# Tag Info

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Did model rocketry for a while... from a practical standpoint this isn't feasible... unless you're talking about deploying something that drops at a rate you know at the time of the parachute deploying (like a weighted streamer. See the link below). Your chute deploys and you don't know the descent rate of it, so there's no way to find the vertical distance. ...

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If $\Delta S = r cos\theta$ then $dl=ds=rd\theta$ and $F_g=mg$ $$W_{g}=\int_0^\pi mgrcos\theta d\theta$$ If you're taking the angle from the center of the circle (which you are, since you said that $\Delta S = r cos\theta$, then the initial position of the ball is $-R$, since displacement is a vector quantity (and the final ...

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You can decouple the horizontal and vertical motion of your rocket. In the vertical direction you have vertical thrust and gravity and horizontally you only have thrust (I ignore air resistance here). As you are interested in the altitude only, we only look at the vertical problem. All kinetic energy in the vertical direction is converted to potential energy ...

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The upright pole is in a position of unstable equilibrium. If the pole deviates from the vertical by an angle $\theta$ then the torque rotating the pole away from the vertical is: $$T = mg \frac{\ell}{2} \sin\theta$$ The moment of inertia of a pole about one end is $m\ell^2/3$, so the angular acceleration will be: $$\frac{d^2\theta}{dt^2} = \frac{3 ... 0 Not sure if I'm getting the question right, still let me answer the way I understood it. In this system you have two equilibrium points: the first is trivial (stick hanging down as in the left picture) [stable equilibrium], the second is as in your left picture with the stick "standing" [unstable equilibrium]. How do these two points differ? If you move ... 3 I am only going to leave a brief answer, seeing that the comments are very accurate. The paradox can simply be resolved by considering the elastic nature of all the objects. How so ever instantaneous might the dt or the time of collision seem to the human eye, actually it occurs over a small duration, based on the elasticity of both the objects involved in ... 0 Buoyancy is based on the principle that water pressure increases with depth. If something is submerged, the pressure acting upwards on it will be slightly larger than that acting down on it. If this imbalance is larger than the acceleration of gravity the object floats. Buoyancy is also proportional to the volume of the object but acceleration is related to ... -1 No, for the buoyancy force also act for the balloon that calmly float, or to push you up when you swim under the sea level. Buoyancy is a volumetric floating effect, it does not relate to surface or dynamics. -1 like the quantum mechanics if you could not measure something so that not exists/ how could you detect a mass if there is no force even if you get close to that to touch it you have a mass yourself so the force would appear and if not you are contradicting newtons law of gravity so the argument is not that simple mass and force are defined by each ... 0 For the system the Hamiltonian is : $$\notag H=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}+\frac{K}{2}\left(x_2-x_1\right)^2 \ .$$ You write down the Hamilton equation of motion: \notag \begin{cases} \dot{x_1}=\frac{p_1}{m_1}\\ \dot{x_2}=\frac{p_2}{m_2}\\ \dot{p_1}=k(x_2-x_1)\\ \dot{p_2}=k(x_1-x_2) \end{cases} ... 0 Your video is appropriate, the sun is in place and all the rest, including the fascinating scenery. Remember, if you're thinking of adding anything else, do not move the object hitting the sun, whatsoever. The sun's gravity is unimaginably powerful, so making bouncing effects would seem unrealistic. Also, as the sun has strong heat, (I know you mentioned ... 1 Let me discuss a simpler version of your rocket-question: one where there is no gravity, so that we don't have to worry about gravitational potential energy. Consider a rocket in free space (vacuum), and consider that the rocket is at rest. Now the rocket fires it's engine for a short time. The engine accelerates the rocket. The rocket now has kinetic ... 1 By using work energy theorem it can be solved. The velocity of the 2kg object till it reaches the 6kg object is given by$$\sqrt{2*g*5}=9.90m/s^2$$apply the conservation of momentum for plastic impact.$$m_1u_1+m_2u_2=m_1m_2V2*9.90+6*0=8*VV=2.475m/s^2$$work energy theorem$$\frac{1}{2}*8*2.475^2=\frac{1}{2}*72*(-x)^2+8*g*xx=1.801472656m$$... 1 when people refer to a "powerful" car ... they actually mean acceleration. This means Torque (which gets translated to Force at the end of the drivetrain). And Force = m x a ... so for a given mass, Torque == Force == acceleration. Unfortunately, the technical definition of the term Power is defined as Torque x Revs. This means the Power curve is sort of ... 0 In simplest terms and using Newton's mathematics: F = m * a. or Force F = mass m * acceleration a. Example #1 - A body on Earth. Now on the planet Earth, the gravitational acceleration "g" is about equal to 9.8 meters/second^2. So let's substitute a=g in the above equation. Then the force required to keep an object of mass m AT REST near the surface of ... 0 It's important to consider exactly where the forces of friction are being applied in a typical rising elevator. In most cases (I would hope), the elevator car isn't just scraping along the walls of the elevator shaft on its way up. That would make for a very noisy and expensive ride to the top, I would think. Elevators are more complex from a kinematics ... 0 Yes, friction force is F_r=\mu N, where N is the normal force exerted by the floor on the object, Here N=m(g + a) So yes friction force also increases 2 I gather that the large source of error you are worried about is the ability of the experimenter to accurately hit the start/stop button on the stopwatch at the start/stop of the ball's journey down the ramp. What is the approximate magnitude of error we'd expect? Before I directly answer your question, let's estimate how bad the experimental error will be ... 0 Now, one g is equal to the acceleration due to gravity by Earth on its surface. Force by washing machine is the centrifugal force as clothes try to go along straight path but the resultant provided by the walls of container of machine which provides the required centripetal force. Now, according to Newton's third law of motion , clothes will also give equal ... 14 The simplest formula for the centrifugal acceleration is$$ a = r\omega^2 $$Here, r is the radius which is 0.25 meters in your case. \omega is the angular velocity which is 2\pi times the frequency f. Your f is 1500 revolutions per minute which is 1500/60=25 revolutions per second. In the SI units, we have$$ a = 0.25\times 4\pi^2 \times 25^2 = ...

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Yes, Newtonian Physics works on a galactic scale. Still, for long distance interactions on fast objects you might want to take into account the finite speed of gravity, but I don't think it is necessary for ordinary galaxies simulations. Conversly a lot of phenomena occur that impact the galactic material: writting a decent simulation is not easy.

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Given an initial velocity vi, mass of the cue ball, coefficient of friction between the ball and surface, and the radius from the center as where the cue has struck, how would I determine the change in velocity. Unfortunately it is not enough to know where the cue has struck the cue ball, as the the spin on the cueball depends on many more factors (for ...

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For these kinds of system we often define a pair of quantities, one which is characteristic of objects or systems and one which is characteristic of interactions. Examples of these pairs are work (interaction) and energy (system) or impulse (interaction) and momentum (system). There is no commonly applied name for the interaction quantity that pairs with ...

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As Q had upward motion with velocity v, it keeps on moving up due to inertia of motion at that instant when P had striked and rebounded.May be due to elastic collision P moves up with v velocity.

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force exerted by the heavy steal ball on air partical is very very small. If a ball moves with even as much as 1000kmps may exert a very small force of 0.000001N on a body. Force exerted on a air particles by steel ball is very small although it has great speed. Force is rate of change of momentum . A very small particle of air cannot reduce its momentum to ...

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The third law of motion: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body. You ask : Then why does'nt it bounce back when it hits a water surface ? Bouncing back depends on elasticity of surfaces deformations and dissipative motions of the ...

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If a steel ball hits an iron surface it bounces back due to the Newton's third law of motion. Then why does'nt it bounce back when it hits a water surface ? or for that matter even air ? The molecules of water / air should apply the force of equal magnitude on the ball as the ball applies to the water molecules. Can anyone please explain this? A simple ...

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Yes. The tear is initiated at stress concentrations around the holes, where stress is highest. After initiation, the tear continues to propagate along the line of highest stress. Stress is a function of force and geometry ($\sigma_{n} = \frac {F}{A_{n}}$). In a piece of paper without holes, the stress is uniform, and the paper tears when stress exceeds ...

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As you've already figured out, roll the sphere down a frictionless hill with zero slipping and use $$\frac{1}{2}I\omega^2 + \frac{1}{2}m v^2 = mgh.$$ To use this equation, you will need to understand the relation between $v$ and $\omega$ and then you will need some way to measure either $v$ or $\omega$. How you measure the velocity will depend on the ...

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Experimentally measuring a sphere's moment of inertia See: Measurement of the Moment of Inertia of the Rotating Platform and Attached Cylinder. Once you build the platform and determine its moment of inertia you put a sphere on the platform and determine the moment of inertia for the system sphere - platform and then by subtraction the moment of the ...

3

There is a mutual attraction from gravity, and we generally only consider the smaller object here on earth because the earth is so massive, the acceleration of the earth is negligible. This is because $a = F/m$, and with equal $F$ between the two objects, the acceleration will scale as $a\propto 1/m$. For the earth, this leaves $a$ ridiculously small, but ...

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Does the bigger mass EVER move towards the smaller mass? Yes. $F = KMm/r^2$ $M*a_{M}=F$ $m*a_{m}=F$ As you see the smaller the mass the higher the acceleration and in consequence the higher the traveled distance in a given time t. If the above is true, can we technically move the Earth by us(human population) jumping indefinitely? No. Each ...

1

yes, the earth will accelerate towards you , however the Earth's acceleration will be so small for all practical purposes that you usually do not consider it. Earth's acceleration is small because the mutual forces between you and the earth are the same, but the masses are different, so this results in different accelerations (remember: $F=ma$). Now if you ...

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In all cases, the two objects move towards one another. In fact they experience exactly the same gravitational force. However, because acceleration equals force over mass $$\mathbf{a} = \frac{\mathbf{F}}{m}$$ that equal forces causes the heavier object to accelerate much less than the lighter one. But technically, the Earth does move towards you very ...

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If you accelerate relative to other objects, you will see them accelerate towards you. according to Newtons laws if a object is accelerated some force must be acting on it. but this acceleration is not causes by a real force, it is only caused by your own acceleration. Newtons laws then should no longer valid if you are accelerating. However, you can make a ...

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The total energy as a function of time is a constant and it is always equal to the potential energy in the point where $x=x_{max}$ and $v = 0$. The total energy is the work done to bring the free end of the spring from $0$ to $x_{max}$. So $E(t) = E = Work = kx_{max}^2/2 + ax_{max}^4/4$ You simply integrate $f(x)$ from $0$ to $x_{max}$

1

Hint: Use $$m\ddot{x}=-kx-x^3 \\\ddot{x}=v\frac{dv}{dx} \\-\frac{kx^2}{2}-\frac{ax^4}{4}=\frac{m}{2}\left(\frac{dx}{dt}\right)^2$$ It will reduce to a form $$\frac{dx}{dt}=ix\sqrt{c^2+x^2}$$ This is a standard integral, and can be solved, then use $$U=-\int f(x) dx \\T=\frac{1}{2}m\dot{x}^2$$ Total energy $E=T+U\; .$

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I believe there is no definite answer to this question and it has puzzled physicists for a long time. In fact, it has puzzled them so much that after the grand work of Newton's, many people still sought different formulations. The Cartesian school for instance thought that "only pre-existing motion can trigger a change of motion" akin to what people witness ...

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The beam will be accelerating downwards after one guy lets go, and it will be rotating. I would ask myself - how fast is it accelerating? How fast is it rotating? There will be a torque $\Gamma$ due to the remaining force $F$ of the one man $\Gamma=F\ell/2$ resulting in angular acceleration $\dot \omega = \Gamma/I$; and a vertical acceleration of the ...

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Newton's first law states that there exist very special reference frames (that we are going to call inertial henceforth) where any point particle not subject to external forces (interactions) moves in straight lines, i. e. the equation of motion is $\dot{\textbf{p}}(t)=0$. Newton's second law states that, in the above mentioned reference frames (and only in ...

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Suppose a force $F$, acting on object $A$, produces acceleration $a_A$, and the same force $F$, acting on object $B$, produces acceleration $a_B$. Suppose a different force $G$ produces accelerations $a_A'$ and $a_B'$. Then Newton's law implies $a_A/a_B=a_A'/a_B'$, which is not tautological.

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Let's look at this problem from the point of view of equations of motion, see diagram below: Firstly let's make a few assumptions. Ball and cube are of same weight ($mg$) and same size. Simple friction model $F_f=\mu F_n$ holds and $\mu$ is independent of speed. Both objects are completely stationary (no sliding, rolling or tumbling) at $t=0$, at which ...

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$$T_1/\sin \psi=T_2/\sin\psi=W_t/\sin\theta$$ $$10/\sin\psi=10/\sin\psi=W_t/\sin\theta$$ Therefore $$10/\sin\psi=W_t/\sin\theta$$ $$W_t=F$$ $$\theta=360-2\psi$$ $$W_t= \frac{10\sin\theta}{\sin\psi}\tag{using lami's theorem}$$ $$=\frac{\sin (360-2\psi)*10}{\sin\psi}$$ we know that $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ so \sin (360-\psi-\psi)=-sin ...

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This is a very common question asked by students like me in mechanical engineering. The friction is irrespective of area means the friction generated, the vector F (The letter with which we denote) will form whether area in contact is less or more But the ability to stop is determined by the Number of friction vectors developed per area of contact. On the ...

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where spring's energy goes It does not go. From the conservation of energy, $\frac12Kx_\text{max}^2=\frac12mv_\text{max}^2$, and $m = 0$ you get $v_\text{max}=\infty$. In conclusion, your spring will oscillate with $\text{Amplitude} = x_\text{max}\;,v_\text{max}=\infty \; \&\; \omega = \infty\; .$ For more information see: ...

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There are no ideal springs. Therefore the paradox with the infinite acceleration is not a physical one, but an artefact of the mathematical modelling. Conservation of energy does obviously not hold, when there are objects of mass zero in a system (because the kinetic energy will always be 0). So your setting simply does not fulfil the requirements for energy ...

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Normally it is said that the spring has no mass, but in problems where the spring is attached to bodies with larger mass compared to the spring mass. So you pull the string from the side where you have a body of mass $m$ (tha other side is attached to the wall). Then when you release this body with potential $\frac{1}{2} K x^2$ it will be converted in ...

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In all the gas laws, it was assumed that the vector sum of momentum of all those molecules is zero.That is a pretty strong assumption in that case. But here obviously, its not zero. Due to this reason, the force on 'The walls of a container' per unit area, depends on the 'wall' you are choosing, which also depends on the alignment of initial velocity vector ...

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Whenever one applies a sideways force trough the center of gravity of an object, that force has two components: 1) a direct force that tries to overcome friction and slide the object, and 2) a torque that uses friction to produce a rotation of the object by lifting its center of gravity over the leading edge. A short, flat object will tend to slide ...

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