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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

Physics
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|dw:1352058857282:dw| Where does the 0.25m circle come into this? I don't quite get the question
Sorry, I get it now

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Other answers:

:-)
i found angle
|dw:1352059091446:dw|\[T_y=mg\]
|dw:1352059120155:dw|
and the angle is 12
after that i dont get
Using MY angle theta (so I can copy/paste pictures easier later), \[0.05g=Tsin(\theta)\]
Get it?
|dw:1352059242987:dw|
As it does not move up or down
The mass, that is
mass is .05kg
what is the equation actually
Do you get why the equality is correct?
|dw:1352059579894:dw|
\[T_y-mg=0\]I think, as the NET force in the vertical axis is 0, as it doesn't move up or down.
it is circular motion so that must be= mv^2/r
In the horizontal axis yes, but we are resolving in the verical axis now|dw:1352059672600:dw|
\[T=\frac{0.05 *9.8}{\sin78}\approx0.5 N\] \[T_x=Tcos(\theta)=0.5\cos(78)\approx0.1N\]
\[T_x=\frac{mv^2}{r}\]
\[v=\sqrt{\frac{0.1*0.25}{0.05}} \approx 0.7 m/s\]
|dw:1352059964786:dw|
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/
now i will explain my method can u correct me
if i go wrong
plz
开始
OK
|dw:1352060338398:dw|
Tcosx - mg =0
T=mg/cos x
T=6N
\[0.05*9.8/\cos12=0.5\]
Calculator error it seems
lol it is radians
You need to provide dimensions along with your numbers.
where did the .1 came from
0.5 cos78
i got .5N for tension
why did u find that one
because horizontal force?
T=0.5 I got when resolving vertically\[Tsin \theta=mg\]
|dw:1352061108396:dw|
got u thx

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