Recent content by undividable

Is the following logic correct?: If you have an hamiltonian, that has time has a variable explicitly, and you get the lagrangian,L, from it, and then you get an equivalent L', since L has the total time derivate of a function, both lagrangians will lead to the same equations euler-lagrange...

is there a way for me see EVERY solution from goldstein's book? i already have some solutions, but not every one

so, as the mouse moves upward, the axle exerts and external force (normal downward force) on the turntable so it stays with v=0? and if the axle was a part of the system? the linear momentum should still change, but now the external net force would be is 0

A pet mouse sleeps near the eastern edge of a stationary, horizontal turntable that is supported by a frictionless, vertical axle through its center. The mouse wakes up and starts to walk north on the turntable. Is the momentum of the system constant? i understand that the initial momentum is...

so the component that is constant is equal to R and is perpendicular to the momentum vector, and the other vector component is time-dependent and parallel to the momentum, so the vector product will equal a vector with components R and 0?

well, i can't figure out r→

i did, and my question remains, why does the text in the image say that m1vR and m2vR are the angular momentum for m1 and m2 respectively, R is not the position vector, even if we consider the center of mass of m1 and m2 as what the position vector is referring to

i have, but the problem says that in the analyzing part it says that m1vR and m2vR are the angular momentum for m1 and m2 respectively, and i can not understand why we use R, that is not the position vector.

Homework Statement it is given in the image i uploaded Homework Equations L→=r→ × p→[/B] The Attempt at a Solution If the angular momentum of a particle is the cross product of the position vector of the particle from the axis and its linear momentum, how can the angular momentum of m1 and...

But shouldn't every choice of coordinate system give us a correct awnser?

My question is simple, in the problem on the picture that i uploaded, why is -fk =ma I understand the friction is in the negative direction, só it is negative, but the netforce, and the aceleration, are also in the negative direction ,só why are they positive? Shouldt it be -fk=-ma ?

Homework Statement I uploaded a picture with the problem from the textbook and its solution Homework Equations The relevante equations are in the attsmpt at a solution The Attempt at a Solution First of all, i understood the solution from the solution manual, but in my solution the only...

shouldn't it be ma(-iy)=mg(-iy) since the net force and the gravitational force are in the negative direction?, as in T(iy)=Fg(-iy), where they are in opposite directions

substituting the vector components here we get Tz=Fgz, right? this component equation is the same as yours ∑Fz=Tz-Fgz=0 but in the following step you did things differently so in the vector equation ∑F=Fg, , as in the vector equation T=-Fg, when we put in the vector components we get...

but why didn't you use unit vector on ma? in ma=mg(-iy), isn't the net force also a vector?