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 theorm to find the tension?
Ex:
Fnety = Tsinθ - mg
Fnetx = Tcosθ
Tysinθ - mg = mv2/r
Txcosθ = mv2/r
Ty...
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...