New answers tagged

1

You won't be able to find the potential because of the presence of drag and the forcing term. In other words, this is a non-conservative field and therefore it is impossible to define a potential. Luckily, though, you don't need that because you have an explicit expression of the acceleration and that's all you need to use the Verlet or velocity Verlet ...


0

A force $F$ compresses the springs by an amount $X$ where $F = (k_1+k_2)X$ and the same force compresses spring $1$ by an amount $x$ where $F = k_1 x$.


2

One more nasty factor: What is the expansion speed of your propellant. Take the Jules Verne approach and your spacecraft falls far short no matter how much powder you put in the gun because the expansion velocity is too low. Your craft will never exceed the expansion velocity of the propellant. Note, however, that you don't have to use explosives (or ...


1

If the planet was at a much greater distance from the binary than the distance between the stars themselves (say 10 times) its motion would would be similar to a Keplerian orbit around a star equal to the combined mass of the stars and situated at their centre of mass. It has been shown that such an orbit is stable indefinitely. There would be very small ...


1

Find the center of mass (CM) of the two connected particles. Then, determine the distance (r) from the center of mass of the part of the spring where the third particle hit it. Then, if you could allow a collision time and force while in contact, and assume the collision to be frictionless (this is might be hard for point particles and thin spring, cause ...


0

The terminology you use seems a little loose, air resistance, a type of friction, is a force acting opposite to the relative motion of any object moving with respect to a surrounding fluid, so it is not measured in mph but in force units. But to answer your question, when the airflow leaves the amplifier it produces a force or a change in momentum onto ...


3

The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it ...


0

I want to focus on one thing, here. The nature of the normal force. You write Doesn't normal force oppose any other force? which is a easy impression to get when you are introduced to the normal force in the context of things sitting on other things in a gravitational field, but that's not the best way to think about it. The normal force keeps ...


0

Your method is correct. However, since the observer is outside the lift. The velocity of the lift must be taken into account as well. v(lift) can be found by kinematic equations. And, velocity of the bolt would be its velocity(one that you found out)-velocity of the lift after it has reached a height h.


0

The compression pulse that propagates through the metal spheres of Newton's cradle are not ordinary sound waves. They are approximate solitons (a nonlinear wave form that balances dispersion against nonlinearity). It is this property of soliton pulses that is responsible for the observed behavior. More Details: Newton's cradle is a physical manifestation ...


1

The simple answer is that not only momentum but also energy needs to be conserved. This puts constraints on the number of balls that can be activated in the cradle. Note that this does not always give a unique solution either. But it enforces that $ n $ balls to $ n $ balls is a "stable" solution.


1

Wanted to give a different angle to thinking about this. Since you mentioned you wanted to consider this from the inertial frame (non rotating), then there is indeed no (fictitious) force. But in that frame, we can consider the water is indeed falling; however, the rate at which the bucket is also "falling" is such that the two stay together - in other ...


1

In the bucket experiment when the bucket reaches the top of the circle why will it have a normal force acting on the water downwards? The normal contact force is exerted by the bottom of the bucket as explicitly mentioned by the author. So, it is acting downward when the bucket is inverted. Doesn't normal force oppose any other force? There is no ...


1

Imagine a scenario where the bucket is rotated at just the right speed so that the centripetal acceleration required to keep the water on a circular path is exactly 9.81 $ms^{-1}$. Then at the top of the rotation, all the centripetal acceleration is supplied by gravity. However the bucket might be rotating faster in any given scenario but it still rotates ...


0

A simplified answer is obtained by ignoring all possible effects and dependencies other that the mass of the earth (M), its angular velocity ($w$), the mass of the impacting object($m$,) and its velocity ($v$). The rotational energy of the earth is $I w^2/2$, the kinetic energy of the object is $mv^2/2$. To stop the rotation and then make it go in ...


46

Other answers don't mention the fact that no single impulse (e.g, like being fired from a gun) can launch a projectile into orbit. A purely ballistic projectile fired from a gun must either crash back into the planet, or it must escape from the planet altogether. In order to achieve orbit, at least two impulses must be applied to the projectile. The first ...


0

we can understand simply how the time period of a pendulum increase and decrease in an elevator. 1- when we go downward in an elevator(downwards accelerated elevator ) . we feel defect in our wight. we know wight $W=mg$. $W$ our wight is decreasing as elevator going downward . our mass is constant . so $g$ acceleration due to gravity is deceasing . ...


1

In figure A the satellite is orbiting in a circular orbit. By the satellite increasing speed it would turn the orbit into a more egg shaped orbit as shown in B. This would rather increase the time for orbit revolution rather than decreasing it When the satellite is approaching B.1 it's velocity will begin to slow down as earths gravity begins to pull it. ...


10

I think the heart of the question is whether one could arrange a continuous combustion of propellant along the length of the barrel. In that way the acceleration occurs along the length of the barrel in a more gentle way. Since the expanding gases from the propellant in a shell casing expand and the pressure of the expanding gases declines along the way it ...


14

If there is no atmosphere, and the station is a relatively smooth cylinder, you can indeed float there as the exterior walls spin around you (in the middle, or just above a wall, or anywhere). Now, suppose you start drifting towards a wall (maybe you threw your shoe the other way). You move towards the wall, but do not accelerate due to the rotation of the ...


27

Anything launched into orbit by such a gun needs to travel at orbital velocity (in fact above orbital velocity) in the lower atmosphere. That's generally undesirable, to put it mildly: there will be really serious heating.


0

Let's say you have got such gun. Next logical step will be to install on the satellite a smaller gun that would shot-back several small shells and so accelerate the satellite, indeed? If this small on-board gun would use really many small shells (size of molecula) then your are getting just a traditional rocket. Apparently it is not much difference for ...


15

Aside from the interior ballistic aspects of these various projects, it was quickly realized that any satellites launched by gun would have to withstand high g-loadings during firing of the gun and the size and mass of the satellite would be greatly constrained by the dimensions of the bore of the gun and the maximum impulse which could be provided by the ...


29

You are correct in that if the astronaut is undergoing no translational or rotational motion relative to the centre of rotation of the space station the astronaut will feel weightless as in diagram $A$ and will not touch the space station. This is equivalent to jumping onto a rotating turntable with no friction acting. That feeling of weightlessness is due ...


43

Put a stationary astronaut in a small room inside a large spinning cylinder. After an instant walls of that room will hit him, and suddenly he will have the same velocity as the room. Due to angular motion, the room accelerates towards the axis of the cylinder. Subsequently, through the support force from the floor (the floor is at the surface of the ...


7

All you written is correct. Go further to the next point 4: once he comes to the cylinder wall and stands on it, he will get same angular speed as the cylinder, then he will also get centrifugal force and rotational gravity as in a film.


0

I'm not quite sure how he gets a value of $36.36$ for that constant, because it depends a bit on exactly how he does his approximations and what values he assumes for different constants. But basically it's this: For a spherically symmetric mass distribution like the one considered here, the gravitational acceleration is simply, from Newton's law for the ...


0

Anna, you are more or less right. It is pressure from matter at higher densities that can stop the gravitational collapse. It depends on the state of the matter, and in a simplistic description the equation of state. As it collapses as a hot gas after it exhausts it nuclear fuel, and maybe after a supernova explosion, it'll collapse. The first point at ...


4

There are already good answers here, but I'm afraid that to the best of my knowledge, Diracology's (and indeed Halliday-Resnik-Krane's) expression of the potential energy is not correct. I would like to point to this paper by Lior M. Burko which focusses on the subtleties of the derivation of the kinetic and potential energy of the string as a whole and ...


0

The definition of force is $F = m a = m \frac{dv}{dt}$. The average force is $<F> = \frac{1}{\Delta t} \int_0^{\Delta t} F(t) dt$. Write $F= m \frac{d v}{d t}$ to find $<F> = \frac{1}{\Delta t} \times$ momentum change.


1

If you see on the cylinder as a wheel then note that its center moves twice slower than top point attached to the block. Same is for acceleration. In each moment the cylinder rotates around the point of its touch to the table, so radius from touch-point to the center is twice less than radius to the cylinder top point.


0

It's because of the product rule of derivatives, that states d(fg(t))/dt=f(t)(dg/dt)+g(t)(df/dt). In this case, let's call this part: (-isinθ+jcosθ)=g(t), and solve it: d(rωg)/dt=r(dωg/dt)=r(ω*(dg/dt)+g(dω/dt)). If you substitute g for (-i*sinθ+jcosθ), you get the equation for acceleration your book presents.


1

The electron energy (the portion that changes at least) will be mainly due to kinetic energy (translational) and potential energy due to the potential difference between the cathode and the anode. The electron does have a "spin", but this spin isn't like that of a spinning sphere. The reason for the name spin is simply that the electron spin describes the ...


0

To most likely win tug of war,put the heavier people in the back because when they lean back,the rope moves further back.Now if you have someone that is only like 4 feet 10 inches and they are strong,still put them in the front.Heaviest to lightest would be good.Also putting the non heavy people but strong in the front is good to.


0

The answer should be d). For Simple harmonic motion, the period (and frequency) are independent of the amplitude and the initial phase of the motion. Note: the period of oscillation IS the inverse of the frequency


2

Using Torricelli's law (https://en.wikipedia.org/wiki/Torricelli%27s_law) you get $v=\sqrt{2gh}$ irregardless of how big the opening is. Now you can calculate how much mass is leaving the tank at any time by multiplying the volume that is leaving the tank by its density: $$\Delta m=A\Delta s\cdot\rho$$ with $\Delta m$ the mass leaving the tank at any second, ...


0

If by "solve" you mean to get an answer that is independent of θ, the answer is... Nope. You need θ, it doesn't cancel out. However, here's a slightly similar problem that you CAN solve: You've got a similar hillside, and a sled at the top of it. The height of the hillside is h, the friction coefficient is μ, and the angle is θ. You let the sled slide ...


1

For a particular setup, the equations may get very simple: the tank should be massless ($M=0$) and the hole is all the way at the bottom of the container. Then $h$ is proportional to the mass of the water in the container: $m=m_0 h/h_0$, with $h_0$ the initial height. It's straightforward to derive that the force generated by the water jet is $F=2\rho g h ...


1

The Pistons apply the force on one side of the cylinder wall. Which side depends on the rotational direction of that particular engine.


1

Where is the spring balance here? It's not present. It's a statement that anywhere you find a force, you can put a spring balance along with it to measure the magnitude (at least conceptually). In this case, we might imagine that there is a solid radius present (like a rod), and a spring balance attached to the radius, free to move radially, but ...


1

Here's the illustration that I would use1: The blue spheres represent blobs of material, where a blob could be any number of atoms or molecules. The red lines represent the bonding force between the blobs. Young's modulus $E$ represents the strength of the bonding force, i.e. the strength of exactly one bond. When you stretch the material, all of the ...


4

Multiply both sides of the equation by $\Delta l$ to obtain: $$k\Delta l=EA\frac{\Delta l}{l_0}$$ The relationship on the left is the tensile force F. The quantity $\frac{\Delta l}{l_0}$ is the tensile strain. The Young's modulus E times the tensile strain is equal to the tensile stress $\sigma$:$$\sigma=E\frac{\Delta l}{l_0}$$ The tensile stress times ...


2

Just to more fully answer your question, here is an example of what differences in distances can mean mean as far as they affect gravitational forces. The planet Jupiter is extremely massive, and one side of one of it's moons, Io, feels a slightly larger gravitational pull than the opposite side. This difference in distance results in a gravitational force ...


1

If the strings have length d=0.08 m, then the distance from the center to each mass is $r=d/\sqrt{2}=0.056$ m, which you have noted yourself. The centripetal force is $F_c=m\omega^2r=0.678$ N. But the centripetal force vector is the sum of the tensile force vectors of the two strings, each of which is at 45 deg. Therefore, you have to divide $F_c$ by $\sqrt ...


4

That formula holds for a simple pendulum of length $L$ in a gravitational field $g$ and released from rest with an amplitude $\theta_0$. Since this system is conservative its mechanical energy is constant and equals the gravitational potential energy when it is released. Setting the zero of potential energy at the fixed point of the pendulum, the mechanical ...


4

Yes: I've done that. I used to have a device for the purpose, commonly called "rollers". It's like a treadmill for bicycles.


1

Now the real problem I'm having is trying to decide what the forces are acting on the system in order to come up with my differential equation? As there are no external forces the linear momentum of the system will be constant of motion and the constant can be taken as zero as well. If m1 and m2 are the masses at any time described by position ...


1

To do the equations of motion you need positions from an inertial reference frame. Say a wall far away. Call positions of the two masses $x_1$ and $x_2$. The spring force (tension is positive) is $$ F = k (x_2-x_1 ) $$ and the two equations of motion $$\begin{align} F & = m_1 \ddot{x}_1 \\ -F & = m_2 \ddot{x}_2 \end{align} $$ All this is trivial. ...


0

Lower mass means lower weight, faster acceleration, and faster speed. You didn't understand the problem correctly. A bike moving in a straight line is kept in an unstable equilibrium. The greater the weight, the less stable the equilibrium.


2

Yes. All objects are gravitationally attracted to one another. Even people. Therefore, the earth will draw you to itself no matter how far out you are.



Top 50 recent answers are included