5.6 角速度

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  角速度

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  Introduction
  ::导言

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  Do you remember riding on a merry-go-round when you were younger?  If two people are riding on the outer edge, the  speed s at which they're moving should be the same. But what if one person is close to the center, and the other is on the edge? They are on the same object, but their speed is not the same. 
  ::你还记得你年轻时骑在旋转木马上吗?如果两个人骑在外缘,他们移动的速度应该是一样的。但如果一个人靠近中心,而另一个人在边缘呢?他们是在同一物体上,但他们的速度不同。

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  Suppose you are  standing 2.5 feet from the center, and a friend is on the outside edge 7 feet from the center. If it takes  6 seconds to complete a rotation, w hat is the speed that each of you is moving?
  ::假设你站在距离中心2.5英尺的地方, 朋友在距离中心7英尺的外部边缘。 如果完成一个旋转需要6秒时间, 你们每个人移动的速度是多少?

  

   

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  Angular  Speed
  ::角速度

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  Imagine the point on the larger circle is the person on the edge of the merry-go-round, and the point on the smaller circle is the person towards the middle. If the merry-go-round spins exactly once, then both individuals will also make one complete revolution in the same amount of time.
  ::想象一下大圆圈的点是旋转木马边缘的人,而小圆圈的点是向中间方向的人。 如果旋转木马旋转一圈,那么这两个人也会在同一时间进行一场完整的革命。

  

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  However, it is obvious that the person in the center did not travel as far as the person on the edge. The circumference (and of course the radius) of that circle is much smaller, and therefore the person who traveled a greater distance in the same amount of time is actually traveling faster, even though they are on the same rotating object. The person on the edge has a greater linear speed . But there is something about the two individuals traveling around that is the same: They both rotate at the same speed. This type of speed, measuring the angle of rotation over a given amount of time, is called the angular speed .
  ::但是,很明显,中间的人没有远到边缘的人。圆圈的环绕(当然还有半径)小得多,因此,在同一时间段中走远的人实际上走得更快,即使他们在同一旋转对象上。边缘的人有更大的线性速度。但这两个人环绕的线性速度是相同的:他们都以同样的速度旋转。这种速度在一定时间段上测量旋转角度,称为角性速度。

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     Angular  Speed
  ::角速度

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  $$\begin{array}{r}\omega =\frac{\theta }{t},\end{array}$$

  
  ::T, 来来来来来来去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去。

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  where omega  $\begin{array}{r}\omega \end{array}$ is the symbol for angular speed ,  $\begin{array}{r}\theta \end{array}$ is the angle of rotation expressed in radians, and $\begin{array}{r}t\end{array}$ is the time to complete the rotation.  
  ::以弧度表示的旋转角度。 t是完成旋转的时间。

   

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  In this drawing, $\begin{array}{r}\theta \end{array}$ is exactly one radian, or the length of the radius bent around the circle. If it took a particle at point $\begin{array}{r}A\end{array}$ exactly 2 seconds to rotate through the angle, the angular speed  of it would be:
  ::在此绘图中, __ 完全是一个弧度, 或圆圆周围半径的长度。 如果在 A 点 精确的 2 秒 上用一个粒子旋转角, 其角速将是 :

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  $$\begin{array}{rl}\omega & =\frac{\theta }{t}\\ \omega & =\frac{1}{2}\text{radians per second}.\end{array}$$

  
  ::12弧度每秒。

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  Learn, Play, and Explore with Angular Speed: 
  ::以角速度学习、玩耍和探索:

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  Linear Speed
  ::线性速度

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  In order to know the linear speed of the particle, we need the actual distance—that is, the length of the radius. Assume t he radius is 5 cm.
  ::为了了解粒子的线性速度,我们需要实际距离——即半径的长度。假定半径为5厘米。

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  If linear speed  is $\begin{array}{r}v=\frac{d}{t}\end{array}$ , then  $\begin{array}{r}v=\frac{5}{2}\end{array}$ or 2.5 cm per second.
  ::如果线性速度为 v=dt, 那么 v=52 或 2.5 厘米/ 秒 。

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  If the angle were not exactly 1 radian, then the distance traveled by the point on the circle is the length of the arc  $\begin{array}{r}s=r\theta ,\end{array}$  or the radius length times the measure of the angle in radians.
  ::如果角度不完全为1弧度,则圆圆点所穿行的距离是弧 s=r的长度,或者半径长度是角的弧度的倍数。

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  Substituting into the formula for linear speed  gives  $\begin{array}{r}v=\frac{r\theta }{t}\end{array}$ or $\begin{array}{r}v=r\cdot \frac{\theta }{t}\end{array}$ .
  ::将 v=rt 或 v=rt 改为线性速度的公式。

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  Look back at the formula for angular speed. Substituting $\begin{array}{r}\omega \end{array}$ gives the following relationship between linear and angular speed:   $\begin{array}{r}v=r\omega \end{array}$ . So the linear speed  is equal to the radius times the angular speed .
  ::回看角速度的公式。 替代 \\\ 给出线性和角速度之间的以下关系 : v=r 。 所以线性速度等于角速度的半径乘数 。

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  Remember in a unit circle the radius is 1 unit, so in this case the linear speed is the same as the angular speed .
  ::记住单位圆圆的半径为 1 单位, 在这种情况下, 线性速度与角速度相同 。

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  $$\begin{array}{rl}v& =r\omega \\ v& =1\times \omega \\ v& =\omega \end{array}$$

  
  ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}哦!

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  Here, the distance traveled around the circle is the same for a given unit of time as the angle of rotation, measured in radians.
  ::在这里,圆圆周围的距离与以弧度测量的旋转角度相同。

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     Linear  Speed
  ::线性速度

  $$\begin{array}{r}v=r\omega ,\end{array}$$

      where  $\begin{array}{r}v\end{array}$  is the linear speed, $\begin{array}{r}r\end{array}$  is the radius, and $\begin{array}{r}\omega \end{array}$  is the angular speed. 

    

   

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  Examples
  ::实例

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  Example 1
  ::例1

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  A bug is standing near the outside edge of a compact disk that is rotating. H is radius from the center of the disc is 6 cm. He notices that he has traveled $\begin{array}{r}\pi \end{array}$ radians in 2 seconds. What is his angular speed? What is his linear speed?
  ::窃听器正站在旋转的光盘的外部边缘。 他从盘中半径为 6 厘米。 他注意到他在两秒内已经行驶了 弧度。 他的角速是多少? 他的线性速度是多少 ?

  

  
  

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  Solution:
  ::解决方案 :

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  Step 1: The equation for angular speed is
  ::第1步:角速度方程式是

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  $\begin{array}{r}\omega =\frac{\theta }{t}=\frac{\pi }{2}\end{array}$ radians per second.
  ::2弧度每秒。

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  Step 2: Use the given equation to find his linear speed:
  ::第2步:使用给定方程式查找他的线性速度 :

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  $$\begin{array}{r}v=r\omega =(6)\left(\frac{\pi }{2}\right)=3\pi \approx 9.42\text{cm per sec}.\end{array}$$

  
  ::v=r(6)(2)=39.42厘米/秒。

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  Example 2
  ::例2

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  How long does it take the bug in Example 1 to go through 2 complete turns?
  ::例1中的错误需要多久才能经过2个完整转弯?

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  Solution:
  ::解决方案 :

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  Step 1:  Since the angular speed of the bug is $\begin{array}{r}\frac{\pi }{2}\end{array}$ radians per second, use the equation for angular speed and solve for time:
  ::第1步:由于虫子的角速度为 2 弧度/秒, 使用方程式的角速度和解析时间 :

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  $$\begin{array}{r}\omega =\frac{\theta }{t}\end{array}$$

  
  ::

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  $$\begin{array}{r}t=\frac{\theta }{\omega }\end{array}$$

  
  ::来来来来来去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去去

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  Step 2: Since there are $\begin{array}{r}4\pi \end{array}$ radians in 2 complete turns of the disc, use this for the value of $\begin{array}{r}\theta \end{array}$ :
  ::第2步:由于圆盘的2个完整转弯中有4弧度,因此用它来表示 :

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  $$\begin{array}{r}t=\frac{4\pi }{\frac{\pi }{2}}=4\pi \times \frac{2}{\pi }=8\text{ seconds}\end{array}$$

  
  ::t=424228秒

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  Example 3
  ::例3

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  Doris and Lois go for a ride on a carousel. Doris rides on one of the outside horses, and Lois rides on a smaller one near the center. Lois’s horse is 3 m from the center of the carousel, and Doris’ horse is 7 m farther away from the center than Lois’s. When the carousel starts, it takes them 12 seconds to complete a rotation.
  ::多丽丝和露意丝乘旋转木马去兜风。 多丽丝骑着外面的一匹马,露意丝骑着靠近中间的一匹小马。 露意丝的马距离旋转木马中心3米,多丽丝的马距离中间距离中心7米远。 当旋转木马开始时,他们需要12秒才能完成旋转。

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  Calculate the linear speed of each girl. Calculate the angular speed of the horses on the carousel.
  ::计算每个女孩的线性速度。计算旋转木马的角速度。

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  Solution:
  ::解决方案 :

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  Step 1: C alculate the angular speed 1st. $\begin{array}{r}\omega =\frac{2\pi }{12}=\frac{\pi }{6}\end{array}$ , so the angular speed is $\begin{array}{r}\frac{\pi }{6}\text{ radians}\end{array}$ , or $\begin{array}{r}0.524\end{array}$ . Because the linear speed depends on the radius, each girl has her own.
  ::第一步 1: 计算角速度 1st. 2126, 角速度为 6 弧度, 或 0. 524。 因为线性速度取决于半径, 每个女孩都有自己的 。

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  Step 2: Lois: $\begin{array}{r}v=r\omega =3\cdot \frac{\pi }{6}=\frac{\pi }{2}\approx 1.57\text{ m/sec}\end{array}$
  ::第2步:露意丝:v=r=r=%36221.57 m/sec

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  Step 3: Doris: $\begin{array}{r}v=r\omega =10\cdot \frac{\pi }{6}=\frac{5\pi }{3}\approx 5.24\text{ m/sec}\end{array}$
  ::步骤3:多丽丝:v=r106=535.24 m/sec

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  Example 4
  ::例4

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  Looking back to the Introduction, what is the angular speed of you and your friend?
  ::回想一下导言 你和你朋友的角速是多少?

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  Solution:
  ::解决方案 :

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  It takes 6 seconds to complete a rotation. A complete rotation is the same as $\begin{array}{r}2\pi \end{array}$ radians. So the angular speed is
  ::完成旋转需要6秒完成。 完全旋转与 2+ 弧度相同。 所以角速度是

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  $$\begin{array}{r}\omega =\frac{\theta }{t}=\frac{2\pi }{6}=\frac{\pi }{3}\end{array}$$

  
  ::T=263

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  radians per second, which is slightly more than 1 (about 1.05) radian per second. Because both of you cover the same angle of rotation in the same amount of time, your angular speed is the same. In this case you rotate through approximately 60 degrees of the circle every second.
  ::每秒的弧度略高于1(约1.05) 弧度每秒。 因为两位的旋转角度相同, 时间相同, 您的角速度相同 。 在此情况下, 您每秒旋转大约 60 度 。

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  Example 5
  ::例5

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  Return again to the problem in the Introduction. You are standing 2.5 feet from the center, and your friend is riding on the outside edge, 7 feet from the center. If it takes 6 seconds to complete a rotation, what is the speed of each person?
  ::回到引言中的问题。 您距离中心站着2.5英尺, 您的朋友正骑在外边缘, 距离中心7英尺。 如果完成旋转需要6秒时间, 那么每个人的速度是多少 ?

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  Solution:
  ::解决方案 :

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  As discussed previously, the linear speeds of you and your friend are different. Using the formula, your linear speed is
  ::正如先前讨论过的,你和你的朋友的线性速度不同。使用公式,你的线性速度是

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  $$\begin{array}{r}v=r\omega =(2.5)\left(\frac{\pi }{3}\right)\approx 2.6\text{ft per sec}.\end{array}$$

  
  ::v=r(2.5)(3) 2.6英尺每秒。

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  Your friend's  linear speed is
  ::你的朋友的直线速度是

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  $$\begin{array}{r}v=r\omega =(7)\left(\frac{\pi }{3}\right)\approx 7.3\text{ft per sec}.\end{array}$$

  
  ::v=r(7)(3) 7.3英尺/秒。

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    Example 6
  ::例6

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  The Large Hadron Collider (LHC) near Geneva, Switzerland began operation in 2008 and is designed to perform experiments that physicists hope will provide important information about the underlying structure of the universe. The LHC is circular with a circumference of approximately 27,000 m. Protons will be accelerated to a speed that is very close to the speed of light ( $\begin{array}{r}\approx 3\times {10}^8\end{array}$ meters per second).
  ::瑞士日内瓦附近的大型强子对撞机(LHC)于2008年开始运作,目的是进行物理学家希望能够提供关于宇宙基本结构的重要信息实验。 LHC环绕环绕约27 000米。 质子的加速速度将非常接近光速(每秒3×108米)。

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  How long does it take a proton to make a complete rotation around the collider? What is the approximate (to the nearest meter per second) angular speed of a proton traveling around the collider? Approximately how many times would a proton travel around the collider in one full second?
  ::质子在对撞器周围完全旋转需要多长时间?质子在对撞器周围飞行的角速是多少(接近每秒的米)?质子在对撞器周围飞行的角速是多少?质子在一整秒内绕对撞机飞行多少次?

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  Solution:
  ::解决方案 :

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  Part  1 : 
  ::第1部分:

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  $$\begin{array}{rl}v& =\frac{d}{t}\\ 3\times {10}^8& =\frac{27,000}{t}\\ t& =\frac{2.7\times {10}^4}{3\times {10}^8}=0.9\times {10}^{-4}=9\times {10}^{-5}\text{ or }0.00009\text{ seconds.}\end{array}$$

  
  ::v=dt3x108=27,000tt=2.7x1043x108=0.9x10-4=9x10-5或0.0009秒。

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  The proton rotates around once in 0.00009 seconds.
  ::质子在0.00009秒内旋转一次

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  Part 2:
  ::第2部分:

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  $$\begin{array}{r}\omega =\frac{\theta }{t}=\frac{2\pi }{0.00009}\approx 69,813\text{ rad/sec}\end{array}$$

  
  ::t=20000969,813 rad/sec

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  Part 3 :
  ::第3部分:

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  In one second it will rotate around the LHC,
  

  $$\begin{array}{r}1÷0.00009\approx 11,111.11\end{array}$$

  
  ::一秒钟后,它将环绕LHC旋转, 10000911,111.11。

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  times, or just over 11,111 rotations.
  ::或超过11,111个旋转。

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  Example 7
  ::例7

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  Ted is standing 2 meters from the center of a merry-go-round. If his linear speed is 6 m/s, what is his angular speed?
  ::Ted站在距旋转木马中心2米处 如果他的直线速度是6米/秒,他的角速度是多少?

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  Solution:
  ::解决方案 :

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  Since the equation relating linear and angular speed is given by $\begin{array}{r}v=r\omega \end{array}$ , we can solve for omega:

  $$\begin{array}{r}\omega =\frac{v}{r}=\frac{6\text{ m/s}}{2\text{ m}}=3\text{ radians per second}.\end{array}$$

  
  ::由于 v=r给定的直线速度和角速度等方程,我们可以为 omega 解析 : @vr= 6 m/s2 m= 3 弧度/秒 。

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  Summary
  ::摘要

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  • Angular speed:  $\begin{array}{r}\omega =\frac{\theta }{t}\end{array}$  where omega  $\begin{array}{r}\omega \end{array}$ is the symbol for angular speed,  $\begin{array}{r}\theta \end{array}$ is the angle of rotation expressed in radian measure, and $\begin{array}{r}t\end{array}$ is the time to complete the rotation.  
    ::角速度 : 是角速度的符号, 是以弧度量表示的旋转角度, t 是完成旋转的时间 。
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  • Linear speed:  $\begin{array}{r}v=r\omega \end{array}$  where  $\begin{array}{r}v\end{array}$  is the linear speed,  $\begin{array}{r}r\end{array}$  is the radius, and  $\begin{array}{r}\omega \end{array}$  is the angular speed.
    ::线性速度 : v=r = 表示线性速度, r 表示半径, r 表示角速度 。

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  Review
  ::回顾

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  Beth and Steve are on a carousel. Beth is 7 feet from the center and Steve is right on the edge, 7 feet further from the center than Beth. Use this information and the following picture to answer questions 1-6.
  ::贝丝和史蒂夫在旋转木马上。贝丝离中间有7英尺,史蒂夫就在边缘,距离中间有7英尺,比贝丝更远。用这些信息和下图回答问题1至6。

  

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  1. The carousel makes a complete revolution in 12 seconds. How far did Beth go in one revolution? How far did Steve go in one revolution?
     ::旋转木马在12秒内彻底革命。贝丝在一次革命中走多远?史蒂夫在一次革命中走多远?
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  2. If the carousel continues making revolutions every 12 seconds, what is the angular speed of the carousel?
     ::如果旋转木马继续每12秒进行革命 旋转木马的角速是多少?
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  3. What are Beth and Steve's linear speeds?
     ::贝丝和史蒂夫的线性速度是多少?
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  4. How far away from the center would Beth have to be in order to have a linear speed of $\begin{array}{r}\pi \end{array}$ feet per second?
     ::贝丝离中心有多远 才能达到每秒一英尺一英尺的线性速度?
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  5. The carousel changes to a new angular speed of $\begin{array}{r}\frac{\pi }{3}\end{array}$ radians per second. Now how long does it take to make a complete revolution?
     ::旋转木马变换成新的角速为每秒 ++3 弧度。 现在要进行彻底的革命需要多长时间 ?
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  6. With the carousel's new speed, what are Beth and Steve's new linear speeds?
     ::用旋转木马的新速度 贝丝和史蒂夫的新线性速度是什么?
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  7. Beth and Steve go on another carousel that has an angular speed of $\begin{array}{r}\frac{\pi }{8}\end{array}$ radians per second. Beth's linear speed is $\begin{array}{r}2\pi \end{array}$ feet per second. How far is she standing from the center of the carousel?
     ::贝丝和史蒂夫上另一只旋转木马,其角速为每秒8弧度。贝丝的线性速度是每秒2英尺。她距离旋转木马中心多远?
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  8. Steve's linear speed is only $\begin{array}{r}\frac{\pi }{3}\end{array}$ feet per second. How far is he standing from the center of the carousel?
     ::史蒂夫的直线速度只有每秒3英尺
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  9. What is the angular speed of the minute hand on a clock (in radians per minute)?
     ::时钟上的分钟手角速度(以每分钟弧度计)是多少?
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  10. What is the angular speed of the hour hand on a clock (in radians per minute)?
      ::时钟的小时手角速度(以每分钟弧度计)是多少?
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  11. A certain clock has a radius of 1 foot. What is the linear speed of the tip of the minute hand? (Assume the tip of the minute hand is 1 foot away from the center.) 
      ::某个时钟半径为 1 英尺。 分钟手端的线性速度是多少 ? ( 假定分钟手端距离中间 1 英尺 。 )
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  12. What is the linear speed of the tip of the hour hand if it is 1 foot from the center of another clock?
      ::时钟手尖的线性速度是多少? 如果距离另一个时钟中心1英尺,那么时间手尖的线性速度是多少?
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  13. The tip of the minute hand on a clock has a linear speed of 2 inches per minute. What is the radius of the clock?
      ::时钟上的时钟一角的时钟的线性速度为每分钟2英寸。时钟的半径是多少?
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  14. What is the angular speed of the second hand on a clock (in radians per minute)?
      ::时钟上的二手角速度(以每分钟弧度计)是多少?
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  15. The tip of the second hand on a clock has a linear speed of 2 feet per minute. What is the radius of the clock? ( Assume the second hand has the same length  as the radius .) 
      ::时钟上的二手端的线性速度为每分钟2英尺。 时钟的半径是多少? (假定二手的长度与半径相同 。)

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  Review ( Answers)
  ::回顾(答复)

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  Please see the Appendix.
  ::请参看附录。