通知动议

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Weather satellites, like the one shown above, are found miles above the earth's surface.  Satellites can be polar orbiting, meaning they cover the entire Earth asynchronously, or geos tationary, in which they hover over the same spot .
::与上述卫星一样,气象卫星在地球表面上方几英里处被发现。卫星可能是极地轨道,也就是说它们覆盖整个地球,或地球静止,它们在同一地点盘旋。

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Arthur C. Clarke was the first to propose that satellites be placed in "geosynchronous orbit" around the Earth. These satellites orbit with the Earth over a time period of 24 hours, causing them to appear to stand still from Earth. Many of the fixed satellite dishes we see today, such as the ones used at home for satellite TV, are aimed at satellites in geosynchronous orbit. Learn more about the  circular motion of a satellite by exploring the Clarke’s Dream simulation below:
::Arthur C. Clarke是第一个提议将卫星置于环绕地球的“地球同步轨道”的建议者。这些卫星在地球轨道上运行了24小时,它们似乎从地球站立不动。 我们今天看到的许多固定卫星天线,例如用于卫星电视的固定天线,都以地球同步轨道上的卫星为目标。 通过探索以下克拉克的梦幻模拟,更多地了解卫星的循环运动:

 

 

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Circular Motion
::通知动议

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The earth is a sphere. If you draw a horizontal straight line from a point on the surface of the earth, the surface of the earth drops away from the line. The that the earth drops away from the horizontal line is very small – so small, in fact, that we cannot represent it well in a drawing. In the sketch below, if the blue line is 1600 m, the amount of drop (the red line) would be 0.20 m.  If the sketch were drawn to scale, the red line would be too short to see.
::地球是一个球体。 如果你从地球表面的一个点上画一条水平直线, 地球表面就会从线上掉下来。 地球从水平线上掉下来的线非常小 — — 事实上,如此小,我们无法在绘图中很好地代表它。 在下面的草图中,如果蓝线为1600米,那么投落量(红线)将是0.20米。 如果草图被拉到宽幅上, 红线会太短, 看不到。

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When an object is launched exactly horizontally in , it travels some distance horizontally before it strikes the ground. In the present discussion, we wish to imagine a projectile fired horizontally on the surface of the earth such that while traveling 1600 m horizontally, the object would fall exactly 0.20 m. If this could occur, then the object would fall exactly the amount necessary during its horizontal motion to remain at the surface of the earth, but not touching it. In such a case, the object would travel all the way around the earth continuously and circle the earth, assuming there were no obstacles, such as mountains.
::当一个物体完全水平发射, 它在撞击地面之前水平移动一些距离。 在目前的讨论中, 我们希望想象一个射弹在地球表面水平发射, 这样在水平飘移1600米时, 该物体会完全跌落0. 20米。 如果可能发生, 那么该物体会完全跌落到其水平运动所需的数量, 以便留在地球表面, 而不是触碰它。 在这种情况下, 该物体会连续地绕着地球, 环绕地球, 假设没有障碍, 如山。

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What initial horizontal would be necessary for this to occur? We first calculate the time to fall the 0.20 m:
::发生这种情况需要什么样的初始水平? 我们首先计算0.20米的跌落时间:

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  $\begin{array}{r}t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{(2)(0.20\text{m})}{9.80{\text{m/s}}^2}}=0.20\text{s}\end{array}$
::t=2da=(2)(0.20米)9.80米/s2=0.20秒

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The horizontal velocity necessary to travel 1600 m in 0.20 s is 8000 m/s. Thus, the necessary initial horizontal velocity is 8000 m/s.
::0.20秒内旅行1600米所需的水平速度为8,000米/秒。因此,必要的初始水平速度为8,000米/秒。

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In order to keep an object traveling in a circular path, there must be an toward the center of the circle. This acceleration is called .  In the case of satellites orbiting the earth, the centripetal acceleration is caused by gravity. If you were swinging an object around your head on a string, the centripetal acceleration would be caused by your hand pulling on the string toward the center of the circle.
::为了保持一个物体在圆形路径中运行, 圆形中心必须有一个物体。 这种加速度被称为 。 在卫星环绕地球运行的情况下, 子宫加速度是由重力造成的。 如果您在一根绳子上将一个物体绕在头部上, 则子宫加速度将因你的手拉向圆形中心而导致 。

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It is important to note that the object traveling in a circle has a constant but does not have a constant velocity. This is because direction is part of velocity; when an object changes its direction, it is changing its velocity. Hence the object's acceleration. The acceleration in the case of uniform circular motion  is the change in the direction of the velocity, but not its magnitude.
::必须指出,在圆圈中飞行的物体有一个常数,但没有一个常数速度。 这是因为方向是速度的一部分; 当物体改变方向时, 它正在改变其速度。 因此, 对象的加速度。 在统一旋转的情况下, 加速度是速度方向的变化, 而不是其大小 。

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For an object traveling in a circular path, the centripetal acceleration is directly related to the square of the velocity of the object and inversely related to the radius of the circle. 
::对于环绕路径飞行的物体,子宫加速度与物体速度的正方形直接相关,与圆半径反向相关。

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::ac=v2r

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Taking a moment to consider the validity of this equation can help to clarify what it means. Imagine a yo-yo. Instead of using it normally, let it fall to the end of the string, and then spin it around above your head. If we were to increase the speed at which we rotate our hand, we increase the velocity of the yo-yo - it is spinning faster. As it spins faster, it also changes direction faster. The acceleration increases. Now let's think about the bottom of the equation: the radius. If we halve the length of the yo-yo string (bring the yo-yo closer to us), we make the yo-yo's velocity greater. Again, it moves faster, which increases the acceleration. If we make the string longer again, this decreases the acceleration. We now understand why the relationship between the radius and the acceleration is an inverse relationship - as we decrease the radius, the acceleration increases, and visa versa.
::考虑一下这个方程的有效性可以帮助澄清它的含义。 想象一个 Yo-yo。 与其正常使用它, 让它掉到弦的尽头, 然后在你的头上旋转。 如果我们提高我们手旋转的速度, 我们就会提高yoyo的速度—— 它的旋转速度会更快。 当它转得更快时, 它也会更快地改变方向。 加速度会加快。 现在让我们想想方程的底部: 半径。 如果我们把Yo-yo字符串的长度( 离我们更近的 yo-yo ) 减半, 我们就会把Yo-yo 的速调更大。 再说一遍, 它会加快, 加速速度会加快。 如果我们再使弦更长, 这样会降低加速速度。 我们现在理解为什么半径与加速之间的关系是反向的- 当我们减少半径, 加速度会增加, 并且反之签证 。

 

 

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Examples
::实例

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Example 1  
::例1

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A ball at the end of a string is swinging in a horizontal circle of radius 1.15 m. The ball makes exactly 2.00 revolutions per second. What is its centripetal acceleration?
::字符串尾端的球在半径为1.15米的水平圆里摇摆着。球每秒精确地产生2.00次革命。它的子宫加速度是多少?

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We first determine the velocity of the ball using the facts that the circumference of the circle is  $\begin{array}{r}2\pi r\end{array}$ and the ball goes around exactly twice per second.
::我们首先利用以下事实来决定球的速度:圆环的环绕是2°r,球每秒转动精确两倍。

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$\begin{array}{r}v=\frac{(2)(2\pi r)}{t}=\frac{(2)(2)(3.14)(1.15\text{m})}{1.00\text{s}}=14.4\text{m/s}\end{array}$
:v)(2)(2-2)(r)t=(2)(2)(2)(3.14)(1.15米)1.00s=14.4毫秒/秒

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We then use the velocity and radius in the centripetal acceleration equation.
::然后我们使用子宫加速度方程中的速度和半径。

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$\begin{array}{r}{a}_c=\frac{v^2}{r}=\frac{(14.4\text{m/s}{)}^2}{1.15\text{m}}=180.{\text{m/s}}^2\end{array}$
::ac=v2r=( 14.4 m/s) 21.15 m=180. m/s2

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Example 2  
::例2

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The moon’s nearly circular orbit around the earth has a radius of about 385,000 km and a period of 27.3 days. Calculate the acceleration of the moon toward the earth.
::月球几乎环绕地球的环绕轨道的半径约为385,000公里,周期为27.3天。 计算月球向地球的加速度。 月球向地球的加速度。

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$\begin{array}{r}v=\frac{2\pi r}{T}=\frac{(2)(3.14)(3.85\times {10}^8\text{m})}{(27.3\text{d})(24.0\text{h/d})(3600\text{s/h})}=1020\text{m/s}\end{array}$
:二)rT=(2)(3)14(3.85×108米)(27.3 d)(240,000 h/d)(3600 s/h)=1020 m/s)

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$\begin{array}{r}{a}_c=\frac{v^2}{r}=\frac{(1020\text{m/s}{)}^2}{3.85\times {10}^8\text{m}}=0.00273{\text{m/s}}^2\end{array}$
::ac=v2r=(1020 m/s) 23.85×108 m=0.00273 m/s2

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As shown in the previous example, the velocity of an object traveling in a circle can be calculated by
::如上一个例子所示,一个物体在圆圆内飞行的速度可以由

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::v=2rT

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Where  $\begin{array}{r}r\end{array}$ is the radius of the circle and  $\begin{array}{r}T\end{array}$ is the period (time required for one revolution).
::r 是圆的半径, T 是周期( 一次革命所需的时间) 。

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This equation can be incorporated into the equation for centripetal acceleration as shown below.
::如下文所示,这一方程式可纳入子宫加速度方程。

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$\begin{array}{r}{a}_c=\frac{v^2}{r}=\frac{{\left(\frac{2\pi r}{T}\right)}^2}{r}=\frac{4{\pi }^{2}r}{T^2}\end{array}$
::ac=v2r= (2°rT)2r=4°2rT2

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Summary
::摘要

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• In order to keep an object traveling in a circular path, there must be an acceleration toward the center of the circle. This acceleration is called centripetal acceleration . 
  ::为了保持物体在圆形路径中飞行, 必须在圆心中心加速。 这种加速被称为子宫加速 。
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• The acceleration in the case of uniform circular motion changes the direction of the velocity but not its magnitude.
  ::在统一循环运动的情况下加速改变速度方向,但不改变速度的大小。
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• Formulas for centripetal acceleration are $\begin{array}{r}{a}_c=\frac{v^2}{r}\end{array}$  and $\begin{array}{r}{a}_c=\frac{4{\pi }^{2}r}{T^2}\end{array}$ .
  ::子宫加速度的公式为ac=v2r和ac=42rT2。

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Review
::回顾

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1. An automobile rounds a curve of radius 50.0 m on a flat road at a speed of 14 m/s. What centripetal acceleration is necessary to keep the car on the curve?
   ::汽车在平坦公路上以14米/秒的速度绕过半径50米的半径。 需要哪种子宫加速度才能使汽车保持曲线?
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2. An object is swung in a horizontal circle on a length of string that is 0.93 m long. If the object goes around once in 1.18 s, what is the centripetal acceleration?
   ::对象在0.93米长的字符串长度的水平圆上旋转。如果对象在1.18秒内旋转一次,则子宫加速度是多少?

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Explore More
::探索更多

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Use this resource to answer the questions that follow.
::使用此资源回答下面的问题 。

 

 

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1. What does centripetal mean?
   ::子宫胎是什么意思?
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2. What is uniform circular motion?
   ::什么是统一的循环动议?
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3. Why is centripetal acceleration always towards the center?
   ::为什么子宫加速 总是向中心?