ksaimouli
  • ksaimouli
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ksaimouli
  • ksaimouli
@henpen
anonymous
  • anonymous
|dw:1352058857282:dw| Where does the 0.25m circle come into this? I don't quite get the question
anonymous
  • anonymous
Sorry, I get it now

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

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