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a conical pendulum is formed by attaching a 50g mass to a 1.2m string. The mass swings around the circle of radius 25cm (a) calculate the speed of the mass
 one year ago
 one year ago
a conical pendulum is formed by attaching a 50g mass to a 1.2m string. The mass swings around the circle of radius 25cm (a) calculate the speed of the mass
 one year ago
 one year ago

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henpenBest ResponseYou've already chosen the best response.1
dw:1352058857282:dw Where does the 0.25m circle come into this? I don't quite get the question
 one year ago

henpenBest ResponseYou've already chosen the best response.1
dw:1352059091446:dw\[T_y=mg\]
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
dw:1352059120155:dw
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
after that i dont get
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Using MY angle theta (so I can copy/paste pictures easier later), \[0.05g=Tsin(\theta)\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
As it does not move up or down
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
what is the equation actually
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Do you get why the equality is correct?
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
dw:1352059579894:dw
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[T_ymg=0\]I think, as the NET force in the vertical axis is 0, as it doesn't move up or down.
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
it is circular motion so that must be= mv^2/r
 one year ago

henpenBest ResponseYou've already chosen the best response.1
In the horizontal axis yes, but we are resolving in the verical axis nowdw:1352059672600:dw
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[T=\frac{0.05 *9.8}{\sin78}\approx0.5 N\] \[T_x=Tcos(\theta)=0.5\cos(78)\approx0.1N\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[v=\sqrt{\frac{0.1*0.25}{0.05}} \approx 0.7 m/s\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Yes I know that centrifugal is fictitious, but it's required if you want to use Newton's Laws in rotating systems. http://xkcd.com/123/
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
now i will explain my method can u correct me
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
dw:1352060338398:dw
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[0.05*9.8/\cos12=0.5\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Calculator error it seems
 one year ago

katragaddasaichandraBest ResponseYou've already chosen the best response.0
You need to provide dimensions along with your numbers.
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
where did the .1 came from
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
i got .5N for tension
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
why did u find that one
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.2
because horizontal force?
 one year ago

henpenBest ResponseYou've already chosen the best response.1
T=0.5 I got when resolving vertically\[Tsin \theta=mg\]
 one year ago
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