2.8 角速率

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  角速度

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  To find a particular song on your MP3/MP4 Player, you may use a scroll wheel. This involves moving your finger around the wheel in a circular motion. Unfortunately for you, the song you want is near the very bottom of your songs list. Since these media players can often hold over 1,000 songs, you have to scroll fast! As you are moving your finger in a circle, you might wonder if you could measure how fast your finger is covering the distance around the circle.
  ::要在您的 MP3/ MP4 播放器中找到一首特定的歌曲, 您可以使用滚动轮。 这涉及到在旋转动作中将手指在方向盘上移动。 不幸的是, 您想要的歌曲离您的歌曲列表的底部很近 。 由于这些媒体播放器通常可以持有1,000多首歌曲, 您必须快速滚动 。 当您在圆圈中移动手指时, 您可能会怀疑您是否能够测量您的手指在圆圈周围的距离有多快 。

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  Watching your finger, you realize that your finger is moving around the circle twice every second. If the radius of the scroll wheel is 2 cm, what is the angular velocity of your finger as you scroll through your songs list? What is the linear velocity ?
  ::看着你的手指, 你意识到你的手指每秒都在环绕圆圈移动两次。 如果滚动轮的半径是 2 厘米, 在滚动在歌曲列表中时你的手指的角速度是多少? 线性速度是什么 ?

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  Angular Velocity
  ::角速度

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  You may already be familiar with the measurement of speed as the relationship of an object's distance traveled to the time it has been in motion. However, this relationship is for objects that are moving in a straight line. What about objects that are traveling on a circular path?
  ::您可能已经熟悉速度的测量, 因为一个对象的距离一直到它运动的时间。 但是, 这种关系是针对正以直线移动的物体。 那么在圆形路径上运行的物体呢 ?

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  Do you remember playing on a merry-go-round when you were younger?
  ::你还记得小时候玩旋转木马吗?

  

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  If two people are riding on the outer edge, their velocities should be the same. But, what if one person is close to the center and the other person is on the edge? They are on the same object, but their speed is actually not the same.
  ::如果两个人骑在外缘,他们的速度应该是一样的。但是,如果一个人靠近中间,而另一个人在边缘呢?他们是在同一物体上,但他们的速度实际上是不一样的。

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  Look at the following drawing.
  ::看看下面的图画

  

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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,小圆圈的点是向中间方向的人。 如果旋转木马的旋转正好是一次,那么两人也将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 nearly as far. 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 object. So the person on the edge has a greater linear velocity (recall that linear velocity is found using $\begin{array}{r}\text{distance}=\text{rate}\cdot \text{time}\end{array}$ ). If you have ever actually ridden on a merry-go-round, you know this already because it is much more fun to be on the edge than in the center! But, there is something about the two individuals traveling around that is the same. They will both cover the same rotation in the same period of time. This type of speed, measuring the angle of rotation over a given amount of time is called the angular velocity.
  ::然而,很明显, 中心的人没有近距离旅行。 该圆圈的环绕( 当然还有半径) 比较小得多, 因此在相同时间里走远的人实际上走得更快, 即使他们在同一天体上。 因此, 边缘的人具有更大的线性速度( 提醒注意线性速度是用距离=时间来发现的 ) 。 如果您曾经实际骑过旋转的轨道, 您已经知道这一点了, 因为它在边缘比在中间要有趣得多 。 但是, 有两个人环绕这个时间行走的人有些相同。 他们将在同一个时间段进行相同的旋转。 这种速度, 测量一个特定时间的旋转角度, 叫做角速度 。

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  The formula for angular velocity is:
  ::角速度的公式是:

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

  
  ::

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  $\begin{array}{r}\omega \end{array}$ is the last letter in the Greek alphabet, omega, and is commonly used as the symbol for angular velocity. $\begin{array}{r}\theta \end{array}$ is the angle of rotation expressed in , and $\begin{array}{r}t\end{array}$ is the time to complete the rotation.
  ::=================================================================================================================================================================时================================时完成旋转时=====

  

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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 point $\begin{array}{r}A\end{array}$ exactly 2 seconds to rotate through the angle, the angular velocity of $\begin{array}{r}A\end{array}$ would be:
  ::在这一绘图中, __ 完全是一个弧度, 或圆周周围半径的长度。 如果从 A 点 精确的 2 秒 旋转过角, A 的角速度将是 :

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

  
  ::t12 弧度/ 秒

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  In order to know the linear speed of the particle, we would have to know the actual distance, that is, the length of the radius. Let’s say that the radius is 5 cm.
  ::为了了解粒子的线性速度,我们必须知道实际距离,即半径的长度。让我们假设半径是5厘米。

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  If linear velocity 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, 则 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的长度,S=r,或半径长度乘以角的弧度。

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  Substituting into the formula for linear velocity gives: $\begin{array}{r}v=\frac{r\theta }{t}\end{array}$ or $\begin{array}{r}v=r\cdot \frac{\theta }{t}\end{array}$ .
  ::替换线性速度给定的公式: v=rt 或 v=rt 。

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  Look back at the formula for angular velocity. Substituting $\begin{array}{r}\omega \end{array}$ gives the following relationship between linear and angular velocity, $\begin{array}{r}v=r\omega \end{array}$ . So, the linear velocity is equal to the radius times the angular velocity.
  ::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 velocity is the same as the angular velocity.
  ::记住单位圆,半径为1单位,在此情况下,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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  Calculating the Linear and Angular Velocity
  ::计算线形和角速率

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  1. Lindsay and Megan are riding on a Merry-go-round. Megan is standing 2.5 feet from the center and Lindsay is riding on the outside edge 7 feet from the center. It takes them 6 seconds to complete a rotation. Calculate the linear and angular velocity of each girl.
  ::1. Lindsay和Megan骑着一辆旋转木马,Megan站在距离中心2.5英尺处,Lindsay在距离中心7英尺处的外部边缘,他们需要6秒钟才能完成旋转。计算每个女孩的直线速度和角速度。

  

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  We are told that 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 velocity is:
  ::我们被告知完成一个旋转需要6秒时间。 完全旋转与 2 弧度相同。 所以角速度是 :

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  $\begin{array}{r}\omega =\frac{\theta }{t}=\frac{2\pi }{6}=\frac{\pi }{3}\end{array}$ radians per second, which is slightly more than 1 (about 1.05), radian per second. Because both girls cover the same angle of rotation in the same amount of time, their angular speed is the same. In this case they rotate through approximately 60 degrees of the circle every second.
  ::t=263 弧度/秒,略高于1(约1.05),弧度/秒。因为两个女孩的旋转角度相同,时间相同,其角速相同。在这种情况下,她们每秒旋转大约60度。

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  As we discussed previously, their linear velocities are different. Using the formula, Megan’s linear velocity is:
  ::正如我们先前所讨论的那样,它们的线性速度是不同的。 使用公式,梅根的线性速度是:

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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/ 秒

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  Lindsay’s linear velocity is:
  ::Lindsay的线性速度是:

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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)(7)(%3) 7.3英尺/秒

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  2. A bug is standing near the outside edge of a compact disk (so that his radius from the center of the disc is 6 cm) that is rotating. He notices that he has traveled $\begin{array}{r}\pi \end{array}$ radians in two seconds. What is his angular velocity? What is his linear velocity?
  ::2. 虫子站在旋转的光盘的外部边缘(其半径从盘中半径为6厘米),他注意到两秒钟内已漂移弧度。他的角速度是多少?他的直线速度是多少?

  

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  We know that the equation for angular velocity is
  ::我们知道角速度的方程是

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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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  We can use the given equation to find his linear velocity:
  ::我们可以用给定方程 找到他的直线速度:

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

  
  ::v=r(6)(2)9.42厘米/秒

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  Solving for Unknown Values 
  ::解决未知值

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  How long does it take the bug in the previous problem  to go through two complete turns?
  ::前一个问题中的错误经过两个完整转弯需要多长时间?

  播放段落

  Since the angular velocity of the bug is $\begin{array}{r}\frac{\pi }{2}\end{array}$ radians per second, we can use the equation for angular velocity and solve for time:
  ::由于虫子的角速度是 2 弧度每秒, 我们可以使用方程式的角速度和解析时间 :

  播放段落

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

  
  ::

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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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  Since there are $\begin{array}{r}4\pi \end{array}$ radians in two complete turns of the disc, we can use this for the value of $\begin{array}{r}\theta \end{array}$ :
  ::由于圆盘两个完整转弯处有4弧度,我们可以用它来表示 :

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

  
  ::t=42=4228秒

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to find the angular and linear velocity of your finger. 
  ::早些时候,你被要求找到 手指的角速和直线速度。

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  As you found out in this section, the angular velocity is the change in angle divided by the change in time. Since you sweep around the circle twice in a second, this becomes:
  ::正如您在本节中发现的那样,角速度是角的改变,而角则随着时间的改变而除以。由于您在一秒内环绕圆圈翻了两次,这便变成:

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  $\begin{array}{r}\omega =\frac{4\pi }{1}=4\pi \end{array}$ rad/sec
  ::41=4rad/sec

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  Further, you can find the linear velocity with the equation:
  ::此外,你可以找到公式的线性速度:

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  $\begin{array}{r}v=r\omega =(2)(4\pi )=8\pi \approx 25.132cm/s\end{array}$
  :2)(4)=825.132厘米/秒

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  Example 2
  ::例2

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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 one of the smaller horses near the center. Lois’ horse is 3 m from the center of the carousel, and Doris’ horse is 7 m farther away from the center than Lois’. When the carousel starts, it takes them 12 seconds to complete a rotation.
  ::多丽丝和露意丝骑着外面的一匹马,露意丝骑着中间附近的一匹小马。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距离中心7米远。 当旋转木马开始时,他们需要12秒时间完成旋转。

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  Calculate the linear velocity of each girl. Calculate the angular velocity of the horses on the carousel. 
  ::计算每个女孩的直线速度。计算旋转木马的角速度。

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  It is actually easier to calculate the angular velocity first. $\begin{array}{r}\omega =\frac{2\pi }{12}=\frac{\pi }{6}\end{array}$ , so the angular velocity is $\begin{array}{r}\frac{\pi }{6}rad\end{array}$ , or $\begin{array}{r}0.524\end{array}$ . Because the linear velocity depends on the radius, each girl has her own.
  ::事实上,首先计算角速度比较容易。2126,所以角速度是6 rad,或0.524。由于线性速度取决于半径,每个女孩都有自己的。

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  Lois: $\begin{array}{r}v=r\omega =3\cdot \frac{\pi }{6}=\frac{\pi }{2}\text{or}1.57m/sec\end{array}$
  ::露意丝: v=r362 或1.57 m/sec

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  Doris: $\begin{array}{r}v=r\omega =10\cdot \frac{\pi }{6}=\frac{5\pi }{3}\text{or}5.24m/sec\end{array}$
  ::多丽丝: v=r106=53 或 5.24 m/sec

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  Example 3
  ::例3

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  The Large Hadron Collider 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于2008年开始运行,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的实验。 LHC环绕环绕约27 000米。 质子的速度将加速到非常接近光速的速度(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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    $\begin{array}{r}v=\frac{d}{t}\to 3\times {10}^8=\frac{27,000}{t}\to t=\frac{2.7\times {10}^4}{3\times {10}^8}=0.9\times {10}^{-4}=9\times {10}^{-5}\end{array}$ or 0.00009 seconds. $\begin{array}{r}\omega =\frac{\theta }{t}=\frac{2\pi }{0.00009}\approx 69,813rad/sec\end{array}$ The proton rotates around once in 0.00009 seconds. So, in one second it will rotate around the LHC $\begin{array}{r}1÷0.00009=11,111.11\end{array}$ times, or just over 11,111 rotations.
  ::v=dt3x108=27000tt=2.7x1043x108=0.9x10-4=9x10-5=9x10-5或0.009秒。t=200009}69,813 rad/sec 质子在0.00009秒内旋转一次。那么,一秒之内,质子将围绕10009=11111.111倍的LHC旋转,或略高于11,111次的旋转。

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  Example 4
  ::例4

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  Ted is standing 2 meters from the center of a merry go round. If his linear velocity is 6 m/s, what is his angular velocity?
  ::Ted正站在距离转弯曲的中间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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  Since the equation relating linear and angular velocity is given by $\begin{array}{r}v=r\omega \end{array}$ , we can solve for omega: $\begin{array}{r}\omega =\frac{v}{r}=\frac{6}{2}=3\end{array}$
  ::由于 v=r 给出的直线速度和角速度等式是由 v=r 提供的, 我们可以解析 omega : @ vr= 62= 3

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  Review
  ::回顾

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  Beth and Steve are on a carousel. Beth is 7 ft from the center and Steve is right on the edge, 7 ft 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。贝丝距离中心7英尺,史蒂夫就在边缘,距离中心7英尺远,比贝丝更远。用这些信息和下图解答问题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 velocity 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 velocities?
     ::贝丝和史蒂夫的线性速度是什么?
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  4. How far away from the center would Beth have to be in order to have a linear velocity of $\begin{array}{r}\pi \end{array}$ ft per second.
     ::贝丝离中心有多远 才能达到每秒 方英尺的线性速度
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  5. The carousel changes to a new angular velocity of $\begin{array}{r}\frac{\pi }{3}\end{array}$ radians per second. How long does it take to make a complete revolution now?
     ::旋转木马变换成新的角速为每秒 ++3 弧度。 现在要彻底革命需要多久?
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  6. With the carousel's new velocity, what are Beth and Steve's new linear velocities?
     ::与旋转木马的新速度, 什么是贝丝和史蒂夫的新线性速度?
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  7. Beth and Steve go on another carousel that has an angular velocity of $\begin{array}{r}\frac{\pi }{8}\end{array}$ radians per second. Beth's linear velocity 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个旋转木马,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 velocity 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 velocity of the minute hand on a clock? (in radians per minute)
     ::时钟上的分钟手角速度是多少? (以每分钟弧度计)
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  10. What is the angular velocity of the hour hand on a clock? (in radians per minute)
      ::时钟的小时手角速度是多少? (以每分钟弧度计)
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  11. A certain clock has a radius of 1 ft. What is the linear velocity of the tip of the minute hand in radians per minute?
      ::某个时钟的半径为 1 英尺。 分钟手端的直线速度是多少? 以每分钟弧度表示 ?
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  12. On the same clock, what is the linear velocity of the tip of the hour hand in radians per minute?
      ::在同一时钟上,小时手尖的直线速度是多少? 以每分钟的弧度表示?
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  13. The tip of the minute hand on another clock has a linear velocity of 2 inches per minute. What is the radius of the clock?
      ::另一个时钟的时钟的时钟一角线性速度为每分钟2英寸。时钟的半径是多少?
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  14. What is the angular velocity of the second hand on a clock? (in radians per minute)
      ::钟表上的二手角速度是多少? (以每分钟弧度计)
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  15. The tip of the second hand on a clock has a linear velocity of 2 feet per minute. What is the radius of the clock?
      ::时钟上的二手尖线性速度为每分钟2英尺。 时钟的半径是多少?

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  Review (Answers)
  ::回顾(答复)

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  ::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。