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Jun 1 at 11:30 comment added Ján Lalinský @ChrisChristopherson Frame of the small block is frame fixed to the block, thus center of the coordinate system remains in the same point of the block all of the time.
Jun 1 at 1:34 comment added Chris Christopherson @JánLalinský How is the small block's speed zero if we consider the frame of only the small block? I feel like im missing something crucial.
Mar 25 at 11:17 history reopened gandalf61
Vincent Thacker
Michael Seifert
Mar 25 at 9:41 comment added gandalf61 Voting to reopen. Obviously a conceptual question rather than a "do my homework for me" question.
Mar 25 at 9:40 review Reopen votes
Mar 25 at 11:17
Mar 23 at 12:43 history closed Bob D
Farcher
Hyperon
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Mar 23 at 3:57 history became hot network question
Mar 23 at 3:41 history edited Qmechanic CC BY-SA 4.0
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Mar 23 at 0:09 comment added Pato Galmarini @JánLalinský yes, that is what I meant, conservation of mechanical energy takes that into account implicitly. I guess I was not that clear. Also, by "frame of the small block" I assumed "I analyze teh block in isolation", otherwise, as you stated, v=0
Mar 22 at 23:57 comment added Ján Lalinský @PatoGalmarini But you're right that in the frame of the small block, the equation $mgR=\frac{1}{2}mv^2$ is wrong. This is because in that frame $v=0$.
Mar 22 at 23:54 comment added Ján Lalinský @PatoGalmarini normal forces between the big and small block are internal forces to the system, but this does not imply we can ignore them; they can and do work in this case, most visibly on the larger block. The reason we do not need to consider these internal forces when finding the final velocities is that we can find the solution just from the conservation laws, without careful consideration of the internal forces.
Mar 22 at 22:02 vote accept Chris Christopherson
Mar 22 at 21:46 review Close votes
Mar 23 at 12:43
Mar 22 at 21:29 history edited Bob D
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Mar 22 at 20:52 answer added Albertus Magnus timeline score: 2
Mar 22 at 20:48 comment added Pato Galmarini The equation for the small mass is wrong, because you are ignoring the work made by the normal force. you need to consider the conservation of mechanical energy for the entire system, and if so you can ignore the normal force as an internal force.
Mar 22 at 19:57 history asked Chris Christopherson CC BY-SA 4.0