# Tag Info

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These two, by the great Jacob Pieter Den Hartog, are just unbeatable. Classical applied mechanics (a.k.a. engineering). Wonderful introductions, easy to read and pedagogical! Mechanics Mecahnical vibrations

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Go for Concepts in Physics by HC Verma

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Actually, the reasoning that the force in circular motion is perpendicular to the direction of motion does work, and is in fact the correct answer. You appear to be saying that this violates conservation of energy by changing the object's momentum. However, the centripetal force only changes the direction of the momentum, not the magnitude, and thus ...

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You have to distinguish between the distance the man swims, relative to the water around him, and the total distance the man travels, relative to an observer on the river bank. The total distance relative to an observer on the river bank is the distance the man swims measured relative to the water around him combined with the distance the water moves ...

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e=mc2 tells us that, as an object gains velocity it becomes more "massive" (not larger but carries more mass) when approaching near light speeds, you could hit a planet, and it should be like driving through a mm thick wall of water

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I have a quantitative answer which is a thought experiment avoiding all but the simplest equations. An object going from velocity v=0 to v=1 needs to be pushed or pulled in some way. In my explanation I will use the same method to push the object from v=0 to v=1 then from v=1 to v=2, then v=2 to v=3, etc. I will show how the energy of movement embodied in ...

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If this is purely academic exercise then the answer is no, you cannot discern between the different friction forces because they are all acting along the same line of action (through the cg of the part).

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Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough $$E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)}$$ For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order ...

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I thought gravity is uniform acceleration, not increasing acceleration.. Position: $y(t) = \frac{1}{2} g t^2$ Velocity: $y'(t) = gt$; Acceleration: $y''(t) = g$;

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Officially, I completely agree with the other answer given. I would like to offer this answer as a simplistic, intuitive answer to the question. No math involved. I understand where your question comes from. In fact, depending on your current level of education, this question could indicate a high potential for future scientific success. We all know that ...

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In polar coordinates, the velocity being tangent to the circle, it is directed along the $\hat{e}_{\theta}$ vector. The centripetal force is directed along the $\hat{e}_r$ vector. So $\frac{F}{m}\hat{e}_r = \frac{d\vec{v}}{dt} = \frac{d(|v|\hat{e}_\theta)}{dt} \underbrace{=}_{\text{Chain Rule}} \frac{d|v|}{dt}\hat{e}_{\theta} + ... 1 Consider a uniform rod of length$L$pivoting and sliding on a horizontal plane. Kinematics You want to describe the relationship between the coordinates (and their derivatives)$x$and$\theta$and the motion of points A, B, and C. Position Kinematics $$\begin{matrix} \vec{r}_A = x\,\hat{i} & \hat{i} = (1,0,0)\\ \vec{r}_B = \vec{r}_A + ... 0 You haven't told where you have difficulties understanding such simple thing even after doing some research online. I think, Wikipedia has clear description of both. Historically, there were three branches of classical mechanics: Statics: The study of equilibrium and its relation to forces. Kinetics: The study of motion and its relation to forces. ... 0 Begin with equations of motion:$$\Delta x=V\cos\theta*t\Delta y=V\sin\theta*t-\frac{1}{2}gt^2$$We have two unknowns V,t and two equations. From the first equation, we have:$$t=\frac{\Delta x}{V\cos\theta}$$Plug it into the second equation:$$\Delta y=V\sin\theta\left(\frac{\Delta x}{V\cos\theta}\right)-\frac{1}{2}g\left(\frac{\Delta ... 0 Yes. If there is a force in the$x$-direction, then Newton's second law tells us that there will be a corresponding acceleration$a_x$in that direction given by $$a_x = \frac{F_x}{m}$$ where$m$is the mass of the projectile. 5 This is know as the Stick-Slip Phenomenon For the chalk pattern see this. 1 Differentiate$\dfrac{dx}{dt}= \alpha \sqrt{x}$with respect to time again to get: $$\dfrac{d^2x}{dt^2} = \dfrac{\alpha}{2 \sqrt{x}} \dfrac{dx}{dt} = \dfrac{\alpha}{2 \sqrt{x}} \alpha \sqrt{x} = \dfrac{\alpha ^2}{2}$$ Then: $$\dfrac{dx}{dt} = \dfrac{\alpha ^2 t}{2}$$ and: $$x = \dfrac{(\alpha t)^2}{4}$$ for the given initial conditions. 1 Well integration is the basic method. But some observation can help too: $$v=\alpha\sqrt x$$ $$\text{Squaring both sides:}$$ $$v^2=\alpha^2 x$$ We know$v^2\propto x$gives constant acceleration.$\text{Remember} :v^2=u^2+2as.$So, comparing it with this equation we get $$v^2=0+2\frac{\alpha^2} 2 x$$ So, acceleration =$\alpha^2/2$and velocity ... 18 The result you've got would be better known as this: $$\int_0^t\biggl(\int_0^{t'} a\mathrm{d}t''\biggr)\mathrm{d}t' = \frac{1}{2}at^2$$ In other words, it's a derivation of the formula for uniformly accelerated motion. This derivation, or something like it, is one of the first things students in a good calculus-based introductory physics class learn. The ... 0 Well you know that speed like this is usually in kilometers per hour. So if traveled 15km in 30 minutes, it travels twice that in 60 minutes. (1 Hour) So it should be 15km*2 = 30km traveled. Which makes the average speed 30km/h. 0 Given the total height requirement you can find the vertical velocity as$v_y = \sqrt{ 2 (2 h) g}$. Then the time to reach the height$h$is first$t=\sqrt{\frac{h}{g} (2-\sqrt{2})}$and then on the way back down$t=\sqrt{\frac{h}{g} (2+\sqrt{2})}$. If the (horizontal) distance between you and the pole is$d$then for the projectile to pass the pole at the ... 2 If$h$is the height about the earth then $$\ddot{h} = -\frac{G M}{(R+h)^2}$$ $$\ddot{h} = \frac{{\rm d} \dot{h}}{{\rm d}t}= \frac{{\rm d} \dot{h}}{{\rm d}h} \frac{{\rm d} h}{{\rm d}t} = \frac{{\rm d} \dot{h}}{{\rm d}h} \dot{h}$$ $$\int \ddot{h}\; {\rm d} h = \int \dot{h}\; {\rm d} \dot{h} = \frac{1}{2} \dot{h}^2 + K$$ $$\int -\frac{G M}{(R+h)^2}\; ... 4 So you were on the right track with integrating over r and over t. Here's how you could do it: The acceleration at any radius, r (if we assume Earth is a point mass) is:$$a=-{GM\over r^2}$$The minus sign is because the acceleration is anti-radial. Then you can do the following:$$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta vthus$$... 0 You will calculate the velocity for the x, y and z components separately. So if the object has v_x velocity initially, and it spontaneously gains acceleration in the negative y direction, this problem becomes very simple. Over time this object will gain velocity in the y (v_y) direction. It will also always have its' velocity in the x direction, and ... 6 Imagine that you have just two particles with the same mass and same speed, but going in opposite directions. They have opposite momenta, so the total momentum is zero. But they each have energy, and the total energy is not zero. The reason is because kinetic energy is just \frac{1}{2} m v^2. That square means that the kinetic energy can never be ... 2 You do know what the vertical component of the velocity is, because you know how high it got. You can do this in two obvious ways, but I prefer energy conservation. As you threw the rock, it's energy was entirely kinetic: E_0 = \frac{1}{2} m v_v^2 + \frac{1}{2} m v_h^2. At its peak, the energy was partly potential, with a kinetic contribution just from ... 2$$y = x\tan\theta - \frac{g}{2v_0^2 \cos^2\theta}x^2.y=x.\tan\theta-\dfrac{gx^2}{2v_0^2}-\dfrac{gx^2\tan^2\theta}{2v_0^2}A\tan^2\theta+B\tan\theta+C=0$$As \tan\theta\in\Bbb R , so B^2-4AC\ge0 must hold, for the above equation to have real roots for \tan\theta. Use that and you'll get$$y\le \dfrac{v^2}{2g}-\dfrac{g}{2v^2}.x^2$$That ... 2 what exactly does the area under a displacement-time graph denote? I think it just represents what you said: the area under a displacement-time graph. I can't think of any other use for it. There are two main reasons for this: Your quantity, let's call it$f(t)$, retains a memory of where the object has been. That's because the area under the graph ... 1 I don't see this quantity computed very often so I don't think it is that useful, but here is one thing that can be said: If$A$is the area under the graph and$T$is the duration of time over which the integral was preformed then$A/T\$ is the average displacement from the origin during that interval of time.

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