# Tag Info

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 ...

17

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 ...

0

There are several issues occurring here- none of which require a "fictitious force" to resolve. 1) the acceleration on the shot cannot operate in the direction of it's velocity as the shot in the sling maintains a constant distance from the slinger while it is being "slung". 2) there is no "centrifugal" force on the shot... the shot attempts to follow ...

1

I think your confusion with the slingshot is this: when you move your hand in circles to keep the slingshot moving in a circle you need some force, so you feel the weight of the projectile. But you where taught that according to Newton, things want to keep going in a straight line and inertia is this tendency to "not want to move or change direction". So ...

1

Okay, let's lay some ground rules: An object that does not experience any force will fly in a straight line A force applied to an object will change its momentum toward the direction of the force. Now, the trick with circular motion is that both the direction of motion and the direction of the force change simultaneousely, such that the inward cetripetal ...

0

Momentum is a vector, and the job of a force is to change it, in magnitude and direction. The ball has tangential momentum at any point, and when the force is perpendicular only the direction of momentum is affected. You can think of the force as a reaction force needed to keep the ball in a circle, and since it is perpendicular to the direction of motion, ...

1

The direction of acceleration is the same direction for the propagation of inertia usually! This is where you are wrong. Direction of propagation is the direction in which the body is currently moving or rather, changing position. So it is the direction of infinitesimal displacement $d\vec x$ at that instant. Now what makes you think that direction of ...

1

Fictitious forces naturally arise in non-inertial (accelerating) reference frames and you have to be careful with them. In this example, it only leads to confusion. $F = ma$ tells us that when the (total) force is zero, the object will continue in a straight line. It's not the centripetal force, but the absence of a centripetal force that makes the object ...

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