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0

Your first calculation is correct. Nothing wrong with it. The mystery for me is why you go to so much trouble trying to disprove it!


0

At every point the tangential direction is the unit vector of the velocity vector. If you have the velocity components $\boldsymbol{v} = (\dot{x}, \dot{y})$ at every instant, the you decompose this into a magnitude (speed $v$) and direction $\hat{\boldsymbol{e}}$ $$ \begin{align} v & = \sqrt{\dot{x}^2+\dot{y}^2} \\ \hat{\boldsymbol{e}} & = ...


0

The easy way of doing this is to parametrice the trajectory. We have the cartesian definition, so let $ \textbf{r} : \mathbb{R} \rightarrow {\mathbb{R}}^{2} $ be: $$ \textbf{r}(t)=(t,f(t)) \quad\quad t\in (-\infty,\infty) $$ So we got the vector position as function of "time". Then, the velocity and acceleration vectors are defined by: $$ ...


1

You have made some errors in you calculation of distances. Let $$ \mathbf{r}^{\prime}_{k}=\mathbf{r}_{k}-\mathbf{r}_{_{CM}} $$ Then \begin{aligned} r^{\prime\;2}_1 &= \Big(\frac{24}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{1872}{121}\quad \text{(upper left particle)}\\ r^{\prime\;2}_2 &= \Big(\frac{20}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 ...


1

shouldn't we use the parallel axis theorem ... to compute the moment of inertia? You could...if you already had the moment of inertia of the object about its center of mass. Since you don't, it's far easier to simply sum the moments of inertia about the $z$ axis.


3

The question is inconsistent. At least one of the numbers (mass, force, stopping distance, or stopping time) is wrong. Your calculation of the acceleration from force and mass is correct, but an acceleration of $24$ m/s$^2$ for $2$ seconds means that the toy car was initially traveling at $48$ m/s. This is over $100$ mph ($160$ kph) and there is no way the ...


1

The approach really fails, because the proposed motion is kind of broken. Including the integration constant, the distance traveled as a function of time becomes: $s(t) = \frac{1600}{(t+C)^2}$ Since we want $s=0$ when $t=0$, the constant C becomes infinity. This means that, the motion can not really start from rest -- it takes infinite time to do so. This ...


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 ?


0

To calculate the minimum energy needed for the reaction the products are assumed to be stationary, i.e. the momentums are zero. With $ \pmb p^2 = E^2 - \vec p^2 = E^2 = m^2$ follows: $$({p_1^\mu}' + {p_2^\mu}' + {p_3^\mu}' + {\bar p_4^\mu}')^2 = (E_1 + E_2 + E_3 + E_4)^2 - (\vec p_1 + \vec p_2 + \vec p_3 + \vec p_4)^2 = (E_1 + E_2 + E_3 + E_4)^2 = (m_p + ...


0

This is a standard projectile motion problem with the complication that the projectile collides elastically with vertical walls. Assuming that the only force exerted on the projectile by the walls is in a direction which is normal to the walls then: the vertical velocity does not change, and the horizontal velocity reverses direction but the magnitude ...


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 ...


0

The loss of energy (efficiency<1) manifests itself in two ways Friction. When doing free body diagrams add frictional forces at the sliding contacts and find which coefficient of friction $\mu$ gives you the efficiency values you expect. Structural damping (hysteresis). This is more difficult to put in the equations of motion, because they assume ...


0

First of all, the transmission of force between any two pairs of mating gears happens due to surface contact between the two and hence through friction. Next, any loss is a dissipation in energy. Though I am not that well-versed in vibration analysis, I know that an periodic dissipation of energy is accounted in the force-equation using a damping ...


0

In order to solve this question, you have to calculate distance travelled in 4 seconds as well as in 3 seconds. And then subtract these two to get the distance travelled from 3rd to 4th second (which is 3rd second). What you did was that you calculated the total distance travelled from $t=0$ to $t=3$ seconds, instead you have to find the distance travelled ...


0

The frictional force is given by $f = \frac 1 3 mg \sin \theta$ when the no slipping condition ($a=r \alpha$) is satisfied. If the component of weight down the slope, $mg \sin \theta$, was equal and opposite to the frictional force then there would no net force on the cylinder and so the centre of mass of the cylinder would not be accelerating ($a=0$). ...


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 ...


0

The simplest approach to a problem like this would assume that the collision is elastic, and that you have some knowledge of the elastic constant. But a collision between car and human is not that. Instead, let us assume that the "elbow sized object" hits the human in the mid section, and that it doesn't simply go right through him. Then the next thing that ...


0

First of all: your solution is correct, the friction is one third of what it would have been for a "non-round", as you call it, body. How does friction know. I like such questions, but I don't see a convincing simple reasoning right now (but there probably is one), so I will use a little of computation. I hope it's different (less computating^^) than your ...


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.


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 ...


-1

$a=6000(1-\dfrac{4}{3}x)$ is a differential equation in $x$. Solve it then substitute $x=0.75m$ to find out $t$.


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 ...


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 ...


0

The critical observation here is that before the magnet is introduced, there are equal numbers of oppositely charged ions moving in opposite directions, exerting no net force on the water around them. The magnet, however, will deflect a positive ion going one way in the same direction as it will deflect a negative ion going the other way. This produces a ...


-1

The electron was the number of electron which has number of charge is $1.6\cdot10^{-18}\,$eV. As the electric field is the region in which charged particles experience the force to perpendicular direction between electrical poles.The moving ions consists of positive charge and negative charge therefore the cations and anions which can be moving ions in ...


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 ...


0

What's wrong with this is the phrase 'suddenly I hit the particle'. What you are assuming by this is that you have some hard-edged and rigid object which you crash into the particle, which is also hard-edged. But you don't have either of those things: what you have is something which is both not rigid and whose surface is actually a little bit of EM field ...


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 ...


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 ...


0

It is dimensionally correct... In the derivation, you take ut-u(t-1) which makes it appear as u, but is actually u*1 second... Hope this clears your doubt...


0

Let $x = x_2-x_1$ and $y= y_2-y_1$ to simplify the algebra. The initial velocity $v$ has two components $v_x$ and $v_y$ and the angle of projection is $\theta$. For vertical motion you have $y = v_y t - \frac 1 2 g t^2$ and for horizontal motion you have $x = v_xt$ Combining these two equations gives you $y = v_y\left (\dfrac {x}{v_x} \right) - \dfrac ...


0

If you ignore air resistance, it can be solved as follows. Assume the initial velocity at $t = 0$ at $(x_1,y_1)$ be $(v_{x0},v_{y0})$. Assume it hits $(x_2,y_2)$ at $t=T$. Solve for $(v_{x0}, v_{y0})$ in terms of $T$. You then get a family of curves when substituting different $T$.


0

I will give you a hint: We can avoid the constant c in the equation by selecting the starting point of parabola as origin. so we have three points $(0,0)(x_1,y_1)(x_2,y_2)$. Now placing it in equation $y=ax^2 + bx$. We can find a and b. Now comparing with the equation of trajectories we can find out the velocity $v_0$. and angle $\theta$ in form of either ...


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.


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 ...


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 ...


0

The movement comes from push, or pull. A push, or pull comes from one of the forces. For example, a push happens when two things/fields can not co-exist at same point of space time and one (or both) of them get displaced. The displacement, is the movement. The forces govern how things behave. Movement is one of the behaviors caused/governed by forces.


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 ...


0

At a basic level, current models of physics describe a force as the exchange of (virtual) bosons. This is just a model, but here's the general gist at what's a very basic level: There are different types of fundamental forces: Strong, electroweak (electromagnetic + weak) and gravitational. On top of this, there is the Higgs Boson. The strong force is ...


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 ...


-1

In reference to the given participants $A$ and $B$ there can be another participant $M$ (uniquely) identified as the "middle between" $A$ and $B$ by the conditions that $M$ and $A$ had been and remained at rest to each other (which can be expressed in terms of spacetime interval values, following the prescription linked in the OP), likewise $M$ and $B$ had ...


2

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


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

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 ...


0

A particle following a prescribed path has its velocity vector parameterized as $$ \vec{v} = \vec{e} \,v $$ where $\vec{e}$ is the tangent vector and $v$ is the speed at that instant. This is kind of obvious. But you use the above to find the tangent vector if you know that radial vector $\vec{r}$. Use $\vec{v} = \frac{{\rm d}}{{\rm d}t} \vec{r} = \vec{e} ...


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 ...


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 ...


0

To move velocities between two points A and B do the following: $$ \vec{v}_B = \vec{v}_A + \vec{\omega} \times (\vec{r}_B- \vec{r}_A) $$ The derivative of the above moves the material accelerations from A to B $$ \vec{a}_B= \vec{a}_A + \vec{\alpha} \times (\vec{r}_B- \vec{r}_A) + \vec{\omega} \times (\vec{v}_B - \vec{v}_A) $$ There is also the spatial ...


0

If acceleration is linear, use the 'kinematic equations'. If not, use differentiation/integration (in this case, you will probably be given a formula which expresses acceleration).


0

The initial answer is exactly right, and your response shows you understand that the dot product of the force and velocity will give you a scaler value that leads to the right answer. However when I read the question, I was impressed at how hard it is to visualize what is really going on, and the incurable teacher in me thought you might be interested in ...



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