# Tag Info

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I think you have a small misconception about rotation sensing. Accelerometers will sense a frame's rotation of constant angular velocity as well as rotation where the angular velocity varies. This is because constant angular velocity rotation is itself an acceleration: something must undergo a centripetal acceleration to follow a curved path at constant ...

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The equations of motion for the position determine the accelerations: they are second-order differential equations in time: $$\vec F = m\vec a = m\ddot{\vec x }$$ So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the $F=ma$ formula ...

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Imagine a body moving with velocity $\vec{v}$ in the $\hat{x}$ direction and now imagine having it losing speed (decelerating, so $\vec{a} \propto -\vec{v}$) the friction force should be opposite the $\vec{a}$ (acceleration) or the $\vec{v}$ (velocity)? if you think carefully about it, you'll convince yourself that it should oppose velocity and not ...

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This is an excellent question which is much more subtle than it first appears. On a first (or even a second) reading, you seem to answer your own question and then ignore your own answer! What I mean is, you clearly understand that while in an accelerating vehicle you experience a 'jerk' which does not happen with uniform motion, and you also acknowledge ...

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Would you decrease your impact impulse by jumping during the fall? Yes When? Soon enough that it's before impact, late enough that you don't hit the ceiling of the elavator. Beyond that I don't think it matters much. Would it help if you jump inside a free falling elevator? Probably not. Indeed I expect it would make things worse. The ...

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The flaw is your assumption that In this [accelerating] frame we don't see any force so the first law of dynamics is respected. In the accelerating reference frame you do see evidence of a force, even though you don't see the effect you are expecting (acceleration of the object). Like an observer standing on the surface of the Earth, acted on by the ...

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Roughly, this means that if the spacecraft travels 1000 inches on the first day, it will slow down enough to travel only 700 inches on the second day, and then slow down enough more to travel only 400 inches on the third day, and then slow down enough more to travel only 100 inches on the fourth day. In other words, it takes the spacecraft one day to slow ...

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The flaw is that you've failed to do an experiment which will tell you whether the frame is inertial or not. If you do such an experiment -- for instance take a test mass, initially at rest with respect to the frame, release it, and see if it remains at rest -- you will immediately discover that the frame is not inertial.

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$h=R(1-\cos\theta)=R(2\sin^2\frac{\theta}{2})$ $v=\sqrt{2gh}=\sqrt{4gR}\sin\frac{\theta}{2}$ $v$ is a maximum at $\theta=\pi$. The tangential acceleration is $\frac{dv}{dt}=\sqrt{4gR}\cos\frac{\theta}{2}.\frac12\frac{d\theta}{dt}=\sqrt{4gR}\cos\frac{\theta}{2}.\frac12.\frac{1}{R}\sqrt{4gR}\sin\frac{\theta}{2}=g\sin\theta$ as expected. This is a maximum at $\... 0 You are way off in your thinking, and your 3 'tries' show that you are blindly guessing what is going on here. I think you need to go back to the basics of torque and rotational motion. The question asks you to find the angular acceleration$\alpha$about the centre of mass (CM). This is related to net torque$\tau$and moment of inertia$I$by$\tau = I\...

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Is vertical uniform circular motion even possible? No, it isn't. Because magnitude of velocity isn't constant and we know that in a uniform circular motion the object moves with constant speed. $\large{\frac {\mathrm d}{\mathrm dt}}v=g\sin\alpha\neq 0$ ($v$ is the speed (magnitude of the velocity vector $\vec v$) of the object) Is this analysis ...

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They ask you for the angular acceleration $\alpha$, not the linear acceleration. I could not understand what you did, but you know that torque $\tau=I \alpha$, to find $\alpha$ only divide the torque you have by the moment of inertia $I=mL^2/12$ which corresponds to the moment of inertia of a rod around its center of mass.

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You're probably used to thinking about the objects at the bottom of ropes pulling downward on ropes because of gravity. But here, it's better to think of the top of the rope pulling up. This is because here's the sequence of events when the bottom of the elevator hits the ground: 1) The ground exerts a large upward force on the elevator floor in order to ...

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The object at the end of the rope has a negative velocity (in the world frame of reference) when the elevator (and it) comes to a halt. When the elevator stops, you have to go from a negative velocity to zero velocity. That requires a positive acceleration.

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We have to be careful what we mean when we say "moving away". Imagine a grid in which we are at the origin and there are light sources located at each of the grid intersections. If the grid stays as it is while the light sources accelerate away from us, their light will appear to be redshifted. If you set up the accelerations such that everything moves ...

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A comment on Jim's answer (I don't have enough reputation to properly 'comment' yet). I think you make a mistake [edit: his answer's been corrected since] when you expand for small angles measured with respect to the vertical, because that is not any more the equilibrium position. In the new, inclined equilibrium position, the effective $g$ is higher, but ...

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Look at the included figure: This shows a simple pendulum consisting of a bob of mass $m$ and length $l$ acted upon by two forces, namely $mg$ (force due to gravity, acting downwards) and the force $ma$ (due to the acceleration of the train) that is perpendicular to the force of gravity.The angular acceleration of the bob is $\frac{d^2 \theta}{dt^2}$ and ...

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A body in an accelerated reference frame (say, a train with acceleration $\mathbf{ A}$) will appear subjected to an inertial force, i.e., it's necessary to add $\mathbf{F}_i=-m\cdot\mathbf{A}$ to the 'real' forces acting on the body for Newton's second law to hold in this reference frame. You can obtain this result from a change of reference frame, which, ...

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The solution is actually strange for me. It assumes that the distance traveled by each car (in the CM frame) during the acceleration (crumpling) is also equal to the crumple length of each car, but is not always the case. Following the assumption of the solution: In the CM frame, the heavier car moves slower than the lighter car. So that at the moment that ...

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This is a simple question of conceptual understanding. The only force acting on the ball after it is released from the hand is that of the ball's weight due to gravity. Since it is the only force, the consequent acceleration is also the only acceleration. Gravity is being taken to be $10 m/s^2$ in a downward direction. For the answer to be correct, our ...

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According to the definition of the work ($\delta W=F\mathrm dx$), it is better that we say friction doesn't do work. Because, at each point of contact area, friction force is fixed and doesn't move. Friction converts some portion of energy used to move the box, to heat.

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This is a trivial kinematic deduction. \begin{align}s(t_2) &=\int_{t_1}^{t_2}~v(t)~\mathrm dt +s(t_1)\\ &= \int_{t_1}^{t_2}~\left\{v(t_1)+\int_{t_1}^{t}~a(t')~\mathrm dt'\right\}\mathrm dt+ s(t_1) \;.\end{align} Integrating this, we would get $$s(t_2)~=~ s(t_1) + v(t_1)\{t_2-t_1\} + a(t_1)\frac{\{t_2-t_1\}^2}{2} + \dot a(t_1)\frac{\{t_2-t_1\}^3}{6}... 3 Indeed your first suggestion is wrong : \Delta x = v_o t + gt  Instead it should be  \Delta x = v_o t + gt^{2} (You can recheck it) Where you are wrong is here: According to your question v is the final velocity since (v=v_{0}+gt) So \Delta x\neq vt but instead it should be \Delta x =v_{average}\times t In uniform acceleration v_{average} ... 1 Let's take the first equation of motion which is : $$v=u+at$$ Integrate this equation to get: $$\int\frac{dx}{dt}dt=\int{u}dt+\int{at} dt$$ this gives: $$x=ut+\frac{1}{2}at^2+x_0$$ The integration constant can be done away by putting the proper limits on x.(Assuming the ... 0 I don't have a conceptual answer as to why it is that way. But mathematically, your first suggestion is wrong:  \Delta x = v_o t + gt . A unit analysis will show you why:$$ meters = \frac{meters}{seconds} seconds + \frac{meters}{seconds^2}seconds meters \not= meters + \frac{meters}{seconds}$$And we can see that the assumption is simply not true ... 0 When the ball is travelling in your hand before you let go, it is being accelerated by the amount of force you exert on it, this increases its velocity before you let go. (The initial velocity) When you let go of the ball, it immediately begins to slow down due to the acceleration of gravity acting downwards. There is no upwards acceleration, as this ... 4 Assuming rolling without slipping, this can easily be solved by means of Energy Conservation. Let's assume the incline has a height h. During the travel down the incline, potential energy U is then converted to kinetic energy K: K=U K=mgh The kinetic energy K is a combination of translational and rotational energy: \frac12mv^2+\frac12 I\... 2 The mass of the wheel is located farther from the axis of rotation, thus the moment of inertia is greater for the wheel. So it rotates slower with the same rotational energy and therefore the ball rolls faster down the slope than the wheel. 0 The comment by Walter is on the right track: The "acceleration" does not refer to the fact that recession speed increases with distance, because this is just a consequence of space expanding everywhere. This is why we measure the expansion in km/s per megaparsec. Today, the expansion rate (the Hubble constant) is H_0 \simeq 70\,\mathrm{km}\,\mathrm{s}^{-1}\,... 1 The FLRW energy equation for the motion of test masses in the universe is$$ \left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G\rho}{3}. $$the scale factor for space is a and its time derivative is \dot a. I derived this from Newtonian dynamics. The density of mass \rho for the case of a quantum vacuum energy level is constant. I now replace this with ... 0 I would argue that the expanding of space cannot and should not be understood adding space into space, nothingness into nothingness. We have no way of observing the space itself as a reason like the one you presented likes. Things seem to get away from us through and the observed mechanism is called redshift, which, in close distances(inside let's say the ... 0 The simplest explanation of all is to glance at the formula:$$a_{rad} =\frac{v^2}{r}$$a_{rad} is the magnitude of the radial (centripetal) acceleration. So the answer is clearly yes, the magnitudes of the radial acceleration is constant because the speed is. Only direction changes all the time in order to always point towards the centre, as you ... 2 When we talk about kinematics of a point particle, mainly the velocity and acceleration vectors, it would be reasonable to use differential geometry to understand what is going on. The motion of a point particle is described fully if it's given its position vector \:\mathbf{r}(t)\: as function of time \:t\:. Then the velocity vector \:\mathbf{v}(t)\: ... 2 For a car going around a corner of a constant radius moving with a constant speed the magnitude of the centripetal acceleration will be constant but the direction of the acceleration will change. So we would have to say that the centripetal acceleration is not constant in the same way that the velocity is not constant. 2 The centripetal acceleration will always be directed towards the center of the circular arc that the car's instantaneous path is a part of. Thus, the direction of the acceleration will be along the line joining car's position and the point about which it is in instantaneous circular motion. We can easily visualize that during a turn, this line is changing ... 0 The maximum velocity v, observer time t, and traveler proper time T calculated on the "Space travel calculator" site are identical to those on John Baez's relativistic rocket page, but the energy requirement and fuel mass calculations are botched. If you check the explanation notes provided, the energy requirement is calculated as e = 2mc^2(\gamma -... 0 You can do this derivation by breaking the position of the orbiting particle down into components. It isn't short, but I think it's useful because it supplements the algebra with concrete physical analogies. I'll organize it into four parts: decomposition, oscillation, energy, and symmetry. Decomposition The position of a particle moving along a circular ... 0 If this is a projectile problem, and assuming no air resistance or buoyancy, the only force is the weight mg acting constantly downwards, and the only acceleration is g. Neither of these is affected by initial velocity. 0 Initial velocity has absolutely no effect on force. Any inertial frame must obey the same laws of physics as any other frame. You asked what you need to know to calculate force and then said F=ma. I'm not entirely sure what you are getting at but you clearly need to know the mass and the acceleration. Now acceleration is the change in velocity over time (a=... -1 I thought I would indicate how I see this problem. This question involves different ways of defining the payload and total mass of the rocket. The four momentum of a body in flat spacetime, such as a rocket, is$$ P~=~(E,~{\bf p}).  The four-momentum has the spatial momentum and the energy. The energy the rocket before it starts to accelerates has the ...

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