New answers tagged

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There are many good and insightful answers here, but I will just add one thing. As stated, the answer to this question is, of course, emphatically no. But I want to come at it a little more from the angle of what relativity actually tells us. First of all, Galilean relativity is not capable of telling us anything about the acceleration at all, except that ...


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The bulk modulus of water is $2\times 10^9$ Pa $= 2\times 10^4$ ba $= 1.97\times 10^4$ atm. The bulk moudulus is defined as the additional Pressure devided by the corresponding rate in volume change: $$ B = - \frac{1}{V} \frac{\Delta P}{\Delta V} = -\frac{\Delta P}{\Delta V /V}. $$ Therefore for a increse in the pressure $\Delta P$ leads to a change in ...


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Water is relatively incompressible but nonetheless not absolutely. Its compressibility depends on its phase as well as temperature and pressure. When you say change with depth, I assume you mean increasing pressure. The way I would think of your description is some volume of liquid water wiith a heat bath beneath it to keep it at constant temperature and ...


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The question is unclear. Perhaps you might want to include all the forces in including gravitational force and then reformulate the question. Nevertheless, let me provide a complete picture of the earth and rocket system. We have an earth that exerts a gravitational force on the rocket $F_{e-r}^g$ and by Newtons law the rocket exerts an exact equal but ...


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You can firstly think about this scenario without the earth at all. Let's say you have a rocket in free space. A chemical reaction inside the rocket can cause a force $\vec{T}$, which acts on the exhaust material and it will be directed downwards, away from the rocket as you showed. Newton's third law tells you then that there will be an equally large but ...


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It's an old question, but let my add one more aspect to the answers. First of all, intuition fails when talking about lower or higher gravity, as most of us don't have any experience with gravities different from earth, so "obvious" things can easily be wrong. Thought experiment There are the so-called zero-G flights where an airplane following a ...


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@joseph h inertia is a natural property of the ball and all material objects This may be seen from a different perspective. There are confusing uses of mass in which it can be interpreted as an emergent behavior (like temperature) or arguably fundamental, such as its association with the Higgs particle. It is not clear, at least to me, that we understand ...


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Because the rolling resistance of the ball is less than the force of acceleration transferred to the ball from the floor of the bus


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When no forces are acting on an object, it is either completely still or moving at a constant velocity at the same direction in a straight line. It is easy to think keeping an object at a constant velocity requires forces acting on it, but that is because we don't live in a vacuum


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@Eisenstein I suspect from some of your replies that @RKleberhoff is closer to the meaning of your question. I think your question is more metaphysical and ontological. The reference of @YKindaichi provides are what I would call descriptive answers of how rather than why. I think we sometimes confuse mathematics with why rather than the how it is. Sometimes ...


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Let's consider an object is at rest. Now when we try to move it, we feel some sort of resistance. To scientifically explain it we call it inertia. Now as to why it happens, this is just my theory. Lets use the textbook example that a bus is moving and when the bus stops the meth goes inside my lungs and I die. So when I am doing this my body doesn't know ...


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If you google it you can find several explanations for this, but I believe using the classical physics, we can clarify it intuitively and simply. According to the Newton's first law, bodies tend to keep their velocities unchanged. Inertia typically refers to this phenomenon. You can check it by some experimental tests. Therefore, if you want to change their ...


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Philosophical aspect Physics typically cannot answer "why"-questions very well. Physics observes phenomena in the world, creates descriptions of the behaviour observed there, and postulates rules (physics laws) stating that under specific circumstances this very same behaviour will happen again. So, the best answer that physics can give go along ...


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A force as an external influence or action on an object that causes the object to change velocity, that is, to accelerate relative to an inertial reference frame. Newton’s second-law statement, like Newton’s first–law statement, can be applied only in inertial reference frames (Galilean invariance). Newton's second law states: The acceleration of an object ...


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It is the generalized momentum conjugate to $\phi$, defined as $$p_\phi=\frac{\partial L}{\partial \dot{\phi}}.$$ With your choice of generalized coordinates, $r$ and $\phi$, it is the angular momentum. As $\phi$ itself does not appear in the Lagrangian, it is said to be an $\textit{ignorable}$ coordinate, so its conjugate momentum is a conserved quantity. ...


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Yes, the term $\mu r^2\dot{\phi}$ represents the angular momentum, and the first Euler equation telling you that the angular momentum is the constant of motion for the given problem. The second Euler equation is just the radial force equation. If you can try to write newton's equation of motion like $$F_r=m(\ddot{r}-r\dot{\theta}^2)\ \ \text{and}\ \ \ F_\...


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Actually, you don't have to include inertia for an intuitive explanation. Imagine an astronaut up in outer space. He is working on the space station from the outside, floating about in free space. He then let's go of his wrench. What would you intuitively expect the wrench to do? You would expect it to just hang there, right? Because there are no forces ...


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This is called an inertial force. And it is due solely to inertia. It is not a real force that opposes the forward motion of the ball. It is actually the fact that the object has inertia. And what happens is that an observer inside the bus will see the ball accelerate backward if the bus accelerates forwards, and this is due to this inertial force. I know ...


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Suppose $v_0\geq \sqrt{5gL}\quad (1)$ $T-w\cos\theta=m{v^2\over L}\Longrightarrow T=w\cos\theta+m{v^2\over L}\quad (2) $ From the conservation of energy: $v^2=v_0^2-2gh\quad (3) $ $\cos\theta=\frac{L-h}{L}\quad (4)$ $(2),(3),(4)\Longrightarrow T={m\over L }\Big[gL+v_0^2-3gh \Big] \quad(5) $ $(1),(5)\Longrightarrow T>0 $ $\ $ so the string doesn't ...


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Because of Newton’s third law, for every action there is an equal but opposite reaction. When your feet push on the ground, the ground pushes back. The force that your foot exerts on a flat ground is mainly determined by friction. The higher the friction, the more efficiently you can walk. If the frictional force is decreased, the harder it will be to walk. ...


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The peaks of your curve imply infinite energy which is not physically possible. You will reverse direction at a finite acceleration, which will make your triangle wave like curve rounder.


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Walking is applying force on an object and, that object keeping the same approximate location, being propelled in return in the opposite direction. If said object is mobile, like a box on a slippery floor, when you push it it will move and offer no resistance. This resistance is what makes walking possible. This is why sand or, worse, water, are not ideal to ...


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Soft sand deforms when you step on it. The force applied by your foot to the sand moves through the deformation distance, performing work on the sand which your muscles provide. That work does not assist your propulsion; it is lost to the sand. The sleek flat floor does not deform when you walk on it and therefore does not "steal" work from you as ...


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Another way to show you are right -: $x=ut+\frac{1}{2}at^2 = (v-at) t +\frac{1}{2} at^2 = vt -\frac{1}{2} at^2$


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There are two reasons for friction: Interlocking Bonds or cold welding When an object is at rest it makes bonds with the surface and the irregularities on the surface of both of the objects get interlocked due to which the friction at that time is called static friction and has the highest friction force. But when you apply force and make the object move ...


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Here are the four equations being discussed here: \begin{align} v &= u + at \tag{1} \\ x &= ut + \frac{at^2}{2} \tag{2} \\ v^2 - u^2 &= 2ax \tag{3} \\ x &= vt- \frac{at^2}{2} \tag{4} \end{align} Is this derivation correct? The result is always correct. The derivation is correct, except for dividing by $(v - u)$. It is possible that this is ...


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No, wings are designed to carry the maximum load with a minimum of weight. This results in their being flexible; to make them perfectly rigid would dramatically increase their weight and thereby reduce the payload of the aircraft. Aerodynamic flutter resulting from wing or control surface flexibility is then managed in a way consistent with minimizing the ...


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Not knowing algebra will make studying physics difficult. You can learn a lot of physics with just algebra under your belt (and then a bit more with calculus). When two quantities are proportional, such as $w$ and $m$ as expressed in $w=mg$, then in a sense they carry the same information and are redundant. You can view two proportional quantities as two ...


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@Kami I think I understand the experiment you are performing. I'm guessing you are taking a massless unextended ideal spring (plus no frictional forces) and then adding mass and letting it drop. It drops a total distance of L due to gravity before the restoring force of the spring stops the motion and then the whole thing will keep oscillating around the ...


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Keep everything scalar and discuss components only. Translational momentum is $p = m\,v$ Rotational momentum is $L = d\,p$, where $d$ is the moment arm (the perpendicular distance of the line of motion and origin). Translational speed is $v = \omega\, d$, also with the perpendicular distance Rotational momentum is thus $L = (m d^2) \omega$ The thing inside ...


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I like your proposed definition as an intuitive explanation but not as a definition of angular momentum, primarily because: Angular momentum is more fundamental than torque, so defining it in terms of torque is presenting these in a confusing order. Of conserved quantities in mechanics (momentum, energy, etc) angular momentum is the easiest to observe and ...


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In theory, it takes an infinite amount of time for the object to tip over. In practice though, there will always be a chance the object falls down, due to classical statistical mechanics. It's very hard to predict though when. When the object falls down depends on several conditions and qualities of the object. As there are its mass and form, its surface, ...


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I think you're describing the case where the equilibrium position is the mass hanging at rest, with the spring stretched out somewhat due to gravity, and the mass is pulled farther downward by a distance $L$ and then released. This case resembles a spring-mass-damper problem, in which damping can arise in the mechanical deformation of the spring (termed ...


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Let's suppose an ideal spring and neglect the effect of air resistance. Apply: $\sum F = m \vec{a}$. You have tree equilibrium points ($\sum F = 0$): $x_{0}$ when there is no external force: \begin{equation} F_{g}+F_{k}=0 \hspace{5pt} \Leftrightarrow \hspace{5pt} x_{0}=\dfrac{mg}{k} \end{equation} $x_{i}$ when the spring is streched: \begin{equation} F_{g}+...


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Yes, $v_{0}\geq \sqrt{5gL}$ is the necessary and sufficient condition for a body fastened to the end of a non-extensible string of length $L$ to perform vertical circular motion about A without the string being slack at any point of time.


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If you recall the Coriolis force given by $$\mathbf{F}=-2m\mathbf{\Omega} \times \mathbf{v}$$ If for a minute, We don't get into the detail of the analysis but just see the magnitude of this. $$|\mathbf{F}|=2m|\mathbf{\Omega}_\text{earth}||\mathbf{v}_\text{train}|$$ Let's take the speed of the train to be $100 $ Km/Hour which is about $30$ m/sec and mass to ...


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It does. The fact that it is in contact with the earth (or rather, the rails), just means that it cannot move sideways under the force, because the rail restrict its movement. In the same way, if a train travels down a rail and has wind blowing from one side, there is still a force pushing it, but the rails will exert an opposite force so it can't be moved ...


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Yes, this force acts on all objects in motion on a revolving body. You can see this in action even on a merry-go-round. Please note that the force acts tangentially to the pivot/axis of rotation.


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A string can be seen as a boundary on either side. If center of gravity is above the string, the object will stay on the string in equilibrium. Therefore if center of gravity is above the boundary, object will stay on the table.


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If you approach this problem mathematically and tried to calculate the time it takes for an object to tip over as a function of the overhang distance $x$ of the center of mass over the edge, you will find that $t \rightarrow \infty$ as $x \rightarrow 0$. So does this mean that the object will tip but after an infinite time, or that it won't tip at all. ...


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The best way is to find out, using calculations, Suppose initially mass $m$ is stationary and mass $M$ is moving with speed $u_1$, then if after the elastic collision the velocities are $v_2$ and $v_1$ respectively then $$v_2-v_1=u_1$$ $$Mu_1=Mv_1+mv_2$$ $$v_1=\frac{M+m}{M-m}u_1\ \ \text{and}\ \ v_2=\frac{2M}{M-m}u_1$$ Is it possible for a collision to ...


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It is possible under contrived conditions for all of the momentum to be transferred from a large mass object to a small mass object as you have described. The issue is the energy. In such a collision the total kinetic energy of the system after the collision is greater than the total kinetic energy before the collision. So the collision must involve the ...


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If the centre of gravity of the object is vertically above the edge of the table then the object is in equilibrium. However, this equilibrium position is unstable (like a pencil balanced on its point) because a small tilt of the object will lower the centre of gravity, which will then cause the tilt to increase. This is a positive feedback loop. However, if ...


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Let's say the disk has got a non-zero radial velocity. This then has $2$ possibilities. First, the radial velocity is outward along the string and second, the radial velocity is inward along the string. The First case cannot happen because of the restriction given in the question, the string is inextensible. For the Second case, if the disk has a velocity ...


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The ball dropped for a low height like 20 cm will give a lower rebound height than the higher drop heights until it reaches its terminal velocity when it is the fastest downward force it will have due to air resistance. (not throwing, dropping) Therefore the bounce efficiency will decrease and decrease.


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Just a thought: if the object were to tilt even the slightest bit (perhaps it and/or the surface are a bit pliable?) then the CG would move further away from the table (if only ever so sligjtly) - and this is an unrecoverable process. However, I know there is theory stating that there is an "undefined state" about the direction of bounce of a ball ...


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You are mixing two things: average velocity (that is a vector) and average speed (that is a number). They are the same only if the trajectory is a straight line, and you do not change direction (someone could consider "speed with sign", but here I consider speed as always positive.. sign is reminiscent of being a vector in 1D, and here I hant to ...


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When I learned the kinematic equations in high school, this equation was actually presented to us. There is however a reason it's seldom seen. The main "issue" is that your equation, in order to be valid for all $t$, would be better expressed as $$x(t)=v(t)\cdot t-\dfrac12 at^2$$ This is because when $t$ changes, $v$ changes as well. And if you ...


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Your first approach is wrong. You assume that the work is $F\cdot S$, which is only valid iff $F$ is a constant force. However, the force that satisfies your scenario is not constant. The easiest way to show this would be take the negative gradient of the potential energy which gives the force. If you haven't learned this, then you can try solving for the ...


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Because the cord is inextensible, its tension does no work on the disc (in other words, the disc always moves perpendicular to the cord). Therefore we can model the situation as a linear deceleration under friction. It would be the same if the cylinder's radius were zero and the motion were circular.


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