Recent content by Shaku

Alright, I'm glad I started with the easier set of coordinates. Thank you for all the help!

Alright. One last question, If I had used the vertical axis as my y-axis, and horizontal axis as my x-axis, would I just find the tensionx and tensiony and take the pythagorean theorem to find the tension? Ex: Fnety = Tsinθ - mg Fnetx = Tcosθ Tysinθ - mg = mv2/r Txcosθ = mv2/r Ty...

Fnetx = mat mgCosθ = mat gCosθ = at Would this be the correct usage of the equations?

Okay, that makes sense. For part d, where the problem asks top find the tangential acceleration, I know that there are two equations I can use: at = w/t or at = L*alpha. From the diagram, wouldn't it just be gCosθ? I'm not sure how to actually use these equations to get an answer though, as I...

I was just looking over my work, and I noticed that I didn't quite understand why you said T - mgsinθ was the net force in the Y direction. If my axis was horizontal and vertical, it would be Tsinθ - mg, but I had set my axis is along the path of motion. I'm confused how you got your...

Okay, that clarifies things after making a visualization of the situation. T = (mv2)/L + mgSinθ T = (m(2g*LSinθ)/L) + mgSinθ T = m((2g*Lsinθ)/L + gSinθ)

Ohhh, okay. That makes sense now. Since there is no velocity at any point in the path, that would mean there is also no acceleration, so: Fnety = 0 T - mgSinθ = 0, thus Fnety=mv2/r would just be 0. Since Fnety is 0, Fnetx = Fnet So mgCos = mvx2/L ... But I'm pretty sure I did something...

This is how I have it set up: From this, how do I set up my equation to find the tension without breaking it into components? Forces in the Y direction: T-mgSinθ Forces in the X direction: mgCosθ

I'm trying to use Fnet = mv^2/r. The net force in the x direction is mgCosθ and the net force in the y direction is T - mgSinθ. I plugged each respective net force into the equation to solve for vx and vy, and then took the Pythagorean theorem to find V as a single vector instead of it being...

Alright, cool! So now that I have the velocity, I can solve for the tension by using the same equation as I used to find velocity before, but this time solving it for T (*note, this is with corrections): \sqrt{LgCosθ} = vx \sqrt{L(T-mgSinθ)/m} = vy So, solving for V I'd get: V =...

Ah, thanks for catching that. I had my axis set up with with tension along the y-axis, so I got confused when I determined the sin/cos values. I currently have v = \sqrt{2gL} How would I factor the angle into that? Would it be v = \sqrt{2g*Lsinθ}? (I just looked at the triangle the angle...

I used Fnetx = mgsinθ Fnety = T - mgcosθ plugged those into two separate equations mgsinθ = mvx^2/r and T-mgcosθ=mvy^2/r solved for v and found \sqrt{LgSinθ} = vx \sqrt{L(T-mgCosθ)/m} = vy then used the Pythagorean Theorem to find V. Ki + Ui = Kf + Uf I know that the initial kinetic energy...

1. Homework Statement : A rock of mass m is attached to a string of negligible mass and length L. The rock is released from rest from a horizontal position. When the rock is at point P, the string attached to the rock makes an angle θ with the horizontal. In terms of the quantities, m, L, θ and...

It's negative, but that wouldn't change much in my final answer. Just an extra minus sign out front, and it's saying that that's wrong too. Edit: Actually, it wouldn't change the answer at all as I had already done the calculations as if they were with a negative acceleration (I just wrote it...

I found an example from my notes that is pretty similar to this question, and it had the equation: gSin∅ + MkgCos∅ = a as well, so I'm not sure where I'm going wrong. I fixed my arithmetic (which wasn't actually an error, I just didn't write it up on the post correctly), and I did the exact...