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0

The problem is much more simple than you think. The Coriolis force vanishes in this case (this is simple to see, since the motion of the bug is circular with constant speed), therefore only the centrifugal has an effect on friction. Then, the speed on exit can be easily found if you use the conservation of energy, it is $\sqrt{ 2\sqrt{3} gR ( 1-\mu/2)}$. If ...


0

If you are looking for an intuitive understanding think of it in terms of the two types of motion: translation (depends on $v$) and rotation (depends on $\omega$). An object can have either or both types of motions at a given time. Your first equation is more general; it has both translation and rotation.


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 ...


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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

$$\Delta y = v_0 \sin(\theta) t - \frac12 gt^2$$ 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: $$0=v_0 \sin(\theta) t - \frac12 gt^2$$ $$\require{cancel}\frac12 gt^\cancel{2}=v_0 \sin(\theta) \cancel{t}$$ ...


3

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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Centrifugal force does exist... it's clearly the force that makes a centrifuge work: http://en.wikipedia.org/wiki/Centrifuge Can't spin an object in a centrifuge without a centrigual force, right? =)


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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 ...


3

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 ...


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 ...


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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 ...


-1

Kinetic energy is defined as $\frac{1}{2}mv^2$ (in classical mechanics at least). When the motion of an object is subjected to a physical law that is constant through time (for instance $\ddot{r}=-\frac{GM}{r^2}$ where GM is a constant), then when you integrate both sides with respect to distance and multiply by the mass $m$ of the object you get: ...


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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 ...


0

Your girlfriend's book is wrong. ....is due to the mass of the object resisting the inward centripetal acceleration that the object is experiencing" Centrifugal force is not due to the resistance. The resistance towards acceleration is called "Inertia". Centrifugal force only occurs in non-inertial rotating frame of reference.


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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 ...


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 ...


1

Mathematically, terminal velocity—without considering buoyancy effects—is given by $ V_t= \sqrt{\frac{2mg}{\rho A C_d }}$ where $ V_t$ is terminal velocity, $m$ is the mass of the falling object, $g$ is the acceleration due to gravity, $C_d$ is the drag coefficient, $\rho$ is the density of the fluid through which the object is falling, ...


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

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 ...


0

No, you wouldn't. The moment (which is called torque, by the way - at least, the kind of moment you're talking about) of a force around a reference point is $$\vec{\tau} = \vec{r}\times\vec{F}$$ where $\vec{r}$ is the vector from the reference point to where the force is applied. If the force is zero, then you can tell that the moment (the torque) will be ...


0

Both $\vec p$ and $\vec r_2-\vec r_1$ start from the same point, and the tips of both vectors form line $H$. We know that the moment arm is the perpendicular distance from the axis of rotation to the force, and the perpendicular distance from the origin of both vectors to the line $H$ is obviously the same. You can also see that the cross product of two ...


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A vector is a variable which has 2 attributes - size and direction (unlike whats called scalar, which has only size - e.g energy). In most Newton's 2nd law questions, you need to "break down" the direction attribute into 2 components - $x$ component and $y$ component - usually by a simple sin/cos function. Then you can calculate each component size, and ...


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 $$\mathbb{r}_1 = (x+\sin\theta_1, \cos\theta_1)$$ ...


1

Centrifugal force is a pseudo force; the effects you experience from it are due to centripetal force. If you were to eliminate the centripetal force, you would stop going in circular motion and would also no longer feel any centrifugal "force"


0

Like John Rennie commented, there is much unclear which prohibits you from calculating the force. But making some assumptions like a completely elastic collision and the duration of the collision you can calculate the average force. Assume the weight has mass $m_1$, the object mass $m_2$ and the radius of the circle is $r$. The kinetic energy of the weight ...


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 ...


0

It's not clear what you are really asking, but this sounds a lot like the stick balancing problem used to keep undergrads busy in control systems classes. You can make the rod eventually go to any position you want by moving the bottom point appropriately. The Segway personal transport product is a great example of this principle being used in the real ...


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

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 ...


5

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 ...


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, $$ ...


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 ...


0

I'm going to assume some things before I answer: This golf ball doesn't feel the effects of air resistance, or is in a place where that does not matter. You know some common equations for kinetic energy and gravitational potential energy. We ignore any effects of weather and assume the only things contributing to the ball's motion is the golfer and gravity ...


1

You are correct. Each cylinder has a hydraulic force applied at the bottom that is sufficient to accelerate its mass upward. The one with the lighter mass will accelerate more rapidly and reach the top of travel more quickly, but the other will already be moving.


1

A note about the statement that a maximal gravitational force would occur at r=0: we can at least exclude the case for two distinct fermions with identical quantum numbers , by the Pauli Exclusion Principle.


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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 ...


0

Some hints to get you started: First, consider a Galilean referential in which the problem is simpler and assume some fixed $v_P$, without taking care of its magnitude. Second, use what you know of the distance between a line and a point (projection). Third, minimize the distance between the line and point wrt $v_P$ under the contraint that ...


0

The mass $M$ in the equation for orbital velocity would be equal to the mass of the Earth (assuming that the mass of the satellite can be neglected) and thus constant. The gravitational constant, $G$, like the name suggests is constant as well. But the radius of the (circular) orbit, $r$, can vary. So by using different altitudes at which the satellite would ...


2

As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.


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Let the outside pressure be $P_0$. You can find its variation with altitude here. Using Bernoulli Equation, speed of liquid coming out is : $$\rho gh=\frac{1}{2}\rho v^2+P_0$$ $$v(h)=\sqrt{2\rho gh-P_0/\rho}$$ $h$ is the height of water remaining. It is assumed that $r<<R$ and thus $V$ is very small compared to other terms and is neglected. By ...


0

If you have ever seen a pizza being made by hand, you will know that when the baker throws the disk of dough in the air, he makes it spin. As he does so, the pizza "disk" gets bigger because the dough on the outside experiences a larger centrifugal force (in the rotating frame of reference of the pizza. Don't start on "there is no such thing", you asked for ...


0

It takes zero force to push an object along a frictionless plane with a constant velocity. In fact a force can be defined as the rate of change of linear momentum, and zero force applied to a body yields no change on momentum. When friction is added, only when the applied force exactly matches the friction force the resulting motion is uniform. In ...


2

In the situation you gave, it's immediately clear what is meant, and there's no possibility for misinterpretation, so yes, it's perfectly acceptable. (Remember that torque is mathematically defined as a vector for convenience, but the direction of that vector isn't really physical.) The only issue I can see with that is that as you leave the simple ...


1

There is a wikipedia article which describes the effect http://en.wikipedia.org/wiki/Equatorial_bulge Basically the bulge is caused by the rotation of the Earth. The centripetal force is given by $F=m\omega^2 r$. Therefore the poles feel a lesser force than the equator which wants to spin out into a disc. This is balanced by gravity which wants the Earth to ...


2

A very simple motivation for writing $\ddot{\bf x}(t) = {\bf F}({\bf x}, \dot{\bf x}, t)$, which might shed some light, is the following. We are given ${\bf x}(t)$ and $\dot{\bf x}(t)$ and we desire to calculate ${\bf x}(t + \delta t)$ and $\dot{\bf x}(t + \delta t)$. Now, ${\bf x}(t + \delta t) = {\bf x}(t) + \dot{\bf x}(t) \delta t$, which we may ...


2

A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity. It's impossible. Or, don't ignore friction. When an object moves with constant velocity, the total net force on the object is always zero. If you have applied force, there's another force (or, many forces) like friction to counterbalance it. ...


2

The trajectories are uniquely determined means that the theorem of existence and uniqueness applies (so, the differential equation has to be sufficiently regular). Newton's principle states more: the system is fully determined by the position and the speed, that is, by $2n$ constants, where $n$ is the dimension of the space. As you have $n$ equations (one ...


2

Maybe you are pouring sand on your box.$$F=\frac{dp}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$ $$\text{As, } v=0 ms^{-1}$$$$F=v\frac{dm}{dt}$$ Second possibility : If your box is spherical, By Stokes' Law $$F_{viscous}=6\pi\eta rv$$ where $\eta$ is coefficient of viscosity. Hence, your ball attains terminal velocity. $$F=6\pi\eta rv$$ $$v=\frac{F}{6\pi\eta ...



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