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Physics Help
- Thread starter Gribbix
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Gribbix said:
If you had asked me 11 years ago when I was in AP Physics I'd have been all over it. Today, all I know is that someone here can help you, and probably quickly. This doesn't seem like that difficult a problem. Of course, in AP Physics, our teacher started with "Everything you learned in Physics is a lie, here's friction bitches. Have fun
."The physics problem is all about conservation of angular momentum:
L(disk) + L(boy) = L(disk + boy)
Remember that L = I * omega where I = the moment of inertia about the axis of rotation and omega = the angular velocity. (You can look up I for a solid disk rotating about the central axis -- I want to say that it's (1/2)Mr² where M is the total mass of the disk and r is the radius of the disk, but double-check that.)
What is the angular momentum for the boy? For a particle moving linearly, its angular momentum with respect to some arbitrary "axis of rotation" (even though it's not rotating) is L = mvr, where m = the mass of the particle, v = linear velocity, and r = the minimum distance between the path of travel and the axis of rotation. (In this case, since he's running tangent to the edge of the merry-go-round, the r in that equation should be the same r as the one from the moment of inertia of a solid disk with radius r.)
So, expanding terms and using w for omega (since I'm lazy):
I(disk)w(disk) + mvr = I(disk + boy)w_final
where w_final is the final angular rotation of the combined object (the merry-go-round with the boy on it).
(1/2)Mr²w + mvr = (0.5M + m)r² * w_final
The 0.5M + m term comes from the fact that the moment of inertia of a point particle is mr²...again, assuming my memory still works.
That should get you everything you need to solve the problem.
L(disk) + L(boy) = L(disk + boy)
Remember that L = I * omega where I = the moment of inertia about the axis of rotation and omega = the angular velocity. (You can look up I for a solid disk rotating about the central axis -- I want to say that it's (1/2)Mr² where M is the total mass of the disk and r is the radius of the disk, but double-check that.)
What is the angular momentum for the boy? For a particle moving linearly, its angular momentum with respect to some arbitrary "axis of rotation" (even though it's not rotating) is L = mvr, where m = the mass of the particle, v = linear velocity, and r = the minimum distance between the path of travel and the axis of rotation. (In this case, since he's running tangent to the edge of the merry-go-round, the r in that equation should be the same r as the one from the moment of inertia of a solid disk with radius r.)
So, expanding terms and using w for omega (since I'm lazy):
I(disk)w(disk) + mvr = I(disk + boy)w_final
where w_final is the final angular rotation of the combined object (the merry-go-round with the boy on it).
(1/2)Mr²w + mvr = (0.5M + m)r² * w_final
The 0.5M + m term comes from the fact that the moment of inertia of a point particle is mr²...again, assuming my memory still works.
That should get you everything you need to solve the problem.
Wellington
BAAAALLLINNN'
Fuck, I clicked on this topic all like *IN BEFORE JINX*, scrolled down, and there he was with the solution.
One day I will have my day in the sun, one day.
One day I will have my day in the sun, one day.
Wellington said:Fuck, I clicked on this topic all like *IN BEFORE JINX*, scrolled down, and there he was with the solution.
One day I will have my day in the sun, one day.
You're not alone. I always click the physics and math threads to lend a hand if I can, but -jinx- is a machine. He cannot be stopped!
Duality said:
Hmmm...neither is strongly acidic/basic in the Lewis sense, but both are weakly protic (isopropanol moreso, since it is a secondary alcohol). Loss of the hydroxyl proton will form an anion, which is stabilized via induction by the electron-withdrawing alkyl groups (recall that alkyl groups can act as electron "donators" or "acceptors" based on need; "donator" and "acceptor" are in quotes here because there is no formal transfer of electrons a la Lewis acid/base reactions-- it's more a quasi-resonance phenomenon than anything). So the isopropanol would be the more likely of the pair to donate its loosely held proton due to its secondary nature.
Based on that, it would seem to be possible that the hydroxyl group on the ethanol molecule would become protonated, forming the oxonium ion OH2+, which is a good leaving group (i.e., is weakly basic); from there, the trace isopropoxide anions in the solution would initiate nucleophilic attack of the SN2 variety (as opposed to SN1; this is due to the fact that ethanol is a primary substrate and isopropoxide ions are strong nucleophiles-- the secondary nature of the nucleophile might suggest that steric hindrance would be prohibitive, but ethanol is a small-chain primary substrate, so I'd say the effect would be negligible), causing inversion of configuration about the carbon in question. The OH2+ would be the leaving group, and the isopropoxide ion the attacking species; the product predicted by this theory would be isopropyl-ethyl ether.
It might very well be the case that this reaction would occur in the reverse direction (with ethanol going protic and isopropanol being attacked), but the product would be the same in either case, whether the course of reaction was SN2 or SN1 (since no 1,2 shifts-- methyl or hydride-- would be predicted based on the location of the substituents). Also realize that if this reaction takes place in an aqueous solution, the water would act as a mediator/facilitator, and there would be obtained the above product as well as ethanol and isopropanol again, this time with the hydroxyl group donated from the water molecules (though this could only be differentiated from the starting materials-- ethanol and isopropanol-- by use of an isotopic label such as deuterium).
Keep in mind that this is largely speculation on my part, but I'm fairly confident in saying that if some side reaction occurred, this would likely be it-- though it would definitely be present only in trace amounts, as neither of the starting molecules are protic or reactive enough to force a reaction forward; further, the predicted ether product would likely not persist long enough to be isolated, as it would tend to be cleaved by the weak acids/strong bases present in the system, yielding the original fragments. Hope this helped.
-jinx- said:
Eh, biochem is next year for me, but organic chem is actually very enjoyable. I think you'd really enjoy it if you ever looked into it in your spare time-- it's very concept-oriented, the progression of the material is extremely logical, and it's quite a fulfilling experience once you master a particular set of reactions.
I definitely enjoyed O-chem more than general chem, though I enjoyed both of them quite a bit. There's much less quantitative analysis in O-chem than in general chem, however (at least at the beginning; there's tons of qualitative analysis, though), so that may be somewhat unappealing for someone of your background. Still, it comes highly recommended; it's probably been my favorite course during my collegiate career. If nothing else, it's pretty cool to be able to look at the ingredients in household products (shampoos, detergents etc.) and have a good idea of the structure and nature of many/most of them.
</geek>
TheQueen'sOwn
insert blank space here
Wow! This is some good stuff (for Gr12 Physics and Chem)! *Takes notes*
Thanks for most likely helping me on my exam guys!
Thanks for most likely helping me on my exam guys!
Okay, I'm officially an idiot. In my previous post, it seems I may have made an oversight-- that being that since the isopropanol is a secondary substrate, if protonation via the ethanol occurred (as was conjectured), the more likely reaction would have been dehydration to an alkene (propene).
The hydroxyl group of the isopropanol gets protonated as before, but the good leaving group OH2+ slowly dissociates by way of the E1 mechanism, forming a carbocation; this carbocation does not rearrange, since it's already secondary, and hence stabilized by the inductive effect of the alkyl substituents. The conjugate base, ethoxide, which was formed by donation of its mildly acidic proton, then abstracts a proton from one of the methyl groups on the isopropyl carbocation; the electrons that were being used in the C-H bond of that abstracted proton then fold over onto the carbon bearing the positive charge, forming a pi bond between the methyl group and the central carbon. Hence, an alkene is formed, and it is propene:
CH3CH=CH2
I can't believe I made this oversight earlier, but oh well, what can you do. Again, this is still conjecture, as typically both reactions spoken of (nucleophilic substitution in the previous post and dehydration of alcohols to alkenes here) require protonation of the hydroxyl group, and ethanol/isopropanol are only mildly acidic-- dehydration typically involves an acid catalyst like H2SO4 and the application of heat; I'm not quite certain whether ethanol is acidic enough to promote this reaction. Still, the secondary nature of the substrate suggests that if protonation of the isopropanol did occur, then this would be the product generated (though it may not persist in solution long enough to be isolable, since electrophilic addition to the alkene formed then becomes a competing reaction). Also, protonation of the ethanol by the isopropanol (as opposed to the converse) would still yield the isopropyl-ethyl ether discussed previously, as primary alcohols such as ethanol are generally not susceptible to elimination reactions except under vigorous conditions (e.g., strong acid and high heat).
Hope this helped.
In my previous post, it seems I may have made an oversight-- that being that since the isopropanol is a secondary substrate, if protonation via the ethanol occurred (as was conjectured), the more likely reaction would have been dehydration to an alkene (propene).The hydroxyl group of the isopropanol gets protonated as before, but the good leaving group OH2+ slowly dissociates by way of the E1 mechanism, forming a carbocation; this carbocation does not rearrange, since it's already secondary, and hence stabilized by the inductive effect of the alkyl substituents. The conjugate base, ethoxide, which was formed by donation of its mildly acidic proton, then abstracts a proton from one of the methyl groups on the isopropyl carbocation; the electrons that were being used in the C-H bond of that abstracted proton then fold over onto the carbon bearing the positive charge, forming a pi bond between the methyl group and the central carbon. Hence, an alkene is formed, and it is propene:
CH3CH=CH2
I can't believe I made this oversight earlier, but oh well, what can you do.
Again, this is still conjecture, as typically both reactions spoken of (nucleophilic substitution in the previous post and dehydration of alcohols to alkenes here) require protonation of the hydroxyl group, and ethanol/isopropanol are only mildly acidic-- dehydration typically involves an acid catalyst like H2SO4 and the application of heat; I'm not quite certain whether ethanol is acidic enough to promote this reaction. Still, the secondary nature of the substrate suggests that if protonation of the isopropanol did occur, then this would be the product generated (though it may not persist in solution long enough to be isolable, since electrophilic addition to the alkene formed then becomes a competing reaction). Also, protonation of the ethanol by the isopropanol (as opposed to the converse) would still yield the isopropyl-ethyl ether discussed previously, as primary alcohols such as ethanol are generally not susceptible to elimination reactions except under vigorous conditions (e.g., strong acid and high heat).Hope this helped.
WHOAguitarninja
Member
-jinx- said:
It must be something with us physicists beacase I'm the same way, only I'm not even decent at entry level stuff. Chem and my brain are just incompatible.
morbidaza said:
Yeah, except that -jinx- is only "not good" at it due to lack of exposure/practice-- he would certainly be able to understand chemistry without any problem. You, on the other hand, just suck.
(totally kidding, obviously
)EDIT: And isopropanol is a secondary alcohol, not tertiary; hence the elimination product would be propene, not 2-methylpropene. For some reason, I was mistakenly thinking of tert-butyl alcohol in my mind when doing the problem. Whoops.
It's all been fixed above. - Status