Help me with physics =]

[TecHNooB]

Need to have a good long AIM convo with someone on the workings of the Energy Principle. Thnx ATOT!

 

[CraKaJaX]

 

[Syrch]

mention flux capacitor

 

[Rubycon]

Text

 

[TecHNooB]

Originally posted by: MS Dawn
Text

::heartattack::

 

[BrownTown]

div D = rho
curl E = -dB/dt
div B = 0
curl H = J + dD/dt

there, I just explained all of electromagnetics to you .

 

[TecHNooB]

Originally posted by: BrownTown
div D = rho
curl E = -dB/dt
div B = 0
curl H = J + dD/dt

there, I just explained all of electromagnetics to you .

gimme your s/n so I can ask you more questions T__T

 

[hypn0tik]

Energy cannot be created or destroyed. It can be converted from one form to another.

Everything you need to know about energy is right there.

 

[hypn0tik]

Originally posted by: BrownTown
div D = rho
curl E = -dB/dt
div B = 0
curl H = J + dD/dt

there, I just explained all of electromagnetics to you .

QFT. See Random Variable's sig.

 

[Syrch]

Originally posted by: TecHNooB

Originally posted by: MS Dawn
Text

::heartattack::

QFT

 

[TecHNooB]

omg its hypn0tik! help me with this question http://forums.anandtech.com/messageview.aspx?catid=50&threadid=1949388

 

[hypn0tik]

Heres one of the questions:
You pull straight up on the string of a yo-yo with a force 0.30 N, and while your hand is moving up a distance 0.16 m, the yo-yo moves down a distance 0.27 m. The mass of the yo-yo is 0.061 kg, and it was initially moving downward with speed 2.6 m/s.

(a) What is the increase in the translational kinetic energy of the yo-yo?

(b) What is the new speed of the yo-yo?

(c) What is the increase in the rotational kinetic energy of the yo-yo?

Conservation of energy. Gravitational potential energy is being converted into Kinetic Energy.

 

[BrownTown]

also, work = F*d = energy

 

[TecHNooB]

Still dont see how I should set up the problem

 

[Goosemaster]

Originally posted by: MS Dawn
Text

cyclic-guano-sine monophosphate

:laugh:

 

[hypn0tik]

Originally posted by: TecHNooB
Still dont see how I should set up the problem

Consider this example.

Set h = 0 at the base of a tower. You take a 1kg ball 400m up the tower and drop it (with an initial speed of 0 m/s). How fast is the ball travelling just before it hits the ground?

The potential energy of the ball comes from moving it from the base to the top. The amount of energy it gains is:

E_total = mg?h + 1/2 m*v1^2 = mg?h (since v1 = 0).

At the bottom, E_total remains the same.

E_total = mg?h + 1/2m*v2^2 = 1/2 m*v2^2 (since ?h = 0).

You observe that as the ball is falling, it loses potential energy and gains kinetic energy.

The yo-yo is doing the same thing. You should be able to set up your problem now.

Edit: How fast is the ball travelling when it is half way down the tower? E_total remains the same, but the contribution of KE and PE changes.

 

[TecHNooB]

The guy is pulling on the yoyo though -__-; This problem stresses the point particle principle.

 

[hypn0tik]

Originally posted by: TecHNooB
The guy is pulling on the yoyo though -__-; This problem stresses the point particle principle.

Originally posted by: BrownTown
also, work = F*d = energy

 

[BrownTown]

i'll go ahead and admit it, this problem loosk super easy, but when i try to solve it i can't...

Am I missing something here?, or are we not givien enough info?

 

[hypn0tik]

Originally posted by: BrownTown
i'll go ahead and admit it, this problem loosk super easy, but when i try to solve it i can't...

Am I missing something here?, or are we not givien enough info?

I think you have to realize that when you pull on the string, the force is tangential to the yo-yo. This causes the work done in pulling on the string to be converted solely into rotational energy.

I could be mistaken, but that's what comes to mind intuitively.

 

[TecHNooB]

Originally posted by: hypn0tik

Originally posted by: BrownTown
i'll go ahead and admit it, this problem loosk super easy, but when i try to solve it i can't...

Am I missing something here?, or are we not givien enough info?

I think you have to realize that when you pull on the string, the force is tangential to the yo-yo. This causes the work done in pulling on the string to be converted solely into rotational energy.

I could be mistaken, but that's what comes to mind intuitively.

Nah theres more to it =/

 

[BrownTown]

k, that sounds good. Don't think there is more to it, there isn't enough information for there to be any more to it.

a. .1614J

b. 2.472m/s

c. .048J

EDIT: my math is tragicly bad, so don't actually trust those numbers .

 

[Rubycon]

Imagine doing this with a motorcar wheel! :Q

 

[TecHNooB]

OOOH!!@#

Got it =D If you treat the yo-yo as a point particle system, the actual work done by the force upwards is .30 * -.27 + .061 * 9.8 * .27 <-- positive work done by grav. That gives you K_trans.

Then you find the energy added to the system which is the work done by the person aka .30 * .16.

Set the equation up as K_rot + K_trans = Wext. Finally >=O

 

[BrownTown]

yeah maybe i dunno whatever , sorry, this was like 4 years ago and the whole rotational thing is just messing with me

maybe next time ask me an electrical engineering question or something like that which will make me look smart instead of a bumbeling idiot.