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Verbatim question for GR8677 #37
Mechanics}Statics

Sum over each component. Note that the horizontal direction has a net force proportional to the centripetal acceleration a=v^2/r = \omega^2 r, where v=\omega r. Note that T is the tension.

Solve for T above to get,

Find the magnitude of T and use the Pythagorean theorem,

and thus (E) is the right answer.

(The above should be fairly obvious, but if one is totally clueless, then one can eliminate choice (D) from noting units. The angular velocity has units of 1/s but g^2 has time units proportional to 1/s^4.)

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
lelandr
2011-04-24 12:37:38
I think some users have already posted similar or close solutions, but not quite exactly this:

This problem definitely does not necessitate careful calculations, but rather just some dimensional analysis, limits, and a bit of common sense.

a) as l\Rightarrow \infty this solution approaches 0... qualitatively this is not correct, and (provided the mass of the rope is negligible) this should approach simply mg.

b) \theta = 90 gives us 0... this clearly should not happen

c) wrong units... doesn't depend on g...

d) wrong units (w^2r^2 does not correspond to accelleration^2 as it should

e) units are correct, has correct limiting behavior at w,r=0, g=\infty. this is our correct answer
Alternate Solution - Unverified
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 \omega^{2}r]

(E) must be the answer.
Alternate Solution - Unverified
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. F_g forms one leg pointing down, F_c forms one leg pointing radially outwards, and F_T forms the hypotenuse along the string.

Thus, F_T^2=F_c^2+F_g^2
or F_T=((m\omega^2r)^2+(mg)^2)^{1/2}
F_T=m(\omega^4 r^2+g^2)^{1/2}
Answer E

Alternate Solution - Unverified
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)Alternate Solution - Unverified
Comments
mpdude8
2012-04-15 21:16:36
There's a slew of problems with C -- not only units, but T = infinity if the angle is 0? No way.

In A, if r = 0 (i.e., mass is just hanging there), the tension does not = 0.

D is thrown out by units as well -- you can't add a squared acceleration (m^2 / s^4) to something with units (m^2 / s^2). One might be tempted to go with B, since at angle 0, T reduces to mg, but the tension should definitely not be 0 when the angle approaches 90 degrees.
NEC
lelandr
2011-04-24 12:37:38
I think some users have already posted similar or close solutions, but not quite exactly this:

This problem definitely does not necessitate careful calculations, but rather just some dimensional analysis, limits, and a bit of common sense.

a) as l\Rightarrow \infty this solution approaches 0... qualitatively this is not correct, and (provided the mass of the rope is negligible) this should approach simply mg.

b) \theta = 90 gives us 0... this clearly should not happen

c) wrong units... doesn't depend on g...

d) wrong units (w^2r^2 does not correspond to accelleration^2 as it should

e) units are correct, has correct limiting behavior at w,r=0, g=\infty. this is our correct answer
OptimusPrime
2017-04-08 00:36:49
For (B), plugging in \\\\\\\\theta = 90 degrees gives 90/2 = 45 degrees. Then, cos(45) = \\\\\\\\frac{\\\\\\\\sqrt{2}}{2}. It also has the correct dimensions. How else do we eliminate (B)?
Alternate Solution - Unverified
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 \omega^{2}r]

(E) must be the answer.
flyboy621
2010-11-14 19:49:56
^ this
Alternate Solution - Unverified
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.NEC
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. F_g forms one leg pointing down, F_c forms one leg pointing radially outwards, and F_T forms the hypotenuse along the string.

Thus, F_T^2=F_c^2+F_g^2
or F_T=((m\omega^2r)^2+(mg)^2)^{1/2}
F_T=m(\omega^4 r^2+g^2)^{1/2}
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.
Alternate Solution - Unverified
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)Alternate Solution - Unverified
shashiprakash
2007-07-26 00:11:40
we can eliminate two answers on the basis of units. they are (C) and (D).NEC

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