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Summary Centrifugal force and Coriolis force exist only within a rotating frame of reference and their purpose is to "make Newtonian mechanics work" in such a reference. So your teacher is correct; according to Newtonian mechanics, centrifugal force truly doesn't exist. There is a reason why you can still define and use it, though. For this reason, your ...

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The trick is, centrifugal force is a fictitious force. Centrifugal force exists! To everyone denying it, do this to them: xkcd.com/123. However it is a fictitious force. To quote wikipedia: A fictitious force is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference ...

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As I disagree with all the answers I am going to try to explain some of the fundamentals of science: Science in it's very essence can not explain why things happen the way they do, they simply try to model reality based on observations in the past to predict events in the future. In other words, defining a centrifugal force is possible as for example your ...

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Yes, angular momentum depends very much on the origin you pick. (You can see this most clearly by examining a single particle in free space with fixed velocity - the angular momentum is $0$ only if you pick the origin along the particle's line of movement.) It's true that linear momentum is independent of your choice of origin; however $\vec p$ is still ...

4

Let us consider there is no external force and the tyres are rolling smoothly without slipping. Here friction doesn't come into play let me explain you how. Even if friction is present, this friction is not the answer. The concept of friction most books provide are deficient. The term "rolling friction" is also a misnomer. The correct answer is rolling ...

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To figure this out, you need to know about momentum ($p$). That's a combination of how fast something is moving ($v$, for velocity) and how much it weighs ($m$, for mass). You'll also need to understand algebra, which is just using a letter to mean some number you don't know yet. $$p = m\cdot v$$ Momentum is conserved, which means the momentum from both ...

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In Newtonian physics, objects continue moving in a straight line unless a force acts on them, therefore if an object is not moving in a straight line, a force must be acting on it. Consider planets. Why don't they just fly off into space on a straight line? Because the sun pulls them. Consider a rock at the end of a string. Why does it not fly off when ...

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I'm not sure that there is a good way to figure out exactly what happened unless there is video of the crash, however both statements can be true. There will have been some loss of energy in the collision due to overcoming the kinetic friction of your car (getting it moving from being stopped) as well as the energy absorbed by permanently deforming the ...

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The key to the conundrum is that for the purpose of explaining the apparent forces on someone to whom a rotating frame of reference appears to define stationary, for example all human beings everywhere, centrifugal force may need to be taken into consideration since it appears to be there. Although it may be small depending on the speed of rotation. Which is ...

3

The truck will have in its wake some unknown mass of air almost moving with a speed $v$ comparable to the truck's speed $\bf V$. The pressure behind the truck will be lower than the pressure at the sidewalk because air pressure follows the Bernoulli equation, $$P_\mathbf{P} = P_\text{road} + \frac{1}{2}\rho v^2,$$ where $\rho \approx 1~$kg/m$^3$ is the ...

2

Assume $f$ friction is being applied in direction of F. Direction, now, has no significance as $f$ will come out to be negative if it is opposite direction. $$F+f=ma$$ $$FR-fR=I\alpha$$ symbols have their usual meanings Note that if pure rolling occurs, $f$ is static. Also, $$a=\alpha R$$ You can calculate $f$. If $$|f|> \mu_{static}mg$$ You can ...

2

The Kronecker delta is used in the first term, not the second. In the first term, replace $\sum\limits_i \omega_i^2$ with $\sum\limits_{i,j}\omega_i\omega_j\delta_{ij}$. For the second term, rearranging the summations is a pretty common thing to do. The identity that you are interested in is essentially the distributive property. As a simple example, $$... 2 It's the chain rule. We have$$ r_{ij} = \sqrt{w} $$where I've defined w as all those terms underneath the square root. So$$ \frac{d r_{ij}}{dt} =\frac{d r_{ij}}{dw} \frac{dw}{dt} = \frac{1}{2 \sqrt{w}} \frac{dw}{dt} = \frac{1}{2 r_{ij}} \frac{dw}{dt} \,, $$where to go to the second expression I've used the chain rule (the rest is just ... 2 This is really two tricks in one. Let's look at each one individually. The forks/cork/match set is balancing while being mostly not on top of the cup. This has entirely to do with the center of mass for those objects. The center of mass for those four items appears to be on the lip of the cup. This is why, when the presenter pushes down on it, it "wobbles" ... 2 As you stated in your question, the effect of an external magnetic field on an atom depends on the magnetic dipole moment of this atom. Before the introduction of spin, the only contributor to the magnetic dipole moment was the orbital dipole magnetic moment:$$ \vec{M}=\vec{M_L}=-\frac{e}{2m_e}\cdot\vec{L}$$which does not explain the S-G experiment since ... 1 When you have a sphere, let's say, rolling down a ramp, the gravitational potential energy will be converted into two energies: Rotational kinetic energy and translational kinetic energy. You use energy to keep it spinning (AKA moving angularly) in addition to keeping it moving translationally. Therefore, the sphere's total energy at the bottom of the ramp ... 1$$\Delta y = v_0 \sin(\theta) t - \frac12 gt^2$$This is a projectile, so it will hit its max at t_{max}=\frac12t, where t is the total time in which the projectile flies. The total time is when, as you know, \Delta y=0. Hence we've got:$$0=v_0 \sin(\theta) t - \frac12 gt^2\require{cancel}\frac12 gt^\cancel{2}=v_0 \sin(\theta) \cancel{t}$$... 1 In the frame of the car it is not useful to talk about centripetal force. In the rotating frame, you have two forces: the centrifugal force, and the real force of the of the side of the car pushing against you as the centrifugal force accelerates you toward the outside. Note carefully that this force is not a reaction force to the centrifugal force. You ... 1 Centrifugal force is force that pulls rotating object away from the center of rotation, Centrifugal is part of Newtonian mechanics and it's derived from Newton's Second law$$F=ma$$Where F is force in newtons, m is mass of an object and a is acceleration. In circular motion acceleration is a=\frac{v^2}{r} and full equation for centrifugal force is ... 1 As said before, the answer is no(t always), but there is a simple law which can help you predict whether it will be the case or not, and how the torque is distributed across your solid*. Using simple algebra and \times distributivity, one can easily prove that$$ \vec{\tau}_{\vec{p}}=\vec{\tau}_{\vec{o}}+\vec{op}\times\vec{R} $$where \vec{R} is the ... 1 You would only use a component of a vector only if you are interested in that component. We break vectors into components since it is easier to use the vectors in calculation. For example: If you throw a ball, you might want to see how high it goes, or how far you can throw it. Here we use a component of the vector (either vertical or horizontal) without ... 1 Based on your comments (that the exam question said "switch A is pressed"), the question can be answered - and the tutor was correct. The key is to look closely at the diagram, and observe that the lower halves of the compartments are connected together, as are the upper halves. In this diagram, p_1 represents the driving pressure. Now across the ... 1 Folks are looking for non-Newtonian gravity. The experimental gravity group at U. Washington have really been the leaders in the field over the past ten years; they have some nice review papers available for free. Because of the way that short-range forces work in quantum mechanics, we expect that a short-range gravitational interaction would produce a ... 1 This is actually an interesting problem in classical mechanics, dating back to Huygens. We'll work with the three variables you define in the question, namely (x,\theta_1, \theta_2). Also we'll set your m = l = g = 1 for simplicity. Kinetic Energy The position vector of the first pendulum bob is$$\mathbb{r}_1 = (x+\sin\theta_1, \cos\theta_1) ...

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The tyres of the cycle are rolling and the remaining cycle moves with a velocity same as that the centre of mass of the tyres have. Now the question is which force is responsible to bring the cycle at rest. The answer is Air-friction and Rolling-friction. It should be noted that the static and kinetic friction does not come into the picture because the point ...

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Gravitational force decreases with distance squared. So the deceleration due to the sun is negligible at that distance. Acceleration due to gravity is given by $\frac{GM}{r^2}$, where $G$ is the gravitational constant $6.67\times 10^{-11} \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}$. The mass of sun is $2\times 10^{30} \mathrm{kg}$ and the distance is ...

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