Recent content by troytroy

I am about to graduate soon with a double major in math and physics. I have always wanted to go to graduate school for physics, but recently I did an REU and all I did was program the whole time and hated it. Also, I have been noticing that the overall job outlook seems horrible and many...

Okay, so solving for my equation of state for V, I get V = 10T/(p+50), then this means dV = 10dT/(p+50) since p is constant. Also, I know that Q = du + dw and du = 20dT +50dV = 120J - pdV. This can be rewritten as 20dT +dV(p + 50) = 120J. Then substituting the above dV I get 30dT = 120. Thus...

Thanks Chet! for part (b) I know that the pressure is constant so U = 120J + W = 120J + PV. Then we can get 120J + PV = 20T + 50V + 40J then this simplifies to 80J + PV = 20T +50V. Since from the equation of state PV = 10T - 50V we can substitute and get 80J + 10T - 50V = 20T + 50V. This can...

Homework Statement When 120J of heat are added to this gas, the temperature of the gas changes. Find the change in temperature if the heat is added at (a) constant volume. (b) constant pressure. Homework Equations The equation of state of at the gas is (p + 50)V = 10T The...

and b is just a constant that comes from the given equation y=(-x^3)/b^2 initially I wrote a instead of b but it is just a constant

I got L=T-U= ((1/2)m(x'^2+((-3x'x^2)/b^2)^2))+mg((x^3)/b^2) and then from there I calculated the lagrangian of the system by taking the derivatives according to (∂l/∂x)-((d/dt)(∂l/∂x'))=0

also the second derivative is with respect to t so I could just write x''

It's equal to zero because the lagrangian is set to zero. (∂l/∂x)-((d/dt)(∂l/∂x'))=0

(18(x^3)((dx/dt)^2))+((3gb^2)((dx/dt)^2))-(b^4)((dx^2/dt^2))-(9x^4)(dx^2/dt^2)=0

Okay I get it now, thank you very much for taking the time to help me out I really do appreciate it. I will post my final result for my lagrangian to see if they match up!

Wait are you saying we can just leave dx/dt in as it is? wow I can believe I did not realize it because now the lagrangian L will be in terms of only x and x'(x'=dx/dt) so L(x,x')=T-U

I am confused about the (1/2)m(((dx/dt)^2)+(dy/dt)^2) I understand that we solved for dy/dt by differentiating the f(x) but I don't see what we can plug in for the dx/dt term

To plug it into the lagrangian for the kinetic energy term

would dx/dt just be 1/f(x)'?

so for my dy/dt I got -((3x^2)/(a^2))*(dx/dt) and I understand your example but I don't know why I am have trouble see how to actually solve for (dx/dt)?