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1."The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when θ=π." - Is it because if the rope slacks, then the mg component will pull it inwards and so the particle will not move as a circle? Yes. Gravity (the weight $W=mg$) is then strong enough to pull it back from the "swing". The rope is ...


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Note that it is one dimensional motion. At the end of 2 sec, a force (break) is applied, causing a change of velocity of 4 units in 2 sec which means a decceleration of 2 m/s^2. Decceleration is constant since the Vx vs t graph is a st line. You can use use the appropriate equations to find the displacement in the first sec after the application of the ...


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By simple calculus: Between $t=2s$ and $t=4s$, $v_x$ can be described by the function: $v_x=8-2t$. The definition of $v_x$ is: $v_x=\frac{dx}{dt}$, so that $dx=v_xdt$. Integrating between $t=2$ and $t=3$ we get the distance travelled in that interval: $x=\int_2^3(8-2t)dt=2m$. Add to this the $8m$ travelled in the first $2s$, so total distance is ...


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The final position will be the initial position plus the area under the velocity versus time graph. That is the area between y =0 and the velocity function. I'm assuming you're not familiar with integral calculus, but if you look at the dimensions you arrive at by calculating this area you will find that it is meters. Good luck.


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You can't "simply" calculate the average velocity from the velocity at the end points, unless the velocity graph is a straight line. Which it is between 2 and 3. But not between 0 and 3. So the approach you can take is this: What is the distance after 2 seconds of moving at 4 m/s? And what is the average velocity between t=2 and t=3 seconds (straight ...


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If you want to actually simulate the behavior of the planet as it experiences the (vector) force as it moves around, then you need to find a stepping method and write your velocity vectors and position vector in terms of coordinates. I recommend a Verlet velocity method. Others at this site have their favorites, too. Euler's method is not good enough for ...


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In the comments you have said that this question is part of a project regarding interplanetary travel; and the values you have given seem to indicate that what you're really asking is the velocity of an object that was dropped from $500 \space km$ height. If that's the case you can find it using the principle of conservation of energy. $$E = - G {M_Em \over ...


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By definition the acceleration is first derivative of the velocity over time, i.e. a(t) = dv/dt Then obviously v = ∫a(t)dt. If acceleration a is a constant then v = at (we all know that). Otherwise you need to calculate the integral.


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If you drive along the road on the side of a mountain) there are (at least) two kinds of forces on your car: friction and gravity. If you drive to the top of the mountain and back, the net work done against gravity is zero. This corresponds to the fact that your displacement is zero. On the same trip, you had to overcome friction. That depended on your ...


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Yes your reasoning is completely right but is important to add some information. The diagram you show has no information on the ball size and assumes all the mass is concentrated in the center of mass, which for a spherical and isotropic ball should be in the geometrical center. Now, at the moment of impact $t_1$, the border of the ball will start touching ...


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No, all your reasoning is totally right. The conclusion isn't that the graphs are wrong, it's that the time of impact is less than 0.1 second. In this video, for example, the time of impact is just about 0.01 seconds.


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Because for the x-component you have: $$ma_x=0\to a_x=0, $$ and for the y-component $$ ma_y=-mg \to a_y=-g. $$ In both cases the equations of motions are independent of mass. If you have a particle which seperates into smaller pieces at one point and one of those pieces follows the same trajectory as the initial part, the smaller piece follows the same ...


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At every point along the trajectory you have two velocity components $\left\{v_x(t),v_y(t)\right\}$. You have asked what it means for a projectile to retrace its path. That the original projectile broke up and/or the second part now has a smaller mass is not relevant here since this is a kinematics problem. This is simply $\left\{-v_x(t),-v_y(t)\right\}$. ...


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Since all the velocities are constant working out $\frac{dx^{1}}{dx^{0}}$ is simple division (no calculus required). Note that the entries in your vector in $S$ are the values you're interested in: $$(dx^0, dx^1, 0, 0)=(\gamma(d {x^0}' - v d {x^1}'), \gamma (d {x^1}' - v d {x^0}'), 0, 0)$$. So we have: \begin{equation} \frac{dx^{1}}{dx^{0}} = \frac{d ...


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If you have two vectors you can project one onto the other. It you orthogonally project a spacetime vector onto your unit tangent then the length of that projection is how much time you observe separating the two events. That might be enough to answer your question right there. So to get an average velocity between two events first compute the vector A ...


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I'll just answer the part of your question about terminal velocity (better to call it terminal speed...). The terminal speed of a falling object is not caused by anything special about gravity. It is caused by the fact that for an object falling straight down there are two main forces acting on it: gravity (which points down as is constant to a good ...


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I write this because it is too long for a comment Situation You are putting $k$ watts (=joules/second) of kinetic energy into a body. We will show the acceleration decreases with time. $$E=\frac{1}{2}mv^2$$ so $$v=\sqrt{\frac{2E}{m}}$$ so $$\frac{dv}{dt}=\frac{m}{\sqrt{\frac{2E}m}}\cdot\frac{de}{dt}$$ (I can't be bothered to tidy this up - remember I'm ...


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Weirdly I just wrote about this: Magnetic force storage or amplification question Anyway. For the first one! YES the equations for force due to magnetism and force due to gravity are both very similar, they take the form: $$F=\frac{k}{x^3}$$ where $x$ is the distance from the source of gravity/magnetism. I am a mathematician, so I am completely happy ...


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Lets put the observer at A and the particle at B. The position kinematics are: $\vec{r}_A(t) = (a,0,0)$ $\vec{r}_B(t) = (r \cos \phi,r \sin \phi,0)$ The velocity kinematics are $\vec{v}_A(t) = (0,0)$ $\vec{v}_B(t) = (-r \omega \sin \phi,r \omega \cos \phi)$ The acceleration kinematics are $\vec{a}_A(t) = (0,0)$ $\vec{a}_B(t) = (-r \omega^2 \cos ...


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$v = |\vec{\omega}\times\vec{r}|= \omega r$ is always valid for a rigid rotating body. Here, $r$ refers to the distance of any particular point from a chosen axis of rotation, $\omega$, the angular speed of the body about that chosen axis and $v$, the linear speed of that point perpendicular to the radius vector (or the line joining the axis to that ...


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As Alfred Centauri mentioned, the ring is indeed slipping. Think about a ring whose center is fixed but is at the same time rotating. Clearly $v=0$ but $\omega\neq 0$. In general, a well defined relationship between translational velocity $v$ and rotational velocity $\omega$ exists in the case of a ring that is rolling without slipping. Othewise, there is ...


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Further to udrv's answer, and, the technicalities he raises aside, there are two ways to argue the reciprocity relationship that the boost from observer $A$ to $B$ is the boost from $B$ to $A$ but with $v\mapsto-v$. By the detailed arguments in the afterword, we find that the transformations between inertial frames form a group and that group acts linearly ...


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I think udrv's answer hits the nail on the head, but I'll expand a little bit on the intuitive way to look at it. In your example, you have a ship leaving earth flying towards Neptune at relativistic speed. While you might think that the ship is moving fast and the Earth is essentially still (though in truth, the Earth is moving around the sun and around ...


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Your question seems to be: if the observer on Earth sees the spaceship moving at velocity v, how do we know that the observer on the spaceship will see Earth moving at velocity -v? This is known as the "reciprocity principle" problem and it is a good one, in the sense that it raises the following issue: "Does the reciprocity principle follow from the basic ...


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You can actually (in principle) do an experiment for one of those. If you had a huge, dense, thin sheet of matter in empty space you could cut a person sized hole through it then attach a cylinder to the hole and put a flat bottom on the far away end. So on one side it looks like a flat tower coming up from a big plane and on the other side it looks like ...


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The key to understanding this somewhat surprising result is that the relativistic velocity addition formula is not applicable to this calculation. As an example of when to apply the velocity addition formula, sssume there is an object with (1D) velocity $\mathbf u$ in some inertial coordinate system. Now, what is the velocity $\mathbf u'$ of that object in ...


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Let me quote the relativistic velocity addition formula for easy reference: $$v_{AB} = \frac{v_A - v_B}{1 + \frac{v_Av_B}{c^2}}\tag{SR}$$ I'm guessing you interpreted these quantities as follows: $v_A$ is the speed of the light beam relative to George $v_B$ is the speed of Gracie relative to George $v_{AB}$ is the speed of the light beam relative to Gracie ...


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You would have to apply a force upwards to stop the body. Stopping the gravity would stop the acceleration but not the speed that it already has. A good place to start to check the effects of g forces in a human body is this wiki Changing the mass won't stop the fall either. You cannot make the mass zero, you can cut legs and arms but I think you'll be ...


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TL;DR You would eventually reach the higher velocity, but the hill ends before you get close. No Gear Changes First lets dismiss the notion that the power and torque requirements to ascend a hill at constant speed are different. To ascend a constant grade hill at a constant speed requires the same constant force $F$ regardless of velocity $V$. Let's ...


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How did you integrate acceleration to get velocity? Note that $\Delta v = \int_{t_i}^{t_f} a(t) dt$ But you have an acceleration that is a function of position, not time. So you can't naively integrate this and get velocity. There is a trick. Notice that you can rewrite acceleration as $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v ...


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To get velocity from acceleration, you need to integrate with respect to time. But your expression of acceleration is given with respect to position. Thus, your current calculation is not correct. You need to figure out how to convert the position-dependent information to time-dependent information. Since they give you the solution and you just have to ...


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If $$ \overrightarrow{r}=r_{x}\widehat{i}+r_{y}\widehat{j} $$ then $$ \left | \overrightarrow{r} \right |=\sqrt{r_{x}^{2}+r_{y}^{2}} $$ and $$ d\left | \overrightarrow{r} \right |=\frac{r_{x}dr_{x}+r_{y}dr_{y}}{\sqrt{r_{x}^{2}+r_{y}^{2}}} $$ on the other hand $$ d\overrightarrow{r}=dr_{x}\widehat{i}+dr_{y}\widehat{j} $$ and $$ \left | d\overrightarrow{r} ...


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As shown in the diagram $|dr|$ represents the magnitude of the vector difference(that involves the laws of vector addition/subtraction) between $\vec{r_2}\quad \& \quad \vec{r_1}$ while $d|r|$ represents the difference between magnitudes of two vectors which is simply the difference in their lengths.


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Using polar coordinates it holds $ |d{\bf r}| = \sqrt{(d|{\bf r}|)^2 + |{\bf r}|^2(d{\bf \phi})^2}$. From this equation you can see, the two expressions you are asking about are actually only equal (in absolute value) for a straight line through the origin, thus otherwise different. For them to be exactly equal, the $d|{\bf r}|$ should be moreover pointing ...


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The Earth as a rest frame is arbitrary. So saying that two cars have velocities toward each other is just a choice of reference frame. You are right that you can consider yourself to be at rest while the car coming toward you has some speed which is twice what the Earth frame would say it has. As it comes toward you, you observe it to have a contracted ...


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Consider a molecule of oxygen in a balloon. You know that at nonzero temperature all those molecules are bouncing around in all directions. Of course, the mass of air doesn't have any net motion in any direction. Indeed, the average velocity of each molecule is zero: $$\langle v \rangle = 0 \, .$$ Of course, the average energy of any particular molecule is ...


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Another important example: the Pythagorean Sum defined by $f(x) = \sqrt{x}$. As Doetoe says, you're simply transporting the binary operation + to the domain of $f$ if $f$ is e.g. bijective. If your domain is a real interval, your question actually describes the following: Which operation is calculated by the addition of affine co-ordinates implemented by ...


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Let $X$ be a set with a binary operation $\star:X\times X\to X$, which could be a group structure or whatever you want. Let $f:X\to Y$ be a mapping to some other set and let $F(x,y) := f(x\star y)$ (your middle expression is this one with the operation +). If $f$ is sufficiently nice, e.g. a bijection, then your construction transports the binary operation ...



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