Recent content by PFStudent

Hey, Well, since we're beginning at, {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}} {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}} {{z}^{\prime}} = {{z}-{{{v}_{z}}{t}}} {{t}^{\prime}} = {t} We can let, {\vec{v}} = {{\vec{v}}_{\rho}} {{\vec{v}}_{r}} = {{\vec{v}}_{x}}+{{\vec{v}}_{y}}...

Hey, Any help on this, still not sure if I'm on the right path here. Thanks, -PFStudent

Hey, Well, I assume this part is right, {{x}^{\prime}} = {{x}-{{{v}_{x}}{t}}} {{y}^{\prime}} = {{y}-{{{v}_{y}}{t}}} {{z}^{\prime}} = {{x}-{{{v}_{z}}{t}}} {{t}^{\prime}} = {t} so where do I go from here? Thanks, -PFStudent

Hey, Well, in the 1-D case we are only considering motion along one axis, where {\vec{v}} is a constant, hence the following Galilean Transformation, {{x}^{\prime}} = {{x}-{vt}} {{y}^{\prime}} = {y} {{z}^{\prime}} = {z} {{t}^{\prime}} = {t} So, my guess is that by beginning with the...

Homework Statement Show that the form of Newton's Second Law is invariant under: (a). a Galilean Transformation (GT) in 1-Dimension. (b). a Galilean Transformation (GT) in 2-Dimensions. (c). a Galilean Transformation (GT) in 3-Dimensions. Homework Equations Newton's Second Law...

Hey, Ok, we can build on this and establish that the Galilean Transformation (GT) for: 1, 2, and 3; dimensions is the following, GT for 1-D {{x}^{\prime}} = {{x}-{vt}} {{y}^{\prime}} = {y} {{z}^{\prime}} = {z} {{t}^{\prime}} = {t} GT for 2-D (Converted via using Polar Coordinates...

Homework Statement Conventionally, the Galilean Transformation relates two reference frames that begin at the same location and time with one reference frame moving at a constant velocity {\vec{v}} along a positive {x}-axis (which is common to both reference frames) with respect to the other...

Hey, An interesting problem, anyone have a solution to it? Thanks, -PFStudent

Hey, A more substantial and detailed reply would help, anyone? Thanks, -PFStudent

Hey, Still stuck on these questions, a little bit of help would be nice. Thanks, -PFStudent

Hey, 1. Homework Statement . I was reading through the Differential Equation portion of my textbook and didn't quite understand the following paragraph. The above paragraph seemed a little confusing since, conventionally, {y} = {f(x)} Where, {{\frac...

Hey, Not quite, from my understanding, the displacement {{\vec{r}}_{s}} is NOT the same as the displacement {\vec{r}} used when calculating the Work done by the spring. The reason for this is because in, {{\vec{F}}_{s}} = {{-k}{{\vec{r}}_{s}}} the spring's displacement {{\vec{r}}_{s}} is...

Hey, I see. Ok, so let me see if I understand this correctly. When considering the formal equation for the Force due to a spring, {{\vec{F}}_{s}} = {{-k}{\vec{r}}} it is understood that, {\vec{r}} is defined as the displacement and, {\vec{r}} = {{\Delta}{\vec{r}}} where...

Hey, In that case how would one ever derive, {{W}_{s}} = {\frac{k}{2}\Biggl({{{{r}_{i}}^{2}}{-}{{{r}_{f}}^{2}}}\Biggr)} if we could only evaluate our integral, to find the Work done by a spring, from {0} to {{\vec{r}}_{f}}? Instead of the above, if we are limited to evaluating from...

Hey, Ahh, that makes sense. Since we always evaluate the force due to a spring through a linear displacement where the initial distance is the relaxed position (and because so is necessarily zero) and the final distance is the distance from the relaxed position. However, the above leads to an...