I am launching a high-altitude balloon as a part of a physics project I am working on. I know that the amount of helium I need corresponds to about 40 newtons of lift for the launch, which is all I need to know for practical purposes. But if I wanted to account for the ever-changing buoyant force on a weather balloon, I would need a calculus-based model, either an integral or some differential equation.  From the derivation on Wikipedia, I have:  $$ F_b = \oint \sigma dA $$  where $\sigma$ is the Cauchy stress tensor and $dA$ is the surface area in contact with the fluid, or of the latex balloon, in my case. I am an undergrad student and have virtually no knowledge of tensor calculus.   Are there any other calculus-based models for buoyancy? If not, is numeric integration an appropriate strategy for this problem?
121885
For a project into balloon simulation I'd like to know how the force on a balloon changes with altitude:  I know that the Buoyant force on a balloon is:  $F = (\rho_{air} - \rho_{helium})gV$  Using changing air density from the international standard atmosphere and changes in gravitational acceleration I can calculate the buoyant force and add this to the (negative) force of gravity mg.  However I know that it isn't this simple, the balloon changes volume with altitude as the air pressure outside the balloon changes.  I'd like to know how the balloon inflates - this would be a great help!  I think it may be something to do with ideal gas laws -  I can calculate air temperature and pressure from the international standard atmosphere but do I use pressure as the air pressure? - or air pressure minus balloon pressure?
52403
I see numerous websites talking about how a small balloon is at higher pressure than a big balloon. This is a fun counter-intuitive factoid... unless it isn't a fact at all.  Young-Laplace gives  $$T = \Delta p \frac R2 $$  Which means that for a given tension, increasing the radius does decrease the pressure. But why would the tension be constant? A latex balloon is like a 2D spring, right? So the tension in a inflated spherical balloon should be proportional to the surface area, correct?  $$T \propto R^2 $$  But this gives  $$\Delta p \propto R$$  Which means the bigger the balloon, the higher the internal pressure. Which, again, would be the normal intuition - but it would burst the bubble of some simplified explanations. (har har)
99382
A hemispherical tank of radius 2 metres is initially full of water and has an outlet of 12 cm$^2$ cross-sectional are at the bottom. The outlet is opened at some instant.  The flow through the outlet is according to the law  $v(t) = 0.6 \sqrt{2gh(t)}$, where $v(t)$ and $h(t)$ are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time $t$, and $g$ is the acceleration due to gravity. Find the time it takes to empty the tank.   I know to write a differential equation by relating the decrease of water level to the outflow in order to solve. But don't know how and as we are not not given the initial height how can we solve it.
42940
I need to calculate the initial velocity required to launch a projectile at a given angle from point A to point B. The only force acting on the projectile after launch will be gravity – zero air resistance. The projectile is launched within a simulated, virtual environment; however, I am asking for help with the physics rather than the simulation itself.  I have had success (with a different equation) when point A and point B are at the same height; however, once point B is at a different height, my calculations become less precise – well, wrong in fact.  I researched the following formula for finding the range of a projectile on uneven ground. The parameters are all available to me, except the initial velocity that I need to solve for.     $d$ : range or distance   $v_i$ : initial velocity   $g$ : gravity   $\theta$ : launch angle   $y_0$ : launch height   $$d = \frac{v_i \cos\theta}{g}(v{_i} \sin \theta + \sqrt{(v_i \sin\theta)^2 + 2gy_0})$$  I attempted to solve for initial velocity (eq: A):  $$v_i = \sqrt{\frac{d^2g}{2\cos\theta^2(y_0+d \tan\theta)}}$$    Using this equation in my simulation I apply the velocity to a normalised displacement vector and launch the projectile. It gets close to its target but:   $x_{final}$ is always correct $z_{final}$ is always incorrect - it is close to the desired $z$ but always offset by a seemingly proportional amount.    I have spent many hours trying to review the equation but I have been unsuccessful. Any advice would be most appreciated:   is this the correct equation? have I solved for $v_i$ correctly? I have looked for an example equation online but I have not found one structured how I need it, i.e. solving for $v_i$ I have spent hours researching online - perhaps I have missed some good (but entry-level) resources. Do you know of any reference material that may help me?   Update - I have now named the equation above for referencing below  Following on from the answer by @Pygmalion, which I am still gratefully working to understand:  I agree that your derived equation is equivalent to mine (A). Using yours (@Pygmalion's) in my simulation, I therefore observe the same failures: the projectile always lands short of the target.  I have simplified the simulation keeping the launch height and target height the same. I still encounter the same problem with equation A; however, when passing the same parameters to the following equation the projectile always hits the target precisely (eq: B):  $$v_i = \sqrt{\frac{dg}{\sin2\theta}}$$  Given the success of equation B when the launch and target heights are the same and the failure of equation A given the same conditions, I question whether equation A is the correct one to solve the problem since I would expect it to work for any combination of launch|target height.  I think the help I need is around why equation B works. Does it cater for conditions that  equation A overlooks? If equation A is unsuitable, are you able to recommend an alternative?  In short, the $v_i$ calculated by equation A (in my simulations) is always less than $v_i$ from equation B given the same inputs - the projectile therefore always falls short of its target.
59135
I realised, reading another Phys.SE question about  that I don't really understand how  works. Particularly concerning, for a SCUBA diver.  The top answers to that question seem to claim that balloons get their "sense of down" from a pressure differential. They continue: when a car accelerates, the air at the back of the car becomes more dense, and at the front less dense, changing the plane of the pressure differential and so also, the balloon's sense of up. I find that extremely hard to credit. However, I realised that I don't really know why less dense things float in more dense things.  I'm fairly sure it's something to do with displacement of heavier things by lighter things, and I think pressure acting on the lighter thing's surface has something to do with it, but that's about it.
24556
I am using two books for my calculus refresher.   Thomas' Calculus  Higher Math for Beginners by Ya. B. Zeldovich   My question is : When applying Integral Calculus for calculation of volumes of solids, generated by curves revolved around an axis, we use slices of 'cylinders' to approximate the volume of the resulting solid and then integrate the sum of those infinitesimal cylinders. However, when we are using the same techniques to calculate the surface area of the surfaces generated by revolving curves around an axis, we consider the 'slope' of the differential length of curve 'dx', calculate the total length of the curve and derive the required surface area.  Are we not supposed to use the same 'slope' for calculating the volumes of the infinitesimal 'cylinders' for calculation of volumes? Shouldn't we use 'sliced portions of 'cones' as the infinitesimal volumes?? When it come to calculation of volumes of solids of revolution, why are we neglecting the slope of the curve for the differential length and simply assume that it is an infinitesimal cylinder??  Ex: Let us say we want to calculate the surface area and the volume of the solid generated when the parabola y = 10 . x^2 is revolved about the y-axis, with limits of x from 0 to 5.  In such cases, when calculating the volume of the solid, we consider infinitesimal 'cylinders', ignoring the 'slope' of the curve for the differential element 'dx', but when calculating the surface area, we consider the 'slope' of the differential element 'dx'.
106534
So I would have thought that this would be how you derive the  on a spring: basically the same way you do with gravity and other contexts, use $$W=\vec{F}\cdot \vec{x}.$$ If you displace a spring by $x$, then it exerts a force $-k x$, so $F=-kx$, since the displacement is $x$.   So $$W=-kx^2.\qquad \leftarrow\text{ (however, apparently wrong!)}$$  I've seen the correct derivation of work in a spring (with an extra half) and don't doubt that it's correct, but also don't see where my logic fails in this alternate derivation.
77836
I am trying to learn physics by myself as I do not have a good teacher at school. I've been reading Feynman Lectures on Physics and I can't figure out how he found  out this number. Here is an excerpt of :      Let us now illustrate the energy principle with a more complicated problem, the screw jack shown in Fig. 4–5. A handle $20$ inches long is used to turn the screw, which has $10$ threads to the inch. We would like to know how much force would be needed at the handle to lift one ton ($2000$ pounds). If we want to lift the ton one inch, say, then we must turn the handle around ten times. When it goes around once it goes approximately $126$ inches. The handle must thus travel $1260$ inches, and if we used various pulleys, etc., we would be lifting our one ton with an unknown smaller weight $W$ applied to the end of the handle. So we find out that W is about $1.6$ pounds. This is a result of the conservation of energy.       If you divide $2000$ pounds by  $1260$ in you get $\frac{100}{63}$, which is $1.58$, a number close enough to $1.6$ pounds, which according to Feynman is the weight that we need to apply to the end of the handle. I believe, however, that this calculation is wrong given a dimensional analysis of such operation.  Here is my dimensional analysis:  $\frac{2000 pounds * 1 in}{ 1260 inches }$ = $\frac{[M] * [L] }{[L]}$ = ${[M]}$ And this is wrong, as we are suppose to get a force, not a mass.  I found this , where you can find what I think is an email from an student reporting the teacher that he believes he got the number diving  $\frac{5}{\pi}$ . Can you also explain me this?   Finally, in a thread in a Physics Forum question,they concluded that $1.6$ is just the force to keep the system at equilibrium, and you need $\text{a force} &gt; 1.6$ to actually lift the ton. After, someone said that Feynman would have said this answer is wrong. Why is that?
54659
How can one analytically calculate the terminal velocity of a settling sphere in 2D? Actually it would be a circular disk. One cannot simple equate boyancy forces minus the drag right? As stated in   the formula $\textbf{U}_s=2/9(\rho_p-\rho_f)\textbf{g}/9\mu$ [1] does not hold!  [1]
12580
The height of a string in a gravitational field in 2-dimensions is bounded by $h(x_0)=h(x_l)=0$ (nails in the wall)  and also $\int_0^l ds= l$. ($h(0)=h(l)=0$, if you take $h$ as a function of arc length) .  What shape does it take?  My try so far: minimise potential energy of the whole string,  $$J(x,h, \dot{h})=\int_0^l gh(x) \rho \frac{ds}{l}=\frac{g \rho }{l}\int_0^l h(x) \sqrt{1+\dot{h}^2} dx$$  With the constraint  $$\int_0^l \sqrt{1+\dot{h}^2} dx- l=0$$ If it helps, it's evident that $\dot{h}(\frac{l}{2})=0$.  Generally, this kind of equation is a case of a constrained variational problem, meaning that the integrand in  $$\int_0^l \frac{g \rho }{l}h(x) \sqrt{1+\dot{h}^2} +\lambda(\int_0^l \sqrt{1+\dot{h}^2} dx- l)dx$$  Must satisfy the Euler Lagrange equation. The constraint must also be satisfied.  But, in truth, by this point I am clueless. $\lambda$ is worked through $\nabla J  = \lambda \nabla(\int_0^l \sqrt{1+\dot{h}^2} dx- l)$. I have tried this , but get nonsensical answers.  Is this method the best? If so, in what ways am I going about it wrongly thusfar?
47682
I'm a mathematics student trying to grasp some basics about wave propagation. A sentence I find very often in introductive physics textbooks is the following:      In a wave, energy is proportional to amplitude squared.   This is something I would like to understand better in the case of mechanical (linear) waves.  The simplest model is the vibrating string of mass density $\mu$ and tension $T$: here an element of string of rest length $dx$ and vertical displacement $y(x, t)$ possesses a kinetic energy $\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2$ and a potential elastic energy $\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2$. So we have the total energy   $$dE=\frac{1}{2}\mu dx \left( \frac{\partial y}{\partial t}\right)^2+\frac{1}{2}T dx \left( \frac{\partial y}{\partial x}\right)^2,$$  which completely explains the sentence. Can we obtain a similar formula for higher-dimensional waves? How should I modify the above formula for a (simplified model of an) elastic membrane, for example? I would expect something like   $$dE=\text{some constant}\left(\frac{\partial z}{\partial t}\right)^2+\text{some other constant}\left\lvert\nabla z\right\rvert^2,$$  where $z=z(x, y, t)$ is the vertical displacement. Am I right?
64718
I suspect an error in my “Classical Dynamics” lecture notes but my lecturer doesn’t agree with me so I need your help!  We assume that in the $n$-body problem, the force between particle $i$ and $j$ can be derived from a potential i.e  $$\vec{F_{ij}}=-\vec{\nabla_i}V_{ij}$$  where $\vec{\nabla_i}$ is the gradient with respect to position vector $\vec{r_i}$   $V$ would usually only depend on $\Vert\vec{r_i}-\vec{r_j}\Vert$  My lecturer then defines the total energy of the $n$-body system as  $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{ij}V_{ij}$$  However, I am convinced that this is wrong since we’d be double counting potentials. My lecturer says we should indeed be double counting potential. In my opinion, the expression should read   $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{1\le{i}\lt{j}\le{n}}V_{ij}$$  For the second equation we also get:  $$\dot{E}=\sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{1\le{i}\lt{j}\le{n}}\vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}}  + \vec{\nabla_j}V_{ij} \bullet \vec{\dot{r_j}} = \sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{i}\sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}} = \sum_i \vec{\dot{r_i}} \bullet \Bigl(m_i \vec{\ddot{r_i}} + \sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \Bigr) = \sum_i \vec{\dot{r_i}} \bullet \Bigl( m_i \vec{\ddot{r_i}} - \vec{F_{ij}^{tot}} \Bigr) = 0 $$  So this seems reasonable to me. Is this actually correct?  We also have an example of the two-body problem in the notes where we only use $V_{12}$ as the potential, not $V_{12}+V_{21}$
112323
A thin-walled container of mass $m$ floats vertically at the separation surface of the two liquids of density $ρ_1$ and $ρ_2$  . The whole mass of the container is concentrated in the part of height $h$.  The question is to determine the immersion depth $h'$of the container in the lower liquid if the bottom of the container has a thickness $h$ and an area $S$ and if the container itself is filled with the liquid of density $ρ_1$.    I applied the fundamental principle of the statics on the container  $Balance$ $Sheet$:  The weight of the container ($-mg$)  The buoyancy force applied by the fluid with density $ρ_2$ ($ρ_2gSh$)  The weight of the fluid with density  $ρ_1$  (-$ρ_2gSh'$)  I summed up the forces and set them to zero and find my $h'$  but then  realised that the fluid with density  $ρ_1$  applied also a buoyoncy force on the top of the container , and the fluid with density  $ρ_2$  applied a buoyoncy force on the other fluid.  Where's the problem ?
42584
Suppose from a height $H$, I throw a ball of mass $M$ and radius $R$ with initial velocity $u$ into a pool of depth $x$ having a liquid with density $\rho$ and coefficient of viscosity $\eta$.  Upto what height will the Water Splash?   This may be a silly question! (Source : my brain) Assume every condition to be identical with no air resistance. Feel free to assume any other parameters if needed.  P.S. $\rightarrow$ I've see similar questions but some of them want an exact answer and some are not so direct while some are incomplete questions...  Small Note : We do not have to consider the highest 'single' drop... Just assuming almost every water droplet jumped upto a height $h$. Ideal conditions...  Feel free to assume the liquid as WATER if required!
12260
I've started reading Peskin and Schroeder on my own time, and I'm a bit confused about how to obtain Maxwell's equations from the (source-free) lagrangian density $L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ (where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is the field tensor).  Substituting in for the definition of the field tensor yields $L = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\mu A_\nu)(\partial^\nu A^\mu)]$.  I know I should be using $A^\mu$ as the dynamical variable in the Euler-Lagrange equations, which become $\frac{\partial L}{\partial A_\mu} - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)} = - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)}$, but I'm confused about how to proceed from here.  I know I should end up with $\partial_\mu F^{\mu\nu} = 0$, but I don't quite see why.  Since $\mu$ and $\nu$ are dummy indices, I should be able to change them: how do the indices in the lagrangian relate to the indices in the derivatives in the Euler-Lagrange equations?
60339
$$Drag = \frac{1}{2}C_d \rho Av^2$$ I understand that the strength of the drag depends on the density of the fluid the body passes through, the reference area of the body, the , and the velocity of the object. I don't, however, understand the 1/2 and the $v^2$ in the equation.
23383
I have a final exam tomorrow for fluid mechanics and I was just looking over the practice exam questions. They do not provide solutions. But pretty much I have to define pressure at a point and also say why pressure is scalar instead of a vector.  I am thinking pressure at a point is $P=\lim_{\delta A \to 0} \frac{\delta F}{\delta A}$. Please let me know if I am wrong.   But I do not know at all why pressure is a scalar instead of a vector. I know it has something to do with $d \mathbf{F}=-Pd \mathbf{A}$