Hot answers tagged

19

Physics is the discipline that studies natural phenomena, and finds mathematical models that fit the measurements and observations and also predict future behavior of the system under study. Mathematics is a discipline which studies numbers with sophisiticated methods, it has axioms and theorems and can prove statements or disprove them absolutely. To use ...


16

One approach to exploring this question is to study the Glider in Conway's Game of Life. Where does its movement "come from"? It is a direct and easily-verified result of the rules of its universe. I highly recommend you take a few minutes with graph paper to verify its behaviour for yourself. What is really important to see is that motion is not a law ...


5

Yes, this statement is true, in the sense that the four-velocity $u^\mu = (\gamma, \gamma \vec{v})$ always satisfies $$u^\mu u_\mu = 1$$ as you can check using the definition of $\gamma$. (I'm setting $c=1$.) Therefore the magnitude of the four-velocity is always equal to the speed of light. However, this statement can be really misleading. It's true that ...


4

"What is the difference between these two points (centers of cylinders)? " 1) there is no difference in the motion of the points (lines that extend down the cylinder, actually). They each translate in space with the center of the cylinder as you might expect. 2) "Can we define rotation for a point about an axis that crosses that point?" I assume your ...


4

This is the setup described in the equation: The acceleration is defined in terms os the displacement of the bow $x$ by: $$ a = 6000 \left(1 - \tfrac{4}{3}x\right) \tag{1} $$ So initially $x=0$ and when we substitute this into equation (1) we get $a = 6000 \text{ms}^{-2}$. When the arrow leaves the bow so $x=\tfrac{3}{4}$ and we get $a=0$. So far so ...


3

Think about this from the perspective of a person in the elevator. No windows, they can't look outside. As far as they are concerned, they live on a small box-like planet where the acceleration due to gravity is 9.8 + 1.2 = 11 m/s$^2$. In a system where the acceleration due to gravity appears to be 11 m/s$^2$, a bolt drops 2.7 m. How long does it take to ...


3

Because the springs are considered massless. So if there were a force difference between the ends, you would get infinite acceleration. And this is valid not only for the ends, but for any two points, if there is no mass between them. The issue is somehow explained in this post: Is the tension in both ends the same (on a massed string)? And indeed, if the ...


3

The velocity does not double if the acceleration is doubled. The relevant SUVAT equation is: $$ v^2 = u^2 + 2as $$ where in this case $u=0$ so we get: $$ v = \sqrt{2as} $$ A doubling of acceleration means that the velocity would double if the travel time was kept constant. However in this case it's the travel distance that is held constant. The greater ...


2

I think you are asking why this argument remains valid when the mass is oscillating, and does not only apply when it is static. I think the answer to this is that the forces in the springs depend only on their extensions (F=kx), not how quickly the extension is changing (F=kx+bx'+cx"). So at any given extension x the tension is the same regardless of the ...


2

It would still be here, it would just be turned inside out.


2

The equation of motion of the particle is $$m \ddot{x}(t) = F(t)$$ where $x(t)$ is the position and $F(t)$ is the force. In the situation you describe, ("suddenly I hit the particle"), the force as a function of time can be written as $F(t) \propto \delta (t)$, with $\delta$ the Dirac distribution. Integrating once, you obtain that $$\dot{x}(t) \propto ...


2

If your textbook actually derives $(2)$ as the motion of a thrown object, throw it away. The general trajectory of an object thrown from $(0,0)$ at angle $\theta$ is $$ y(x) = x\tan(\theta) - \frac{gx^2}{2v^2}(1+\tan(\theta)^2)\tag{1}$$ and now you say you "impose $\Delta = 0$". Let's analyse that "imposing" a bit more carefully, it is equivalent to $$ v^4 ...


2

Friction acts to oppose the two surfaces in contact from sliding past each other. It will be no greater than necessary to do that.


2

Differentiate $v$ with respect to time. On the left hand side you will get the acceleration $a$. On the right hand side you will get an expression which includes $s$ and $\frac {ds}{dt}$ which is $v$ which is ?


1

I think that what you are asking is the position, velocity and acceleration of a space probe relative to the ever changing positions of the components which make up the Solar system? In the article cited below it describes the coordinate system as follows "Calculation of the trajectory of a space probe requires the use of an inertial coordinate system as ...


1

You have $$J = \frac{{\rm d}a}{{\rm d}t} = \frac{{\rm d}a}{{\rm d}v} \frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}a}{{\rm d}v} a$$ With $v_1$ and $a_1$ as initial conditions then: $$\left. \int J {\rm d}v = \int a {\rm d}a = \frac{1}{2} a^2 + C \\ J (v-v_1) = \frac{1}{2} \left( a^2 - a_1^2 \right) \right\} a = \sqrt{a_1^2+2 J (v-v_1)}$$ Position is found ...


1

Let me rephrase your first part, with constant $a$. The standard solutions, involving time, are: $$a(t)=a\\ v(t)=a t +v_0\\ x(t)=\frac{1}{2}a t^2+v_0 t+x_0$$ From this, it is clear how to get your solution. Eliminate $t$ from the last 2 equations, e.g. by completion of squares: $$a x(t)=\frac{1}{2}[a^2 t^2+2 a v_0 t+v_0^2]+[a ...


1

Rotations for infinitesimal points are not defined. A rotation is only defined when you have two or more points as a way to describe the fact that their relative distance remains constant. Also rotational motion is shared for an entire body, meaning that all point on a body rotate the same. The idea of location for rotation only enters when linear velocity ...


1

@ Nick Basically you are confusing two different coordinate systems. In the coordinate system where you decomposed g into two components, there the question of x and y doesn't seem valid because you have transformed the cartesian coordinate system into a different one. You would have new x and new y there and you have to stick with one while solving the ...


1

Since there is no time dependence in the given equation (in other words, you have only the trajectory in terms of some parameter -- $\theta$ -- with no given connection to the time), you cannot arrive at the velocity by differentiating. Of course $k = \cot\alpha$ for a well-chosen $\alpha$ ... ($\alpha = \arctan(k^{-1})$) :) But why does this prove ...


1

As you said, the next derivative of the velocity with respect to time is the acceleration. And the acceleration could in principle have a step somewhere due to a force starting to act on the object.


1

Yes, but by definition. Not by any meaningful physics. Imagine a path through 3-space. You can define the path by a function of time that returns a position. ${\bf f}(t)=(x(t),y(t),z(t))$. Then the velocity as a function of time is ${\bf v}(t)={\bf f}'(t)$. Easy. You could do the same thing through 4-space, by describing a path parameterized by some other ...


1

the total change in momentum is given by the integral of the force in time: $\Delta p = \int _{t_i} ^{t_f} F(t') dt'$ This is the general formula. For simple forces, this can be integrated by hand to get an analytic result, but if the force depends sensitively on the position or velocity, numerical integration may be the easiest route.



Only top voted, non community-wiki answers of a minimum length are eligible