Circular Motion Question (1 Viewer)

Aerath · Oct 18, 2009

Could someone do the circular motion question for me? Thanks.

gurmies · Oct 18, 2009

$Horizontally: \\\\ T_{1}\sin\theta+T_{2}\sin\theta=\frac{mv^{2}}{r} \\\\ Vertically: \\\\ T_{1}\cos\theta-T_{2}\cos\theta=mg \\\\ Now, \,\, T_{1}+T_{2}=\frac{mv^{2}}{r\sin\theta} \,\,[1] \,\, and \,\, T_{1}-T_{2}=\frac{mg}{\cos\theta} \,\, [2] \\\\ 2T_{1}= \frac{mv^{2}}{r\sin\theta}+\frac{mg}{\cos\theta}\Rightarrow T_{1}=\frac{1}{2}(\frac{5mv^{2}}{64}+\frac{5mg}{3}) \\\\ 2T_{2}=\frac{mv^{2}}{r\sin\theta}-\frac{mg}{\cos\theta}\Rightarrow T_{2}=\frac{1}{2}(\frac{5mv^{2}}{64}-\frac{5mg}{3})$

Thinking about the last part...

Amplifiedspeakr · Oct 18, 2009

$\begin{array}{c} Working\, from\, Gurmies\, work, \\ since\, each\, string\, experience\, a\, tension\, then\, we\, have \\ T_1\geq 0\, and\, T_2\geq 0 \\ T_2\geq 0\Rightarrow \frac{1}{2}(\frac{5mv^2}{64}-\frac{5mg}{3})\geq 0 \\ \frac{5mv^2}{64}\geq \frac{5mg}{3} \\ v^2\geq \frac{64g}{3} \\ v\geq \frac{8\sqrt{g}}{\sqrt{3}} \\ ie\, v\geq \frac{4\sqrt{g}}{1.5} \end{array}$

Might have misread the question :S

gurmies · Oct 18, 2009

Amplifiedspeakr said:
$\begin{array}{c} Working\, from\, Gurmies\, work, \\ since\, each\, string\, experience\, a\, tension\, then\, we\, have \\ T_1\geq 0\, and\, T_2\geq 0 \\ T_2\geq 0\Rightarrow \frac{1}{2}(\frac{5mv^2}{64}-\frac{5mg}{3})\geq 0 \\ \frac{5mv^2}{64}\geq \frac{5mg}{3} \\ v^2\geq \frac{64g}{3} \\ v\geq \frac{8\sqrt{g}}{\sqrt{3}} \\ ie\, v\geq \frac{4\sqrt{g}}{1.5} \end{array}$

Might have misread the question :S

I ended up with a similar thing, perhaps answer is wrong?

study-freak · Oct 18, 2009

gurmies said:
I ended up with a similar thing, perhaps answer is wrong?

The answer is correct. Don't forget to change cm to m before using it!
That was the mistake.

$\\ T_2\geq 0 \\\frac{1}{2}(\frac{500mv^2}{64}-\frac{5mg}{3})\geq 0 \\ \frac{500mv^2}{64}\geq \frac{5mg}{3} \\ v^2\geq \frac{64g}{300} \\ v\geq \frac{8\sqrt{g}}{\sqrt{300}} \\ i.e.\, v\geq \frac{4\sqrt{3g}}{15}$
(sorry, Amplifiedspeakr, I stole your work and just fixed up the cm to m conversion.)<SCRIPT type=text/javascript> vbmenu_register("postmenu_4646384", true); </SCRIPT>

Aerath · Oct 18, 2009

Aha, thanks. I kept getting what Gurmatron was getting.

gurmies · Oct 18, 2009

Aerath said:
Aha, thanks. I kept getting what Gurmatron was getting.

xD

Aerath · Oct 18, 2009

gurmies said:
xD

I think I ripped it off Yana or someone.

gurmies · Oct 18, 2009

Aerath said:
I think I ripped it off Yana or someone.

That would be understandable

Circular Motion Question (1 Viewer)

Aerath

Retired

gurmies

Drover

Amplifiedspeakr

New Member

gurmies

Drover

study-freak

Bored of

Aerath

Retired

gurmies

Drover

Aerath

Retired

gurmies

Drover

Users Who Are Viewing This Thread (Users: 0, Guests: 1)