the angular velocity about its new center of mass after collision.
 
Gza said:
That seems like an awful lot of work to simply calculate the velocity of the center of mass after collision. Just use conservation of linear momentum throughout to calculate translational velocity. Use conservation of angular momentum to find the angular velocity about its new center of mass after collision.
Hmm...I thought the rotation was about the center of mass of the rod not about the center of mass of the system after collision. How do you figure?
 
e(ho0n3 said:
Hmm...I thought the rotation was about the center of mass of the rod not about the center of mass of the system after collision. How do you figure?
The motion of a rigid object (such as the composite object "rod + putty") can be described as a translational motion of its center of mass plus a rotation about its center of mass.

You know how to find the speed of the center of mass after the collision: linear momentum is conserved.

To find the rotational speed about the cm, do this. First find where the cm is just at the instant of collision. Then find the angular momentum about the cm prior to the collision. Since it's conserved, that's also the angular momentum about the cm after the collision. Find the rotational inertia of the "rod + putty" about the cm, then use it to find [itex]\omega[/itex].
 
Doc Al said:
The motion of a rigid object (such as the composite object "rod + putty") can be described as a translational motion of its center of mass plus a rotation about its center of mass.
I think you meant "can be BEST describe...", since motion is relative to a reference frame and the point in the frame I'm describing the motion from. When you use the phrase about its center of mass, I imagine the object is rotating about an axis going through the center of mass, but how do I know, in this problem for example, where the rotation axis is (I know it's not fixed in some location in space, but that is all I know).
To find the rotational speed about the cm, do this. First find where the cm is just at the instant of collision. Then find the angular momentum about the cm prior to the collision. Since it's conserved, that's also the angular momentum about the cm after the collision. Find the rotational inertia of the "rod + putty" about the cm, then use it to find [itex]\omega[/itex].
Let me get this clear: I should do all of my calculations (pre- and post-collion) taking the center of mass of the rod + putty as my reference point.
 
translation plus rotation

e(ho0n3 said:
I think you meant "can be BEST describe...", since motion is relative to a reference frame and the point in the frame I'm describing the motion from.
As long as you realize that the motion of a rigid object is a combination of the motion of its cm plus its rotation about its cm.
When you use the phrase about its center of mass, I imagine the object is rotating about an axis going through the center of mass, but how do I know, in this problem for example, where the rotation axis is (I know it's not fixed in some location in space, but that is all I know).
I'm not saying that the object is in pure rotation about its center of mass. It's also translating.
Let me get this clear: I should do all of my calculations (pre- and post-collion) taking the center of mass of the rod + putty as my reference point.
I would, but I'm lazy. (All what caculations? It's just angular momentum.) As long as you realize that the total angular momentum of the object about any axis is the sum of its angular momentum about its center of mass plus the angular momentum of its mass (assumed concentrated at its center of mass), then you can use any axis. So, if you find the initial angular momentum about the cm, then that will equal the final angular momentum about the cm, which would equal [itex]I_{cm}\omega[/itex].
 

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