New answers tagged

-2

centrifugal force, is the force from the tire that causes the mud to fly off, and it will travel in the opposite direction that the wheel is moving. So if you wheel is propelling you North the centrifugal force will be sending the mud South.


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You are looking for a relationship between $v$ and $r$. If you start with $r$, and you want to "go" to $v$, then you multiply by $\dfrac vr$ since $r\cdot\dfrac vr=v$.


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Suppose we have a rigid object rotating at some angular velocity $\vec{\omega}$ about some fixed axis. It can be shown that in general, $\vec{v} = \vec{\omega} \times \vec{r}$, where $\vec{r}$ is the instantaneous position of the object. This more general equation does take the directions of the quantities into account via their cross product. However, it ...


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The direction is implied to be in the $\hat{\phi}$ direction because it is assumed that (eq 8-9) $r$ is not changing with time. Moreover, it's not really implied since it's the tangential velocity and so is in the tangential ($\hat{\phi}$) direction. To clarify, let's work with an object spinning in a circle in two dimensions, to keep things simple. A point ...


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I think all your questions are along the lines of "How can the Earth, Sun, planets, etc. be in the current configuration, when such a configuration is very unlikely, given all the things that have to go right for this to happen?" To explain this, I would like you to think of the example of a lottery ticket. If 1 billion people buy a lottery ticket, ...


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Well over $99\%$ of the mass of the Solar System is in the Sun, so it is not surprising that the planets and most of the other bodies in the Solar System orbit the Sun in fairly stable orbits. However, these orbits are not all circular, or even close to circles. From the major planets, the eccentricity of Mercury’s orbit is over $20\%$ and the eccentricity ...


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Rotational inertia is the resists any change of rotational velocity If you decrease rotational inertia,angukar velocity will increase The reason why when the distribution of mass is closer to the axis of rotation rotational inertia decreases is that the magnitude of rotational inertia is proportional to the perpendicular distance betwen a mass element and ...


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The path of a planet around a star is indeed an ellipse with the star on one of the focal points. For an ellipse with semi-major axis $a$, and semi-minor axis $b$ the focus point is $f=\sqrt{a^2-b^2}$ distance from the center of the ellipse. With polar coordinates $(r \cos \varphi, r \sin \varphi)$ from the focal point the ellipse tracs the following curve $$...


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the equation of ellipse where the axes at (0,0) is $$x=r(\varphi)\,\cos(\varphi)\\ y=r(\varphi)\,\sin(\varphi)$$ where $$r(\varphi)=\frac{a\,\sqrt{1-e^2}}{\sqrt{1-e^2\,\cos^2(\varphi)}}$$ and $$e=\sqrt{1-\frac{b^2}{a^2}}$$ a,b are the half-axis of the ellipse The velocity $$\vec v=v_r\,\vec{e}_r+v_\varphi\,\vec{e}_\varphi$$ with $$v_r=\frac{dr}{dt}=\frac{dr}...


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This would more suitably be a comment, but do you mean a nice equation that describes ellipses? Perhaps you can try polar coordinates - they might be easier to work with.


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You need to relate the a and v to the angle $\alpha$. You have $a=\frac{\Delta v}{\Delta t}$ and from the triangle with sides equal to v, v and $\Delta v $ you have $\Delta v = 2 v \sin(\frac{\Delta \alpha}{2})$ and $$ a= \frac{\Delta v}{\Delta t}= \frac{2 v}{\Delta t} \sin(\frac{\Delta \alpha}{2})$$ The angle $\theta$ is related to $\Delta \alpha $ by $\sin(...


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Starting with 2.: $$\omega = \frac{a_n}{v}$$ considering $a_n = \omega^2 R$ and solving for $\omega$: $$\omega = \frac{v}{R} = v \frac{d \alpha}{ds} = v \frac{d \alpha}{ds} \frac{dt}{dt} = v\frac{dt}{ds} \frac{d \alpha}{dt} = \frac{d \alpha}{dt}, $$ where in the second step we used $ds = R d\alpha$ and in the last step $v= \frac{ds}{dt}$.


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It doesn't make sense that you have a value for the centripetal acceleration $a$ while $\omega$ tends to zero. Those two follow each other. Physically, if you have a centripetal acceleration $a$ causing you to move around a curve, then you also must have a nonzero angular speed $\omega$ around that curve. Otherwise you wouldn't be moving around the curve and ...


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Only relative velocity is physically meaningful.We can choose either satellite or earth to be at rest.So the relative orbital angular velocity is dθ/dt = dθ/dt'


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When the astronaut first begins spinning his arms around he applies force to them to begin their spin. So according to Newton's third law of motion the arms apply an equal and opposite force on his body spinning it in the opposite direction. As long as he keeps rotating his arms in one direction his body keeps rotating in the opposite direction. When he ...


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Notice that in equation 2, as the angular velocity goes to zero, the centripetal acceleration also goes to zero (as does the tangential velocity). Also in circular motion, the linear speed is the magnitude of the tangential velocity (unless the radius is also changing). The components, $v_x$ and $v_y$, both change as a function of time (unless you are ...


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I can't access freely to the PDF document you cite in your question, so I cannot give a complete answer to your question. However, here are some preliminary remarks: The relationships you mention in your question are limited to a uniform circular motion (on a straight line, you have $a$ finite and $w = 0$). In theory, you don't need $\omega $, you can ...


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