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0

The difference is fairly negligible for most purposes, but on the ISS astronauts experience micro-gravity rather than true weightlesness as they are in a sufficiently low altitude that a thin atmosphere causes drag on the station, slowing it down and causing it to lose height. They have to boost the station back up every few weeks. See ...


0

The main question has already been answered, however to cover some of the tangents raised - and give an example where two objects in orbit may not be in free fall relative to each other - there is an excellent short story by Larry Niven that actually addresses the problem of orbiting bodies and what happens to a person inside a ship as it approaches a very ...


2

Paraphrasing Douglas Adams, "Flying is learning how to throw yourself at the ground and miss". This is pretty much how orbits work - you're actually falling all the time, but missing the body you're orbitting. The reason you're experiencing apparent weight-lessness is because every piece of your body is accelerated the same amount (if you could achieve the ...


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These are two different effects. Satellites don't fall down because they are moving on a circular orbit. Actually, they are falling down all the time, since circular motion is accelerated (though the velocity doesn't change absolute value, it changes direction!), so it is kind of "falling around the earth". The second question is, why doesn't an astronaut ...


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? =)


30

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 ...


9

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 ...


39

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 ...


1

As you move from an inertial frame to one non-inertial frame and then to another, you need to invoke various fictitious forces to continue to misapply Newton's Laws to situations where you can't use them. For example, a passenger in a car going around a curve to the left may choose to consider himself as a frame of reference, and explains his constant zero ...


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 ...


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 ...


1

Actually I think "Meaningful" or "Physically Meaningful" in many cases is as good as you're going to get, although the word can split up into finer meanings. If we think of mathematics as a language, then think of words that describe how well the description meets its intended purpose. Does the mathematical description evoke the "right" ideas? So words you ...


1

1 - It is false! If $E = mc^{2}$ is true only for an object that isn’t moving, the mass never changes (is a "Lorentz invariant"). 2 - Can you rephrase it, please? 3 - Energy and mass are not at all the same thing; an object’s energy can change when its motion changes, but its mass remains the same. 4 - In Special Relativity, time can be variable, its ...


2

Critical mass is actually more about 'the right number of nuclei in a specified space'. As we are talking about solid matter this equivalently translates to a given number of atoms (or molecules depending on your matter). And this furthermore translates to our everyday mass. But it's 'not about mass', it is just a practical way to specify the quantity. So ...


1

The bomb doesn't "care" what its mass might look like to an observer in another frame. If you calculate critical mass you don't worry how big it might seem to observers located in billions of other possible frames of reference. Local frame of reference is the only one valid for making calculations concerning the occurence of local phenomena.


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 ...


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 ...


1

While the other answer are all completely correct, I just want to write a more simplified answer. It's much the same as distances. I you walk 1 meter North and 1 meter East, you can add the two distance vectors and get $\sqrt2$m North-East: $$\vec{d}_1=1m[N]=(1,0),~~\vec d_2=1m[E]=(0,1)$$ $$\vec d=\vec d_1+\vec d_2=(1,1)=1m[N]+1m[E]=\sqrt2m[NE]$$ Adding ...


2

This is due to the superposition principle: when several forces act upon a body, the net force is the sum of the individual forces: $$\vec F_{net} = \sum \vec F_i $$ However, this is only true when the relation between the force and the acceleration is linear. Let's take the gravitational force as an example: say you have three bodies and you have already ...


1

So, you are correct, that it is not quite as simple as that. Have you studied Special Relativity (SR) yet? One of the most fundamental ideas is the limiting speed of light. Nothing moves faster than c, and this means that velocities must add differently. Think of the reverse, if you were on the rocket-ship traveling at $0.5c$ and shot a laser-beam at $c$, ...


0

Let's focus on radiating EM waves first and forget about energy. When you jump into accelerating train and see charge accelerate away from you, this is all in a non-inertial frame. In this frame, electromagnetic theory has to be formulated with modified equations and new appropriate boundary conditions. That being said, nothing forbids static field in ...


0

Einstein says: "If, relative to K, K' is a uniformly moving co-ordinate system devoid of rotation, then natural phenomena run their course with respect to K' according to exactly the same general laws as with respect to K." Apparently, "uniform movement", i.e. one without accelerations, is the assumption behind the whole SR Theory that allows us to compare ...


0

Conservation of energy refers to systems looked from the same reference frame, it does not make sense to require that energy of the same system to be the same in different reference frames. As a consequence of time translational symmetry, energy conservation is usually true unless we drive the system externally which may break this symmetry. Similarly, ...


0

Depending on the magnitudes of the separate constant accelerations of the two objects, depending on the angle between their trajectories and depending on their "initial configuration" (initial separation and initial speeds, as determined by members of one suitable inertial system) there are indeed qualitatively distinct outcomes how such two object would ...


0

No. Due to the phenomenon of space expanding, the light from a point 10 light years away will actually take more than 10 years to get here. Therefore, you will actually see the object as it was much longer than 10 years ago. So if the star exploded as a nebula 10 light years away right now, you would see it not after 10 years, but a little after 10 years. ...


-1

The expression "Total accelration" does not fit if the accelrations have different directions. The vector resultant is actually the "net accelration", or the combined effect of these two accelrations, or equivalently, forces. The vector resultant makes sure that only the effective components are added, and the opposing effects cancel out. Maybe an example ...


4

It makes no sense for a point mass to have 2 accelerations. What you might have done is find accelerations due to 2 forces separately. You can add them as when $m= \text{constant}$, $\vec{F}=\vec{F_1}+\vec{F_2}=m(\vec{a_1}+\vec{a_2})$ When using vectors symbol, its automatically takes care of their directions.


0

Consider a lift with its rope snapped. The lift would be falling freely. An observer is inside the lift (tough luck for him!) releasing the ball just at the moment of the free fall. Since the ball and lift would be falling freely the ball would appear to float. Thus, to the observer in the lift, it would seem as if no force is acting upon the ball, using ...


0

Here, because the coin is placed at the center, the centrifugal forces balance each other. Every point mass in the coin has it's conjugate point at the diameter passing through it and on the same distance from the center on the other side. Hence the coin is under equilibrium and does not fly off.


0

The coin will not move. First, to differentiate between centrifugal and centripetal, I'll start by stating the definitions first. Centrifugal force is the apparent force that draws a rotating body away from the center of rotation. It is caused by the inertia of the body as the body's path is continually redirected. Centripetal force is a force that makes ...


0

No. Speed of light in Vacuum isn't dependent on Solar System's motion. It's a constant. It'd be same even if the motion wasn't there. Due to your question type problems, we've even calibrated our scales to create Relativistic Physics.


4

No the speed of light in vaccuum is an absolute constant $c$ = 299 792 458 m/s The way to add up relativistic speeds is: $u' = \frac{u-v}{1-\frac{uv}{c^2}}$ to account for the constancy of the speed of light You cannot simply add them up. Edit: This also applies to normal everyday speeds. The reason we don't use this formula is because the speeds we are ...


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Calculating the effect of acceleration in special relativity is straightforward, but I suspect the algebra is a bit much at high school level. See John Baez's article on the Relativistic Rocket for a summary, or see Chapter 6 of Gravitation by Misner, Thorne and Wheeler for a more detailed analysis. When you're first introduced to SR you tend to be told ...


3

If you are the observer and you are observing a car move in front of you, then the clock in the car will appear to move slower than the clock in your reference frame and the length of the car will appear to be contracted to you in the direction of motion. All these effects get magnified as the velocity of the car approaches that of light. In the same way, ...


5

Everything remains relativistic. However, the variation between the Newtonian and Einstein relativity is of the order of $v^2/c^2$. So when this number drops below observable error, you can't tell the difference.


0

The direction of the inertia propagation is the direction of the velocity. Any acceleration can be applied to change the velocity. If the acceleration has the same direction of the velocity you will change only its modulo, but if you want to change the direction of the velocity (which is your "line of inertia") you need "a force that push from the side" or, ...


0

The centripetal force can actually be measured. If you take the slingshot as an example, while rotating the end-mass you can measure a tension in the stings of the slingshot. If you stop the motion of the end-mass at a certain point in time, you can observe a velocity of the mass, that is tangential to the circular path it is taking over time. The strings ...


0

A particle does not have to move in the direction of the acceleration. Acceleration is change in velocity, so the change in velocity is in the direction of acceleration. Velocity, being a vector, obeys certain Laws of vector addition(see traingle law and parallelogram law of vector addition). For example, if two forces(another vector) of equal magnitude are ...


0

If you leave earth at age 30 and live for 30 years near a black hole, time will pass more slowly for you and normal for the people on the earth. I don't know the exact age you will be when you spend 30 years near a black hole, but when you return to earth you will be younger than others who had the same age as yours when you left. In other words, others at ...



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