That's amazing, that if the local substation is not grounded the current will still flow inside the earth from my home, even though no current will ever travel up to the local station.
I think the reason that this confuses me might be that I think of it as
But the network system is not a...
So in other words, if I connect an ampere meter in the cable that connects the neutral and the ground I will see 0A? Even when there is a fault and the neutral has a current flowing through it?
This is a topic that almost no one can actually explain from what I found on the internet. This answer https://physics.stackexchange.com/questions/74625/does-alternating-current-ac-require-a-complete-circuit/74999#74999 is the closest I found so far.
The idea is how the grounded neutral...
I will do it tomorrow since I'm exhausted. I'm struggling with that problem for 3 days now. Never in my life, I had to deal with a problem for 3 days continuously. I'm glad that I finally figured it out!
How did great scientists try to solve a problem for years and not get crazy?
Found it ##z = L[1-cosθ']## where ##θ' = θ_{max}##
I ended up to ##sinθ' + 2cos(θ') = 2##. Unfortunately, I don't remember how to solve that. I always forget trig formulas...
Edited:
I thought it didn't matter, so I assumed θ=90 where the m mass is indeed L meters from the ground.
So this is a mistake?
Yes at θmax I don't know the actual height so I can't find the potential energy there.
So in order for this to be right, I have to say:
##mgz = \frac{1}{2}mV^2## where z is...
I thought of something else but I got a math error:
V at the lowest point:
##p1+k1 = p2 + k2 <=> mgL + 0 = 0 + \frac{1}{2}mV^2 <=> v = \sqrt{2gL}##
So at the highest point I now that: ##V = 0##, ##Θ(t) = Θ' = max{Θ(t)}##
and for the lowest point I know that: ##Θ(t) = 0## and ##v =...
But now, how can I use the equation ##|a| = \sqrt{a_{T}^{2} + a_{N}^{2}} = \sqrt{g^{2}sin^{2}(θ(t)) + \frac{|V|^4}{L^{2}}}## and that |a1|=|a2| (1 = highest point in the path, 2 = lowest point)?
By the way, since this path is an arc, then ##|V| = ωL##
PS: O MY GOD, I just figured something out... Well if you see the pendulum in a clock, you see it going up and down (oscillating) at the same speed! This means that when the pendulum reaches its maximum height it changes direction instantly! This is why the speed never changes but the direction...
PS: By the way today I had exams in Physics and this problem was the first one I had to solve :p (unlucky) The question was to find the maximum angle θ that the pendulum can reach if we know that the magnitude of the acceleration is the same when the mass is located in the highest and the lowest...
I understand what you are saying, thanks for this note.
I'm not sure about the potential energy when the pendulum is at its maximum angle, because I don't know what the height is there. I saw on the internet people taking the height = L when the angle of the pendulum is 90 degrees but is this...
##T = m[\frac{|V(t)|^2}{L} + gcos(θ(t))]##
And If I'm correct, this is not a conservative force since it depends on the velocity which depends on time, so, if I was using the principle of conservation of mechanical energy (potential energy in highest point + kinetic energy in highest point =...
Can anyone point out my mistake? Maybe the mistake is the way I describe acceleration, it's the only thing I'm not completely sure if it is correct.
Maybe the way that I said ##a_{T} = |a|cos(θ)## ##a_{Ν} = |a|sin(θ)## and then calculated ##|a|## from ##ΣF_{T}=ma_{T}## it's not correct.
And...
Yes it is accelerating but the acceleration chages only the direction of the velocity keeping it's magnitude constant. It's acceleration I believe is ##R \cdot \frac{dω(t)}{dt}##
Edit:
By the way, I noticed something in my way of thinking when you said that these are scalar quantities (in the...
Is this correct about the magnitude of the acceleration?
As you said the magnitude can not be just g. But also tan(θ) can give negative answers, so how is this a correct magnitude? Maybe, because my coordinate system moves with the ball the system flips when the ball is on the right-hand side...
I think I figured it out (If I don't have any mistakes).
But I think I have. I calculated T and ##tan(θ)sin(θ)-cos(θ)## gives a negative number.
But T should always point towards the positive y-axis. Something feels wrong...
Can you tell me my mistake? Thank you!
Yes I know how to reduce it to just Θ and how to use the odd and even properties of sine and cosine to get rid off the 90 degrees as well.
The problem now is acceleration. This is the only thing (for now) that I don't know what it's equal to. I don't even know how to draw it on the diagram...
But this is what I did, didn't I?
Then I calculated Wy and Wx where Wy is in the opposite r-direction as you describe.
T is always parallel with the r-direction which means it has only one component, the other one is zero.
This is as far as I reached. I don't know what to do now since I'm not sure what a1, and a2 are equal to.
PS: I also attached a Pendulum Analysis.pdf file in case you can't read the pictures.
But these equations don't stand in this problem. The magnitude of V can't be Lω(t) since this would be...
Now, what happens at a random point in time? Can you give me an idea on how to approach this problem in order to find out T as a vector? I believe the magnitude of T does not change, only it's direction, that's what my senses are telling me. If this is true, then from the calculations of my...
You are right, when it passes through the vertical position the horizontal component of T is zero. At that point the horizontal component of the net force should also be zero because gravity doesn't have one as well, in fact, gravity is always zero on the x-axis along the whole motion. So...
Hello! I'm trying to understand how this pendulum works. I found this video that explains how to calculate the T force from the rope.
He uses the preservation of kinetic and potential energy in order to find the magnitude of the velocity and then using Newton's second law, he calculates the T...
I definitely did a mistake there (x(t) is not exponential), but you can definitely calculate the velocity as a function of time, and then using ##\frac{dx}{dt}##, you can integrate to find the position function.
This is from a different exercise but it has the same idea (with the one of this post) in order to get s(t) from the s[v(t)] relation:
I'm not sure if I can integrate over ##ds(t)##. My professor indeed does it but he doesn't show the dependency of each variable so I don't know if s depends on...
By the way, I need to do some research, since it's the first time doing these kinds of integration and it seems a bit compilating.
Are you guys sure that this is correct?
##\int_{x[v(0)]}^{x[v(t)]}dx(t) = \int_{v(0)}^{v(t)}dv(t)##
The solution in my notes (professor's solution) does something...
I believe you are right. It's probably easier this way and then using the initial value you can go and calculate the C constant because the bounds gave me a very hard time solving the integral. Plus when you have to use u-substitution to solve the integral things start getting weird since you...
I ended up with the above. I still have time inside this equation, so I'm not sure how to find the distance traveled until it stops. I can set ##v(t) = 0## but what is the time at that point? I do not know... I need to take the derivative, find v(t), and set that to zero in order to figure out...
Wow, I think I'm getting it! So you basically create an association between distance and velocity. In other words, you express the distance as a function of velocity and once you figure that out you will just plug in 0 for velocity, yielding the answer of the exercise immediately. Is this...
I'm actually afraid to plug the time here:
C is also pretty big, not to mention t is also in the exponent as well, I will be lost in the calculations...
I can't believe university physics has so many calculations.
So far I've done, Calculus, Vector Calculus, Complex Analysis (Applied maths)...
I actually figured it out but I believe I still made a mistake somewhere (Probably math-related) because the time that I got, does not seem to be in seconds...
I attached a pdf file since the solution is quite big and I can't write it in latex.
I figured out the following time @ v(t) = 0 (ball...
I just thought it has the same magnitude as the weight force but different angle. So I thought from the diagram that the angle is 90 and just wrote that.
I know it's a bit overkill. I know how to solve it by analysing weight in two vectors and then going only 1D (what we did in high school) but...
Known:
1) The mass of the ball is ##m## (constant ##\frac{dm}{dt} = 0##)
2) ##v(0) = v_{0}##
3) Air drag force magnitude ##| \vec F_{D} | = B \cdot | \vec v(t) |## (##B \in R##)
4) The ramp is frictionless.
5) The magnitude of Earth's acceleration = ##g##
I'm not sure if θ is known or not, and...
When you say ##i = \frac{dq}{dt}## it makes sense since current is the flow of charge over time. But why was voltage defined as
##v = \frac{dw}{dq}## ? What made physicians define it in this way? Is there a mathematical way that can lead to this definition or
did they define voltage just on the...
Yes, I'm an Electrical And Computer Engineer undergraduate student. This is the only physics course about mechanical physics. After that, we have two more physics courses called "Electromagnetic field 1 and Electromagnetic field 2" (maxwells equations, coulombs law, gauss law etc).
The course is called just Physics (We have another one called Electromagnetic Field 1 in the next semester and another one after that (next year) which is called Electromagnetic Field 2).
For prerequisites says only this, nothing else:
I haven't asked the professor since I'm a distant...
Thank you! You totally eased my frustration.
It makes sense to be a complete course on Newtonian physics and only an introduction to SR. When it comes to algebra, linear algebra - vectors, calculus, and diff equations I don't really have any problems. The only problems might be that I'm not...