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Help me with physics =]

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

there, I just explained all of electromagnetics to you 😛.
 
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 😛.
Click to expand...

gimme your s/n so I can ask you more questions T__T
 
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.
 
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 😛.
Click to expand...

QFT. See Random Variable's sig.
 
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?
Click to expand...

Conservation of energy. Gravitational potential energy is being converted into Kinetic Energy.
 
Originally posted by: TecHNooB
Still dont see how I should set up the problem 🙁
Click to expand...

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.
 
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?
 
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?
Click to expand...

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.
 
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?
Click to expand...

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.
Click to expand...

Nah theres more to it =/
 
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 😛.
 
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
 
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.
 
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