Recent content by Mamed

Ok this is how i have reasoned when looking at this. ##\frac{dE}{dt} + P = 0 ## we know how the equation is set up ##E = \frac{J\omega^2}{2}; P = \frac{P_{fr0}}{\omega_0^3}*\omega^3 → a * \omega^3## we replace E and P with their dependency of ##\omega## and cross out one ##\omega##...

Sorry forgot the ##t## there in the nominator. There is supposed to be a ##t## there like this. ## \omega(t) = \frac{\omega_0}{2*P_{fr0}*t/J\omega_0 + 1}##

Ok so ##dE/dt + P = 0## as ##dE/dt## has dimension Joule/s which is the same as Watt. So what we will have is ##\frac{d\omega^2J}{dt2} + a\omega^3 = 0## (not sure if I am doing the next step correct or not) but if we take away one ##\omega## on each side we will get...

Ok if we do that we will get an equation a standard solution that looks something like ##\frac{d\omega}{dt} = -a\omega^3## -> ##\frac{d\omega}{\omega^3} = -adt ## ##\omega(t) = \frac{1}{\sqrt{2at + C_1}} ## where ##C_1 = \omega(0) = 1/w_0 ## And if i plot this as a fuction of time i...

Sorry its more P_{fr} = P_{fr_0} (\frac{w}{w_0})^3 If we look at Newtons second equation we have \frac{}{} J * \frac{dw}{dt} + kw = 0 it gives that \frac{dw}{dt} + \frac{k}{J}w = 0 if we set the solution as w * e^{(k*t /J )} we get \frac{d}{dt}w * e^{(k*t /J...

Yes these two things are know and given. The wheel is running at the grid network of 50Hz and at that speed Pfr is known and constant. I think that the friction losses are decreased with the angular frequency to the power of 2 or 3 but i am not really sure on this. Ie Pfr(w) = Pfr0 * w^3

Homework Statement Hello I need to find a an equation that describes the rundown of a wheel as function of the time. The wheel is running at an constant speed and is then disconnected and runs down to standstill. We know what the angular velocity of the wheel is at time 0, Wm. We know...

lol ok, the r2 is outer radius at temperature T2 and r1 is inner with T1. L is the length. And the resistance is set so that T1----R1-----Tm------R2------T2--->Q If we use Ohms law here, then Tm = T2+Q*R2 We know that Q = (T1-T2)/(R1+R2) -> Tm = T2 + R2*(T1-T2)/(R1+R2) Tm = T2*R1/(R1+R2)...

Those are for thermal conductivity and and the length, when you put them in the equation they are canceled out. Tm is a mean temperature in an lumped parameter cylinder with no internal heat generation. And that's how they arrive at the solution. No further explaining done. And i want to know...

No they have it all expressed as T_m = T_2 \underbrace{\left( \frac{\displaystyle r_2^2}{\displaystyle r_2^2-r_1^2}-\frac{\displaystyle 1}{ \displaystyle 2 \ln{\left( \displaystyle r_2 / r_1 \right) }} \right)}_{k_1} + T_1 \underbrace{\left( \frac{ \displaystyle 1}{\displaystyle 2...

The objective is to find some coefficients by comparing two equations. T_2 \cdot \frac{R_1}{R_1+R_2} + T_1 \cdot \frac{R_2}{R_1+R_2} and T_2 \cdot k_1 + T_1 \cdot k_2 I compare and set k_1 = \frac{R_1}{R_1+R_2} (1) k_2 = \frac{R_2}{R_1+R_2} (2) I expand the...

I think i might have made a mistake when substituting. I have \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 And i substitute by y = sin(\theta) x = cos(\theta) So when i do the substitution should i include that dy = cos(\theta)d\theta or can i just substitute x =...

I think i might have made a mistake when substituting. I have \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 And i substitute by y = sin(\theta) x = cos(\theta) So when i do the substitution should i include that dy = cos(\theta)d\theta or can i just substitute x =...

so there is no way for me to just but in the boundaries? i want to implement this in a MATLAB function, later on does it mean that i have to use a nummerical method to solve it then? and what do you do if you if you have a circle is that also impossible to solve? because the only difference...

Hi Im trying to estimate a semicircle or ellips of a kind with an integral. And right now I'm trying to get the integral of a ellips. I need to integrate the equation x(\theta) = \int_0^{\pi/2} \frac {d\theta}{a\sqrt{1-sin^2(\theta)/b^2}} I tried...