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## New answers tagged newtonian-mechanics

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The two laws are the same. To see this break down your rotating object into a sum of point masses. Then consider one of these masses: The angular momentum of our point mass is given by: $$L = rmv$$ so: $$\frac{dL}{dt} = \frac{d}{dt}(rmv)$$ For circular motion $r$ is constant so we get: $$\frac{dL}{dt} = rm\frac{dv}{dt} = rma$$ But the second ...

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Since it can be derived from the first equation, it's not an independent principle. In fact, you don't really need an angular momentum to make physical predictions. It's just useful to have. I remember when I programmed some elastic bodies simulation - particles connected with springs. I only implemented Newton's laws and equation for the spring. When I ran ...

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This question have two approximate answers: If sun mass decreased by half but suddenly then planet will pull out into interstellar space. If mass of sun decreases slowly into half then planet orbit will be long elliptical because of sun weaker gravitational force.

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The boundary conditions and units (setting $m=1$ and choosing the length $l$ of the path also $1$) obscure the connection to the "lift" definition, but in fact they coincide for the given situation. The formula you derived for the work done against a constant force is $W = a$, which should really read $$W = m\cdot a\cdot l$$ If the force on the body is ...

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Answer Earth will escape from the solar system (It no longer in a) Reason IF you remember the Virial Theorem in order for a planet to have an orbit the average potential energy and average kinetic energy of a planet must have a relationship of $$2K_{avg} = -U_{avg}$$ Where $U$ is a gravitational potential energy, which is given by $$U_{avg} = ... 0 Notice, the static frictional force between tyre & the road will act in the lateral direction normal to the direction of motion of car while taking a turn without slipping on the road. The static friction between tyre & the road will provide the centripetal force to balance the centrifugal force (mr\omega^2) only when a car takes a turn or moves ... 1 If all the mass of a rigid body was squeezed down into a single point (at the center of mass) then the mass moment of inertia would be zero. But since the mass is distributed in space it would take a finite angular momentum to spin the body up about the center of mass. The importance of the center of mass is that it is the location when the mass moment of ... 0 No, an Atwood's machine with masses m and 3m has acceleration of g \frac{3m -m}{3m+m} = g/2, as explained in https://en.wikipedia.org/wiki/Atwood_machine#Equation_for_constant_acceleration 0 The conserved quantities in your problem are: total mechanic energy T+U, total linear momentum \vec P, and total angular momentum \vec L of the N point masses. They are defined as:$$E=T+U \quad\text{ total mechanical energy}U=-G\sum_{i}^N \sum_{j}^N \frac{m_i m_j}{|\vec r_i -\vec r_j|}\quad\text{ total potential energy}$$... 1 Although this has been answered many times already, anywhere on this site, the following holds: First law (existence of inertial reference frames) There exist in the universe some very particular reference frames where a point particle not subject to external forces moves in a straight line, i. e. \dot{\textbf{p}}=0. Second law (equation of motion in ... 0 Apply the formula of power P\ \mathrm{(W)} required for lifting the mass m\ \mathrm{(kg)} with a constant velocity v\ \mathrm{(m/s)}$$\text{power}=\text{(force)}\times (\text{velocity})P=mgv$$remember the starting torque shouldn't exceed the maximum torque of engine or it can't lift the weight. 0 Notice, the normal reaction exerted by the floor on the object is the weight of the object \bullet  \mathbf{\text{Normal reactive force}=N\uparrow } (upwards) \bullet  \mathbf{\text{Gravitational force}=Mg \downarrow } (downwards) \bullet  \mathbf{\text{Resultant force }=Ma\uparrow} (upwards) Now, balancing all the forces acting on the ... 1 It is correct except for one sign, note that the work done by friction is negative (since you move the block in the opposite direction w.r.t. the friction force) and thus it is equal to$$ -\mu mg d \cos \theta $$with this solving for  v gives you 11.49 m/s. 1 Maybe this is a bit beyond what you wanted, but you are implicitly evoking Einstein's equivalence principle, which tells us that gravitational accelerations and inertial accelerations are equivalent. It's only because this principle applies that we can add a gravitational and an inertial acceleration. Gravtitational accelerations are measured relative to a ... 0 The 'weight' in these cases is usually taken to be the normal force between the floor of the lift and the object. Using N2L on the object$$Ma=N-Mg$$Where up is taken to be positive and N is the normal force. Rearranging this gives:$$N=M(a+g)This is the normal force and therefore the 'weight'. Here we have introduced no pseudo forces and have worked ... 0 Consider the very first image, where the object is standing on end. Points G and the point marked later as the "instantaneous center" C are at height r above the table. As the object rolls, they both drop down until C is on the table. So from the kinematical view, neither G nor C can be the center because they are rotating around another point, namely the ... 0 Consider the following diagram: Left is the pool ball being hit by the cue and right an instant after the cue hit. If we hit the pool ball below the centre point of the pool ball, two things will happen: 1) An impulse force causes temporary acceleration to the right, resulting in the pool ball acquiring a rightward translation velocity v. From the ... -2 No, Newton's Second Law of motion is only an approximation and doesn't work on anything larger than a solar system. When you get into the domain where the acceleration is on the order of 10^{-14}\space km\space s^{-2}, then you can see limits of the approximation. Stars at the edges of spiral galaxies travel much too fast to be governed by Newton's ... 1 Aren't circles special cases of ellipses? In general, orbits can be either, but are usually elliptical (at least ideally). 0 Tension is a concept which is introduced in physics. The molecular interpretation of the concept is not requited. Because it is complicated. Your preposition that the tension in the rope the summation of tension in the individual molecule is not correct an the molecular structure may not be linear. Direction of the tension force is always acting away from ... 0 Rolling friction has a complex origin unlike static and sliding friction. During rolling, the surfaces on contact get momentarily a little, and this results in a Donnie are (not a point) of the body being on contact with the surface. The net effect is that the component of the contract force parallel to the surface opposes motion. 0 Tension is a force transmitted by a rope. At an individual point in the rope, if the rope is stationary, there can be no net force so all forces cancel out. In a sense this means there is equal "tension" to the left and to the right; in that interpretation there is no direction (although I would normally say the tension is "along the length). When you have ... 1 Here's how to intuitively understand that a=g. Take a metal ball having mass 1kg and drop it. Its downward acceleration is 9.8m/s^2, right? Now take a second ball and drop it. Same thing, right? Now drop both at the same time. Same? Now connect them together (with a tiny drop of weld metal) into a single 2kg mass, and drop them. Do they suddenly slow ... 1 Acceleration due to gravity remains roughly constant near the surface of the earth. Yes, a = F/M, but as mass increases, the force exerted by gravity increases too( F\ \alpha \ m1m2\over r^2), keeping F/M or a roughly constant around the surface of the earth 1 As the ball swings downward, its gravitational potential energy is converted to kinetic energy. At the bottom of the swing, its velocity will all be in the horizontal direction. From this, you can calculate the velocity with which the ball strikes the block. In a perfectly elastic collision, both kinetic energy and momentum are conserved. From this you can ... -1 Wrong density again... Density of CO2 is 1.8 kg/m3 =1.8 gram/liter Its 100 cc in 1 litre. So the density of CO2 is 0.018 gram/cc. And for air 0.013 gram/cc. Why not recalculate it all? 0 If you resolve the forces vertically and horizontally, it will pose a lot of problems. Just resolve the forces into radial and tangential components, it will work fine. I could not understand the specifications of the problem from the handwriting, so I cannot give a detailed solution. Its not a difficult problem, so i think You will be able to do it yourself ... 1 In the ideal case where there is no friction and no perturbations and the top starts to spin in a perfectly vertical alignment, the two configurations (inverted or not) of the top are completely identical. However, once you have the top start rotating with a tilt from the vertical axis, or consider perturbations that will tilt it even if it wasn't, then the ... 2 Force is not divided, it applied to the first bag, and then the first bag will make a force on the second one, and the second on the third. The first bag feels two forces, the one you apply and the reaction from the second bag, the second bag in turns feels two forces, one from the front bag and one from the rear bag. If the bags are attached trough ropes, ... 1 If all of them feel the same force, they would have an acceleration that would give them a speed and hence a kinetic enegy greater than the work done by the applied force. It would violate the conservation of energy and conservation of linear momentum principles. The force on each bag will depend on their individual masses, you can compute the acceleration ... 1 No the time taken does not depend of the velocity attained by the first ball(if they are ideally rigid) it rather depends on the elasticity or rigidity of the balls. So for ideally rigid bodies, the time taken to transfer approaches 0. Nothing would happen with an increase in distance between the two balls. See: Is the reaction force for a stone hitting a ... 2 This is only from intuitive meaning. Let us find the mathematical meaning of k/m . Here, the equation is \frac{d^2x}{dt^2}=-\frac{k}{m}x suppose \frac{k}{m}=K Then, the solution of this equation is x=A\sin{\sqrt{K}t} quantity inside the sine function is the angle. Thus, \sqrt{K} must be angular velocity. Thus,\sqrt{K}=\omega\\ ...

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The key point here is to find enough expressions for $\dot m$. Here are some hints: 1: Define some constants first (such as mass density of raindrop, mass density of water droplets in space)! 2: Express $\dot m$ in two different ways, one using attributes of spherical geometry, another using the rate at which the raindrop picks up water droplets as it ...

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Since as you're stating humid air ("condensed", even), why not assume net evaporation doesn't occur, so $\dot{m}=0$? Then set up a simple Newtonian equation of motion ($x$-axis is the vertical): $$F_{net}=ma$$ $$mg-\alpha v^2=ma$$ $$mg-\alpha \dot{x}^2=m\frac{d\dot{x}}{dt}$$ Which is separable. And for low velocities you could also use Stokes Law, ...

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I am somewhat not satisfied with the explanation of accepted answer. I found no mention of normal force which is key to understanding this situation. Also the outward force is centrifugal force, as the frame is rotating. I'll make my point real quick. No banking In this case, normal reaction is orthogonal to centrifugal force, so it has no effect on it. ...

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Kinetics: In physics and engineering, kinetics is a term for the branch of classical mechanics that is concerned with the relationship between the motion of bodies and its causes, namely forces and torques. Kinematics: Kinematics is the branch of classical mechanics which describes the motion of points, bodies (objects), and systems of bodies ...

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It is a hypothetical condition as inertial will never let this condition happen. For the sake of argument I am using impulse. faster you stop an object more will be the force. Example using gloves to stop a fast ball in sports. $$F_{impact}*t=mv-mu$$ $$F_{impact}=\frac{mv-mu}{t}$$ $$F_{impact}=\lim_{t \to 0}\frac{mv-mu}{t}$$ According the equation the force ...

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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 ...

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