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