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GR8677 #37
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Alternate Solutions |
nakib
2010-04-02 12:30:08 | T provides for both the circular motion and preventing the mass from falling towards earth. So one must expect two terms in the expression for T: one with an angular acceleration and the other with g. So, eliminate (A) and (B) and (C).
[One can also eliminate (A) and (B) by noting the directions they point at. (C) also has wrong units.]
(D) has wrong units. [centripetal acceleration is  ]
(E) must be the answer. |  | spacebabe47
2007-09-30 08:36:46 | For some reason the concept of tension in force diagrams never made much sense to me.
A different way of looking at the problem. Think of the forces as making a right triangle.  forms one leg pointing down,  forms one leg pointing radially outwards, and  forms the hypotenuse along the string.
Thus, 
or ^2+(mg)^2)^{1/2})
^{1/2})
Answer E
| | eshaghoulian
2007-09-15 05:14:12 | Can be done by limits/units: C/D have incorrect units; in the limit r = theta = 0, A gives you 0 tension (should be mg); B is maximized at theta=0 (gives mg) when, physically speaking, it should be minimized at that point (consider triangle inequality) | |
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Comments |
nakib
2010-04-02 12:30:08 | T provides for both the circular motion and preventing the mass from falling towards earth. So one must expect two terms in the expression for T: one with an angular acceleration and the other with g. So, eliminate (A) and (B) and (C).
[One can also eliminate (A) and (B) by noting the directions they point at. (C) also has wrong units.]
(D) has wrong units. [centripetal acceleration is ]
(E) must be the answer. | | BerkeleyEric
2010-01-12 22:13:16 | C and D have the wrong units, so those can be eliminated immediately. A fails to satisfy the limit of r=0. The expression for B in the limit of theta = pi (with the mass almost hitting the ceiling) gives zero tension, which does not make physical sense. So this leaves E. |  | spacebabe47
2007-09-30 08:36:46 | For some reason the concept of tension in force diagrams never made much sense to me.
A different way of looking at the problem. Think of the forces as making a right triangle. forms one leg pointing down, forms one leg pointing radially outwards, and forms the hypotenuse along the string.
Thus,
or
Answer E
wittensdog
2009-09-28 16:15:35 |
I think this is the fastest way to do this problem, and ETS loves to give useless information. The angle and length of the chord are totally unnecessary. I even hesitated for a minute on this one as a result of those pieces of information, from the power of suggestion throwing me off I suppose. But whenever you know the two perpendicular components of something, never even bother thinking about the angle. It's always true that,
T^2 = Tx^2 + Ty^2
where x and y are some perpendicular axes (in this case the vertical and radial directions are always perpendicular). The only time that angle information should even be considered is if you don't explicitly know Tx and Ty, or you're actually trying to find the angle, in which case, dividing the two expressions,
Tcos(theta/2) = mg,
Tsin(theta/2) = mrw^2
to give the tangent of theta/2 would definitely be the fastest, in my opinion. I personally think that trying to consider units in these problems is over-thinking everything way too much.
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| | eshaghoulian
2007-09-15 05:14:12 | Can be done by limits/units: C/D have incorrect units; in the limit r = theta = 0, A gives you 0 tension (should be mg); B is maximized at theta=0 (gives mg) when, physically speaking, it should be minimized at that point (consider triangle inequality) | | shashiprakash
2007-07-26 00:11:40 | we can eliminate two answers on the basis of units. they are (C) and (D). | |
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