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ksaimouli
 4 years 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
ksaimouli
 4 years 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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352058857282:dw Where does the 0.25m circle come into this? I don't quite get the question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352059091446:dw\[T_y=mg\]

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1352059120155:dw

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2after that i dont get

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using MY angle theta (so I can copy/paste pictures easier later), \[0.05g=Tsin(\theta)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352059242987:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0As it does not move up or down

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2what is the equation actually

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you get why the equality is correct?

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1352059579894:dw

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

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2it is circular motion so that must be= mv^2/r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In the horizontal axis yes, but we are resolving in the verical axis nowdw:1352059672600:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[T=\frac{0.05 *9.8}{\sin78}\approx0.5 N\] \[T_x=Tcos(\theta)=0.5\cos(78)\approx0.1N\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[T_x=\frac{mv^2}{r}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[v=\sqrt{\frac{0.1*0.25}{0.05}} \approx 0.7 m/s\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352059964786:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes 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/

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2now i will explain my method can u correct me

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2dw:1352060338398:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[0.05*9.8/\cos12=0.5\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Calculator error it seems

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You need to provide dimensions along with your numbers.

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2where did the .1 came from

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2i got .5N for tension

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2why did u find that one

ksaimouli
 4 years ago
Best ResponseYou've already chosen the best response.2because horizontal force?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0T=0.5 I got when resolving vertically\[Tsin \theta=mg\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352061108396:dw
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