# Tag Info

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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 ...

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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 ...

0

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

0

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"

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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 ...

3

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, $$... 1 Based on your comments (that the exam question said "switch A is pressed"), the question can be answered - and the tutor was correct. The key is to look closely at the diagram, and observe that the lower halves of the compartments are connected together, as are the upper halves. In this diagram, p_1 represents the driving pressure. Now across the ... 0 I'm going to assume some things before I answer: This golf ball doesn't feel the effects of air resistance, or is in a place where that does not matter. You know some common equations for kinetic energy and gravitational potential energy. We ignore any effects of weather and assume the only things contributing to the ball's motion is the golfer and gravity ... 1 You are correct. Each cylinder has a hydraulic force applied at the bottom that is sufficient to accelerate its mass upward. The one with the lighter mass will accelerate more rapidly and reach the top of travel more quickly, but the other will already be moving. 0 A note about the statement that a maximal gravitational force would occur at r=0: we can at least exclude the case for two distinct fermions with identical quantum numbers , by the Pauli Exclusion Principle. 1 Assume f friction is being applied in direction of F. Direction, now, has no significance as f will come out to be negative if it is opposite direction.$$F+f=maFR-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$$You can ... 0 Some hints to get you started: First, consider a Galilean referential in which the problem is simpler and assume some fixed v_P, without taking care of its magnitude. Second, use what you know of the distance between a line and a point (projection). Third, minimize the distance between the line and point wrt v_P under the contraint that ... 0 The mass M in the equation for orbital velocity would be equal to the mass of the Earth (assuming that the mass of the satellite can be neglected) and thus constant. The gravitational constant, G, like the name suggests is constant as well. But the radius of the (circular) orbit, r, can vary. So by using different altitudes at which the satellite would ... 2 As the formula you wrote suggests, it depends on r, the distance between the satellite and the centre of mass. 3.1\:\mathrm{km/s} is for geostationary orbit, and 8000\:\mathrm{m/s} = 8\:\mathrm{km/s} is for low Earth orbit. Please see Wikipedia for more details. 1 Let the outside pressure be P_0. You can find its variation with altitude here. Using Bernoulli Equation, speed of liquid coming out is :$$\rho gh=\frac{1}{2}\rho v^2+P_0v(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 rvv=\frac{F}{6\pi\eta ...

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If you apply a force on the box, and see no acceleration, then the force you apply is equal to the friction force. Friction is velocity dependent, you cannot say "the friction force is so much" independently of the force you are applying.

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What I understand from the qualitative statement is that of all possible laws that acceleration could "obey," it actually obeys a 2nd ODE! It is irrelevant, whether or not, other functions also obey the same law.

2

By Newton's second law of motion, if there is a nonzero net force there is an acceleration. If there is no acceleration then the net force is zero. In the situation you describe, where the box has no acceleration, there must be another force balancing $F_{app}$ otherwise there will be an acceleration.

0

The problem is in assuming the force is forwards at a contact point right underneath the axis of the wheel, together with the no-slip condition. This leads to zero friction force, since their is no friction when their is no slipping. A solution can only be found when deformation of the wheel is taken into account, allowing the force to act on a different ...

0

This is what the situation looks like: The two radio transmitters behave like a Young's slits experiment, so they will produce a diffraction pattern in the plane of the car's motion. As the car drives north it drives through the diffraction pattern. You have to work out what the diffraction pattern will look like given the separation of the transmitters, ...

0

What I was doing wrong was putting manufacturer-provided motor torque directly into the formulas; actually the overall gear ratio must be taken into account, and it results from data taken around on internet that for electric vehicles it is =~8 (dimensionless). Hence the proper expression to use for $v_f tanh(\frac{F}{mv_f} t)$ is not $v_f ... 1 The problem in yours is that you are taking the net force acting downward to be$(m_2+m_3)g$is incorrect and that led you to take the total mass to be$m_1+m_2+m_3$which is again incorrect because$m_2\neq m_3$. If$m_2=m_3$then the center of mass of$m_2$and$m_3$will lie on the straight vertical line through the center of the pulley B and the force ... 1 The answer to this question depends on whether you're working "in real life" or on a physics problem. On a physics problem, the object doesn't start to move. The net force is exactly 0 at this time, so there is no acceleration. In reality, however, the object would possibly move. There's a variety of reasons for this: The normal force, static ... 1 The equation for gravitational force F=Gm1m2/r^2 gives the force of attraction b/n any 2 bodies with point mass m1 and m2 and separated by a distance 'r'..it means both the objects are attracted towards each other by a force F=Gm1m2/r^2..It is also coherent with newtons 3rd law i.e action and reaction forces are equal.. The expression F=ma or a=F/m ... 0 Yes, this is all correct so far. What you need to remember here is that Force is a vector quantity. That is, it has a direction associated with it. A force pushing you into the ground is the not same as one pushing you up into the sky, like the seat of a flying airplane. So here you need to lable your force$\vec{F}_1$, say, and this would be the force ... 1 If$a$is the acceleration of object 1 (should write as$a_1$), then$m$should be$m_1$. Vice versa. 2 The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of$m_2$and$m_3$(separate from the total system) is given by $$a=\frac{m_3-m_2}{m_2+m_3}g\tag{1}$$ Mass$m_2$is accelerating upwards, hence the acceleration in your case of$a_0-a$; likewise mass$m_3$is ... 0 If it is the mechanical damage just after impact that is of interest, and not the recovery, you are interested in what is felt locally at the scale of a single cell e.g. Then the quantities you may want to calculate are also local: e.g., the energy dissipated in the tissue per unit volume. The energy dissipated in the sample is the kinetic energy of the ... 1 This is because it is assumed that the test charge does not produce any electric field of its own and its magnitude is negligibly small, so it doesn't apply any force on the test charge. 1 First of all, I would like to say that the answer Floris gave is the correct way to do the problem you've set forward, but I thought it worthwhile to note that the result${5 \over 2}(R-r)$is the answer if, rather than rolling a marble down the track, you are sliding a cube. If this problem comes from a textbook, there were some unstated assumptions in ... 3 Your expression for the velocity looks right; but we have to get a few other things taken care of. First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression. On the other hand, you need to take account of the energy of the sphere ... 1 The question seems to be wrong. If they want to know tangential acceleration, they should have given angular acceleration. From given things , 'r' , 'v', and "mu" we can only find centripetal acceleration. And as you said you are getting answer 5 m/s². How it is possible? 0 Sit in the frame of car if you are having problems. Apply tangential and centrifugal pseudo forces. As we are at rest, friction has to act of same magnitude of their resultant and in opposite direction. The answer will be$4ms^{-2}0.5 \times 10=\sqrt{3^2+a_t^2}a_t=4ms^{-2}$0 This is what I did and I think is simple and right : Assume$v$linear speed of centre of mass downwards and$\omega$angular speed around it. Use the fact the bottom point has no vertical speed to find relation between$v$and$\omega\$. And I am done.

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