Suggestions:- Say we define the following: "1" is the time of release; "2" is the time that A impacts arm; "3" is time immediately after A is wedged into arm.
- Note that energy is conserved between 1 and 2 for block A. However, angular momentum is NOT conserved between 1 and 2.
- Note that energy is NOT conserved between 2 and 3 (energy is typically not conserved during an impact). However, angular momentum for the system of arm + A IS conserved.
- Based on the above, consider working the problem in two steps: from 1 to 2 (using conservation of energy for A) and from 2 to 3 (using conservation of angular momentum for A + arm).
6 comments:
what does energy been conserved tell you and what does momentum conserved mean? I can't figure out what that information tells me
How does the momentum of the arm itself come into play? I found the velocity of the plug upon contact and i don't know where to go from there.
--to deborah--
Conservation of energy results from no external, non-conservative work being done on the system. This conservation means that T + V before is equal to T + V afterward. During impact, energy is typically lost to heat, vibrations, etc. and is typically not constant through the impact.
Conservation of angular momentum (about a FIXED point O) results from no moment being exerted about point O. This conservation means that H_O before is equal to H_O afterward. If you find the numerical value for H_O before impact, in this case, it is equal to the numerical value of H_O after impact.
--to jonbrueck--
The rod connecting the sphere and the cup (and pinned to ground at O) is assumed to have mass that is negligible compared to the mass of the rest of the system.
With that assumption, all of the angular momentum of the "arm" comes from the angular momentum of the sphere and cup. That is, write down m*(rxv) for the sphere and cup individually. Add these two angular momentum components together to get the total angular momentum for the arm. The total angular momentum of the system is that of the arm + that of the plug.
Note that both the plug and arm have non-zero angular momentum before and after the impact.
I was wondering if my answer was correct. I'm not sure if I did it correctly. I got omega_2 = -0.304 k rad/sec.
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