# Tag Info

-1

There is only one degree of freedom (the distance between the masses) not two, and only one natural frequency (given by your 1st eqn). Since there are no external forces, the CM of the system does not move. (Sorry but I do not see what all the fuss is about here. Why make such a simple problem so complicated?)

0

Suppose I have two masses m1, m2 connected by one spring of stiffness k The Lagrangian of the system is $$L = \frac{1}{2}m_1\dot q_1^2 + \frac{1}{2}m_2\dot q_2^2 - \frac{1}{2}k(q_1 - q_2)^2$$ where $q_1$ and $q_2$ are the coordinates of $m_1$ and $m_2$ respectively. Now, consider a change of coordinates to the normal coordinates $Q_1$ and $Q_2$ ...

0

Neglecting air resistance, and until any of the fragments of the bomb reach the ground, the centre of mass of the bomb follows the same trajectory as it would if the bomb did not explode - ie part of a parabola. Conservation of linear momentum does not apply here because there is an outside force (gravity) which changes the magnitude and direction of the ...

0

It will stay the same, if we neglect the variation due to gravity (every external force is going to change the momentum). If we assume a uniform distribution of the shrapnels' mass (same size for all shrapnels), the shrapnels going in the direction the bomb was originally going will have, on average, higher velocity. With a great simplification, we can say ...

1

The other natural frequency is indeed zero! Natural frequencies of zero corresponds to vibrational modes of rigid body motion. Rigid body motion is not a vibrational motion in itself, but still arises in the modal analysis of certain systems such as the one above. The reason you get a rigid body mode is because you are able to move the system as a whole, and ...

0

When a top rotates, it rotates about its centre of mass. The centre of the mass is a point on the axis of rotation. Since the axis is also stationary as is the centre of mass, therefore all the points in the axis are eligible to be considered fixed about which the top is rotating. Besides,I would prefer to use the term axis instead of a fixed point.

0

Consider two objects, 1 and 2, colliding for some short time interval $\delta t$. During $\delta t$ let's ignore all forces except the contact force that 1 exerts on 2, $\vec{F}_{12}$ and the contact force that 2 exerts on 1, $\vec{F}_{21}$. As long as the objects touch each other, both of these forces exist, and by the principle of Newton's 3rd Law we know ...

1

I come to think that when a mass collides with another, both of them should always have equal velocities post-collision. To paraphrase Feynmann, no matter how beautiful you may believe your reasoning to be about this, if it doesn't agree with experiment, it is wrong. If the two objects 'stick together' after the collision (the collision is totally ...

1

Often when two object collide it is often represented as an instantaneous impulse exchange. However in reality this happens continuously. Namely both objects are not completely rigid and will deform during the collision, storing energy in the elastic deformation (like a spring) and dissipating energy with any inelastic deformation. During such a collision ...

1

What is conserved during a collision is not velocity, but momentum (mass times velocity). If the collision is elastic, kinetic energy is also conserved: https://en.wikipedia.org/wiki/Elastic_collision. If the collision is inleastic, only momentum will be conserved: https://en.wikipedia.org/wiki/Inelastic_collision The resulting equations (which you can ...

3

You should study Newtonian mechanics before Lagrangian mechanics because Newtonian mechanics is more general than Lagrangian mechanics. In other words, while whenever a system allows a Lagrangian formulation it also allows a Newtonian formulation, the converse is not true; the quintessential case is dynamics in the presence of dissipative forces. Lagrangian ...

7

No, I would highly recommend studying Newtonian mechanics before Lagrangian mechanics. While, yes it is 'possible' to learn about Lagrangian mechanics before Newtonian, a lot of intuition would be lost beginning with one instead of the other which will, in the long run, do no more than harm you or, at best, possibly confuse you. But there are, indeed, many ...

12

It is necessary to study Newtonian mechanics to truly understand Lagrangian mechanics since its underlying foundation is Newtonian mechanics. It is essentially a different formulation of the same thing. In a way when doing Lagrangian mechanics you are still doing Newtonian mechanics just in the way of energy. For example, under Lagrangian mechanics, say we ...

0

There is a demonstration in the following video : https://www.youtube.com/watch?v=jyqOTJOJSoU. It does not have the magnet resting on the Styrofoam as in your photo. But I can see it happening and it is explainable. If your magnet is resting on the Styrofoam when it moves, that is hard to believe - and I cannot see any evidence in your photos that it ...

0

At the beginning I found this question a bit naive. But now I think it is worth to think a little bit about it. They definitely share some properties, at least at the microscopic level. Both have the same microscopic origin: the electromagnetic interaction. The upthrust requires gravity to create a pressure gradient on the fluid, resulting in an upwards net ...

5

If you're thinking about stable orbiting systems the big difference between gravity and the magnetic force is that magnetic monopoles do not exist. The simplest source of a magnetic field is the magnetic dipole. By contrast gravitational monopoles exist but gravitational dipoles do not. The Sun and the Earth are both (approximately) gravitational monopoles, ...

0

What is the significance −ve sign here? And will the work done be +ve? If I want to lift some mass upwards I need to apply a force at least equal to it's weight. This means if I want to pull the mass upwards, with a certain acceleration then an additional force has to be supplied along with it's weight. But the resultant force will be the difference in ...

2

From the definition of work $$W = \int dx F$$ and $$P = \frac{dW}{dt}$$ you can see how we can arrive at $$P = F \frac{dx}{dt} = F v$$ (when considering only the absolute value). To understand it intuitively, imagine the case of a frictionless system in which the car can move at a certain speed without any opposing force. The power required to keep it at ...

0

If you take down as positive then displacement $s = +30$ m, initial velocity $v_i = +8$ ms$^{-1}$ and acceleration $a = +10$ ms$^{-2}$. Using the constant acceleration kinematic equation $s = v_i t + \frac 1 2 a t^2$ where $t$ is the time gives $$(+30) = (+8)t+\frac 1 2 (+10)t^2$$ If you take up as positive then displacement $s = -30$ m, initial velocity ...

31

John Rennie already gave the practical answer considering the atmosphere, noting that without doing anything objects near the ISS will deorbit quickly from drag. But that's letting reality get in the way of a good physics problem. I'll show that while a human can't send a ball crashing into the surface in one orbit, they can come close. The ISS is listed as ...

2

I think your problem is that you didn't change the units in the constant g. It has a value of approximately $9.8ms^{-2}$. Notice that it depends on meters. To obtain the correct result, you should use $980cms^{-2}$. Notice that this constant is off by a factor of 100, so that the result (after the square root) is off by a factor of $\sqrt{100}=10$.

1

Yes you are. If a force is conservative, its work does not depend of any path between any points $A$ and $B$. Since the work integral can depend only on the initial and final points themselves, we define $$W_{A\rightarrow B}=\int_A^B\vec F\cdot d\vec r\equiv U(A)-U(B).$$ Now define the mechanical energy as $E=K+U$ so that $$dE=dK+dU.$$ Suppose there are two ...

0

This is a standard problem in flow through porous media. The equation you are looking for is the Ergun Equation. This is found at the following site: https://en.wikipedia.org/wiki/Ergun_equation. The Ergun equation is used extensively in modeling filtration also. It takes into account the viscous drag in the bed. There is also an extended version of ...

-1

On a practical note, there was a space walk a few years back when they replaced a failed ammonia pump that was the size of a refrigerator. The astronaut simply gave it a swift push away from the station knowing that it would soon deorbit fast enough that it wouldn't be a collision hazard. PS- If you have Kerbal Space Program, this would be a fun thing to ...

1

The relation between velocity v and distance r at which a small body orbits a much larger one of mass M is given by $v^2 = GM(\frac{2}{r} - \frac{1}{a})$ where a is semi-major axis. The perihelion is $p = a(1-e)$ where e is the eccentricity, given by $e^2 = 1 - (\frac{b}{a})^2$ and b is the semi-minor axis. If you don't have values for G and M you can ...

31

You don't need to throw the ball! At the altitude of the ISS the atmosphere is thick enough that it loses 50-100m of altitude every day due to the drag. At that rate over your ten year timescale the ISS would lose 180 to 360km. When you take into account the increased drag at lower altitudes ten years is enough to bring the ISS crashing to a fiery end. So ...

3

The link given by LLlAMnYP for the Ehrenfest paradox gives the classical physics rational : Any rigid object made from real materials that is rotating with a transverse velocity close to the speed of sound in the material must exceed the point of rupture due to centrifugal force, because centrifugal pressure can not exceed the shear modulus of ...

1

To understand why holding objects costs energy even though the work appears to be zero, you have to understand how muscles work. When you are holding an object, your muscles are contracted. The process of muscle contraction consists in a protein filament called Myosin pulling another filament, called Actin. Since this is a dynamical process (the Actin ...

0

It may be 5 years on from the original question, but it's a shame to have only a single accepted answer, which is just plain wrong. Though it's true, that the human body has a "natural intuition for physics", this is only within a pretty wide margin of error (which gets narrower with practice in spear-throwing). It is not a flick of the wrist, which makes ...

1

First, there are a few different methods developed to solve this kind of problem, but it's highly dependent on your background knowledge (calculus), and experience with these kinds of problems. This approach might be over-detailed for some mechanics problems, but this approach is fairly general, so should usually work. Identify any clues in the problem as ...

0

Just the longer version of the answer @xasthor Suppose the acceleration of $m_1$ is $a_0$ towards right. That will also be the downward acceleration of the pulley B because the string connecting $m_1$ and $B$ is constant in length. Also the string connecting $m_2$ and $m_3$ has a constant length. This implies that the decrease in the seperation between ...

0

It is not clear at which level you want to simulate this. Fine grained If you want to do this on a very fine grained level, you just need so simulate Newtonian mechanics and gravity and it should emerge from itself when you have a trajectory set up. This means that you compute the gravitational force to be $$\vec F = - G \frac{Mm}{r^3} \, \vec r \,,$$ ...

3

The question is inconsistent. At least one of the numbers (mass, force, stopping distance, or stopping time) is wrong. Your calculation of the acceleration from force and mass is correct, but an acceleration of $24$ m/s$^2$ for $2$ seconds means that the toy car was initially traveling at $48$ m/s. This is over $100$ mph ($160$ kph) and there is no way the ...

0

static pressure at B can be larger than at A and can be lower. This is not used for estimating stagnation pressure change. If there is no frictional loss, the fact, that the two total pressures are the same, means the energy is conserved. When this is loss, of course energy at A is larger than that at B in order to conserve the energy.

0

As skier slides down, the normal force between skies and dome reduces. It is maximum at top but reduces further as he slides down. It is the normal force which is keeping skier adhered to dome. As the speed of skier increases, there is an increase in tangential speed which gives an increased centrifugal force. Centripetal force tends to skier adhered ...

0

You messed up on the kinematics. The key to solving a pulley problem is to get the kinematics correct. If a is the acceleration of body A to the right, what makes you think that the upward acceleration of body C is a? What makes you think that the upward acceleration of body B is a? If the downward acceleration of mass C relative to pulley c2 is a*, ...

0

In accordance with the homework policy, here are some things to think about in order to solve this. There is a force holding the skier attached to the snow globe: that is gravity. As the angle gets steeper, the component of gravity pointing towards the center of the sphere gets smaller. This force needs to be strong enough to keep the skier moving in a ...

0

There are two primary things that draw the ball into the waterfall (or push the ball upstream), and both of these are much stronger than the two effects mentioned in the question. As a physicist and a white water kayaker, I have a lot of experience with the relevant forces, and they can both be very strong and sometimes life threatening. The first force: ...

0

Phase means the stage in its cycle which a wave has reached. Peak and trough are phases of the cycle, but it is more useful mathematically to describe phase in radians, as though the wave is a point moving round a circular track - the phase is the angle which the current radius makes with the starting radius.

1

$a_0-a$ , here $a$ means the acceleration of two blocks with respect to the pulley. Hence, if $a$ > $a_0$ then also this relation perfectly works. They are assuming $m_2$ to be accelerating upwards because they have preasumed that $m_3$>$m_2$. A very simple way to assume this is that suppose the whole system was at rest. Then the mass $m_2$ would move ...

1

If you take the marked portion of the image then there is a property for pulleys in which forces get magnified. In your case, the tension on the strings that hold B and C is T By the property of magnification, The tension U becomes equal to $$U=T+T=2T$$ So the third equation becomes, $$2T-F_a=4Ma$$ Here is another example of magnification of forces ...

1

Phase is the argument of the wave. This is the definition written in my book and quite hard for beginners (like me). So the question "What is phase?" Phase is the quantity which tells us the status of the wave. In normal x-y grid like x-axis tells us the distance from the origin, in similar way you can think a phase is along x-axis and it gives ...

3

The clue here is " how far apart". The question is asking for distance which must be in terms of the wave's wavelength. Phase measures fractions of wavelength. And you are given information of the wave's speed and the periodic time in which it propagates (frequency). The fundamental "distance = rate * time" applies in terms of the wave speed, wavelength and ...

1

Short Answer - No Refer to the figure below: If you draw free body diagram of the particle at a point beyond C you will notice that there is no force acting towards the surface of sphere. Hence it won't be able to complete the circle beyond point C whatsoever.

1

Under normal conditions and size of the sphere the answer is no because the gravitational pull of the earth would attract the particle towards its centre. If the size of the sphere is bigger than the size of earth then the gravitational pull provided by the sphere will cancel out the g effect of the earth and the net gravitational pull will provide the ...

1

It depends what exactly you mean by "coordinate". If your Lagrangian/Hamiltonian is time-independent, then you may consider time to be purely a parameter parametrizing e.g. the integral curves of the vector field associated to the Hamiltonian on phase space. If your Lagrangian/Hamiltonian is time-dependent, you should indeed properly consider your theory on ...

1

I think this issue is best clarified by closely looking at the way time is mixed into coordinate frame transformations in Classical Mechanics as opposed to Relativistic Mechanics. Let's take the case of an observer, Alice, moving at velocity $v$ in the positive $x$ direction away from her friend Bob. Both Alice and Bob are looking at an object situated at ...

1

If there is no friction the energy you put in initially will be conserved and the flywheel will rotate forever, then you don't need to put in any extra power to keep it running.

0

force is defined as mass * acceleration. From the well known hyperphysics web site: Image credit The net external qualification is crucial.

1

Yes, at the fundamental level all energy terms are normally either kinetic or potential energy. The only demonstration of this that I know of requires a tool called the Lagrangian, which you might not be familiar with. But maybe you can at least get a flavor of how it goes. The Lagrangian, very briefly, is a particularly useful way to represent all the ...

Top 50 recent answers are included