GREPhysics.NET: GR8677 #37

GR8677 #37

Problem

GREPhysics.NET Official Solution

Mechanics $\Rightarrow$ }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.

$\begin{eqnarray} \sum F_x=T \sin(\theta/2)&=&mv^2/r=m\omega^2r\\ \sum F_y=T \cos(\theta/2)-mg&=&0=<br /> \end{eqnarray}$

Solve for T above to get,

$\begin{eqnarray} \sum T \sin(\theta/2)&=&m\omega^2r\\ \sum T \cos(\theta/2)&=&mg<br /> \end{eqnarray}$

Find the magnitude of T and use the Pythagorean theorem,

$||T||=\sqrt{T^2 \cos^2(\theta/2)+T^2 \sin^2(\theta/2)}=m\left(g^2 + \omega^4 r^2 \right)^{1/2},<br />$

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 $1/s^4$ .)

Alternate Solutions

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 Reply to this comment
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) Reply to this comment

Comments

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 Reply to this comment
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) Reply to this comment
shashiprakash 2007-07-26 00:11:40	we can eliminate two answers on the basis of units. they are (C) and (D). Reply to this comment

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we can eliminate two answers on the basis of units. they are (C) and (D).

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