Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
ksaimouli
Group Title
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
 2 years ago
 2 years ago
ksaimouli Group Title
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
 2 years ago
 2 years ago

This Question is Closed

henpen Group TitleBest 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
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Sorry, I get it now
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
i found angle
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1352059091446:dw\[T_y=mg\]
 2 years ago

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

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
and the angle is 12
 2 years ago

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

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

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1352059242987:dw
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
As it does not move up or down
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
The mass, that is
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
mass is .05kg
 2 years ago

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

henpen Group TitleBest ResponseYou've already chosen the best response.1
Do you get why the equality is correct?
 2 years ago

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

henpen Group TitleBest 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.
 2 years ago

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

henpen Group TitleBest ResponseYou've already chosen the best response.1
In the horizontal axis yes, but we are resolving in the verical axis nowdw:1352059672600:dw
 2 years ago

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

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[T_x=\frac{mv^2}{r}\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[v=\sqrt{\frac{0.1*0.25}{0.05}} \approx 0.7 m/s\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1352059964786:dw
 2 years ago

henpen Group TitleBest 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/
 2 years ago

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

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
if i go wrong
 2 years ago

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

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
Tcosx  mg =0
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
T=mg/cos x
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[0.05*9.8/\cos12=0.5\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Calculator error it seems
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
lol it is radians
 2 years ago

katragaddasaichandra Group TitleBest ResponseYou've already chosen the best response.0
You need to provide dimensions along with your numbers.
 2 years ago

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

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

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

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

henpen Group TitleBest ResponseYou've already chosen the best response.1
T=0.5 I got when resolving vertically\[Tsin \theta=mg\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1352061108396:dw
 2 years ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
got u thx
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.