节: 通知动议和内容分析 | 4.2 通知动议和分量分析

章节大纲

Converting between units is essential for mathematics and science in general. Radians are very powerful because they provide a link between linear and angular speed. One radian is an angle that always corresponds to an arc length of one radius. This will allow you to convert the rate at which you pedal a bike to the actual speed you can travel.
::单位之间的转换对于数学和一般科学来说是必不可少的。辐射仪非常强大, 因为它们提供了线性速度和角性速度之间的连接。一个弧度是一个角度, 总是与一个半径的弧长度相对应。这样可以将自行车的踏动速度转换为您能够行驶的实际速度。

\begin{aligned} 1 r e v o l u t i o n = 2 π r \end{aligned}

::1革命=2r

The gear near the pedals on a bike has a radius of 5 inches and spins once every second. It is connected by a chain to a second gear that has a 3 inch radius. If the second wheel is connected to a tire with a 17 inch radius, how fast is the bike moving in miles per hour?
::自行车上踏板附近的齿轮半径为5英寸,每秒旋转一次。它通过链条连接到另一架半径为3英寸的齿轮。如果第二轮与轮胎相连,半径为17英寸,那么自行车每小时以英里计移动的速度有多快?

Circular Motion and Dimensional Analysis
::通知动议和内容分析

Dimensional analysis just means converting from one unit to another. Sometimes it must be done in several steps in which case it is best to write the original amount on the left and then multiply it by all the different required conversions. To convert 3 miles to inches you write:
::度量分析仅仅意味着从一个单位转换为另一个单位。有时它必须在几个步骤中完成。在这样的情况下,最好在左侧写写原数,然后乘以所有不同的所需转换。要将3英里转换为英寸, 您必须写 :

$\begin{aligned} \frac{3 m i l e}{1} \cdot \frac{5280 f e e t}{1 m i l e} \cdot \frac{12 i n c h e s}{1 f o o t} = \frac{3 \cdot 5280 \cdot 12 i n c h e s}{1} = 190080 i n c h e s \end{aligned}$
::3英里1°5280英尺1英里1°12英寸1英尺=3°52°80°12英寸1英寸1 =190080英寸

Notice how miles and feet/foot can be used to convert miles to the desired unit of inches. Units, like numbers, can resolve to one when they exist in both the numerator and denominator ( $\begin{aligned} \frac{3}{3} = 1 \end{aligned}$ and $\begin{aligned} \frac{f e e t}{f e e t} = 1 \end{aligned}$ ). Also notice each conversion factor is the same distance on the numerator and denominator, just written with different units.
::注意如何使用英里和脚/英尺将英里转换为所需的英寸单位。单位,如数字,在分子和分母( 33=1 和 fefeet=1) 中存在时,可以确定一个单位。注意每个转换系数在分子和分母上是相同的距离, 仅用不同的单位写成。

Circular motion refers to the fact that on a spinning wheel points close to the center of the wheel actually travel very slowly and points near the edge of the wheel actually travel much quicker. Angular speed is the ratio of revolutions that occur per unit of time and linear speed is the ratio of distance per unit of time. While the two points have the same angular speed, their linear speed is very different.
::圆形运动是指,在靠近方向盘中心的一个旋转轮点上,转动轮点的实际飞行速度非常慢,而方向盘边缘附近的点实际飞行速度要快得多。角速度是每单位时间发生的革命比率,线性速度是每单位时间的距离比率。这两个点虽然具有相同的角速度,但其线性速度却大不相同。

Examples
::实例

Example 1
::例1

Earlier, you were given the following question: The gear near the pedals on the bike has radius 5 inches and spins once every second. It is connected by a chain to a second gear that has a 3 inch radius. If the second wheel is connected to a tire with a 17 inch radius, how fast is the bike moving in miles per hour?
::早些时候,你得到了以下的问题:自行车上踏板附近的齿轮每秒有5英寸半径和旋转一次。它通过链条连接到半径为3英寸的第二个齿轮。如果第二轮与半径为17英寸的轮胎连接,那么自行车每小时以英里计移动的速度有多快?

A bike has pedals that rotate a gear at a circular speed. The gear translates this speed to a linear speed on the chain. The chain then moves a second gear, which is a conversion to angular speed for the rear tire. This tire then converts the angular speed back to linear speed which is how fast you are moving. Instead of doing all these calculations in one step, it is easier to do each conversion in small pieces.
::自行车的脚踏板以圆形速度旋转齿轮。齿轮将速度转换为链条上的线性速度。链条然后移动第二个齿轮, 这是后轮胎的角速度转换为角速度。该轮胎然后将角速度转换为线性速度, 也就是你移动的速度。与其在一阶内进行所有这些计算, 不如在一阶内进行每一小块的转换。

First convert the original gear into the linear speed of the chain.
::首先将原齿轮转换为链条线性速度。

$\begin{aligned} \frac{1 r e v o l u t i o n}{1 s e c o n d} \cdot \frac{2 π \cdot 5 i n c h e s}{1 r e v o l u t i o n} = 10 π \frac{i n}{s e c} \end{aligned}$
::1革命1秒2英寸5英寸1 革命=10英寸9

Then convert the speed of the chain into angular speed of the back gear which is the same as the angular speed of the rear tire.
::然后将链的速度转换为后齿轮的角速度,与后轮胎的角速度相同。

$\begin{aligned} \frac{10 π i n c h e s}{1 s e c o n d} \cdot \frac{1 r e v o l u t i o n}{2 π \cdot 3 i n c h e s} = \frac{10}{6} \frac{r e v}{s e c} \end{aligned}$
::10英寸1 第二英寸1 革命23 英寸= 106revsec

Lastly convert the angular speed of the rear tire to the linear speed of the tire in miles per hour.
::最后,将后轮胎的角速转换为每小时英里的轮胎线性速度。

\begin{aligned} \frac{10 r e v}{6 s e c} \cdot \frac{2 π \cdot 17 i n}{1 r e v} \cdot \frac{1 f t}{12 i n} \cdot \frac{1 m i l e}{5280 f t} \cdot \frac{60 s e c}{1 m i n} \cdot \frac{60 m i n}{1 h o u r} \\ = \frac{10 \cdot 2 \cdot π \cdot 17 \cdot 60 \cdot 60}{6 \cdot 12 \cdot 5280} \frac{m i l e s}{h o u r} \\ \approx 10.1 \frac{m i l e s}{h o u r} \end{aligned}

::10 rev6 sec217英寸1 rev1 ft12 英寸1英里5280 ft60 秒1 min60 min1 min1 hour= 1021760_606125280 英里10.1 英里小时

Example 2
::例2

Suppose Summit High School has a circular track with two lanes for running. The interior lane is 30 meters from the center of the circle and the lane towards the exterior is 32 meters from the center of the circle. If two people run 4 laps together, how much further does the person on the outside go?
::假设高峰会高中有一个环形轨道,有两条车道可以运行。内陆车道距离圆圈中心30米,通往外部的车道距离圆圈中心32米。如果两个人一起跑四圈,外面的人还能走多远?

Calculate the distance each person ran separately using 1 lap to be 1 circumference and find the difference.
::计算每个人使用1圈的距离, 以1圈的宽度分别运行, 并找出差数。

\begin{aligned} \frac{4 l a p s}{1} \cdot \frac{2 π \cdot 30 m e t e r s}{1 l a p} & \approx 754 m e t e r s \\ \frac{4 l a p s}{1} \cdot \frac{2 π \cdot 32 m e t e r s}{1 l a p} & \approx 804 m e t e r s \end{aligned}

::4圈1230公尺1公尺1公尺 754公尺4公尺1232公尺1公尺804公尺

The person running on the outside of the track ran about 50 more meters.
::跑在赛马场外面的人跑了大约50米

Example 3
::例3

Andres races on a bicycle with tires that have a 17 inch radius. When he is traveling at a speed of 30 feet per second, how fast are the wheels spinning in revolutions per minute?
::Andres骑着自行车赛车,轮胎的半径为17英寸。当他以每秒30英尺的速度行驶时,车轮每分钟在革命中旋转的速度有多快?

Look for ways to convert feet to revolutions and seconds to minutes.
::寻找将脚转换为革命的方法,

$\begin{aligned} \frac{30 f e e t}{1 s e c o n d} \cdot \frac{60 s e c o n d s}{1 m i n u t e} \cdot \frac{12 i n c h e s}{1 f o o t} \cdot \frac{1 r e v o l u t i o n}{2 π \cdot 17 i n c h e s} = \frac{30 \cdot 60 \cdot 12 r e v o l u t i o n s}{2 π \cdot 17 m i n u t e} \approx 202.2 \frac{r e v}{m i n} \end{aligned}$
::30英尺60秒1分12英寸1英寸1英寸1英寸1英寸1英寸1英寸2 革命2 17英寸2 30英寸60英寸12 革命2 17分钟2 17 分钟202.2revmin

Example 4
::例4

When a car travels at 60 miles per hour, how fast are the tires spinning if they have 30 inch diameters?
::当一辆汽车以每小时60英里的速度行驶时,如果轮胎直径为30英寸,车胎的旋转速度有多快?

\begin{aligned} \frac{60 m i l e s}{1 h o u r} \cdot \frac{5280 f e e t}{1 m i l e} \cdot \frac{12 i n c h e s}{1 f o o t} \cdot \frac{1 r e v o l u t i o n}{2 π \cdot 15 i n c h e s} \cdot \frac{1 h o u r}{60 m i n u t e} \\ = \frac{60 \cdot 5280 \cdot 12 r e v o l u t i o n s}{2 π \cdot 15 \cdot 60 m i n u t e} \approx 672.3 \frac{r e v}{m i n} \end{aligned}

::60英里1小时1 5280英尺1英里12英寸1英尺1英寸1英寸1英尺1 革命2 *15英寸1小时60分 =60英里528012 革命2 15英寸60分672.3revmin

Example 5
::例5

Mike rides a bike with tires that have a radius of 15 inches. How many revolutions must Mike make to ride a mile?
::迈克骑着一辆轮胎半径15英寸的自行车。迈克必须进行多少次革命才能骑一英里?

$\begin{aligned} \frac{1 r e v}{2 π \cdot 15 i n} \cdot \frac{12 i n}{1 f t} \cdot \frac{5280 f t}{1 m i} \approx 672.3 \frac{r e v}{m i l e} \end{aligned}$
::1 rev215 in12 英寸 1 ft5280 ft1 mi 672.3revmile

Summary

Dimensional analysis involves converting from one unit to another, sometimes in multiple steps, by multiplying the original amount by required conversions.
::度量分析涉及从一个单位转换为另一个单位,有时采用多种步骤,按所需转换乘以原有数额。

Circular motion refers to the varying linear speeds of points on a spinning wheel, with points closer to the center moving slower than those near the edge.
::圆形运动指旋转轮上各点的线性速度不同,接近中心点的移动速度比靠近边缘点的移动速度慢。

Angular speed is the ratio of revolutions per unit of time, while linear speed is the ratio of distance per unit of time.
::角速度是每个时间单位的革命比率,线性速度是每个时间单位的距离比率。

Review
::回顾

For 1-10, use the given values in each row to find the unknown value $\begin{aligned} (x) \end{aligned}$ in the specified units in the row. Put your answers in terms of $\begin{aligned} π . \end{aligned}$
::对于 1- 10, 使用每行的给定值查找行中指定单位中的未知值 (x)。将答案以 __ 表示。

Problem Number ::问题编号	Radius ::半半径	Angular Speed ::角速度	Linear Speed ::线性速度
1.	$\begin{aligned} 5 i n c h e s \end{aligned}$ ::5 英寸 5 英寸	$\begin{aligned} 60 r p m \end{aligned}$ ::60分钟	$\begin{aligned} x \frac{i n}{m i n} \end{aligned}$ ::xinminxin 进度
2.	$\begin{aligned} x f e e t \end{aligned}$	$\begin{aligned} 20 r p m \end{aligned}$ ::20分钟	$\begin{aligned} 2 \frac{i n}{s e c} \end{aligned}$
3.	$\begin{aligned} 15 c m \end{aligned}$ ::15厘米 15厘米	$\begin{aligned} x r p m \end{aligned}$	$\begin{aligned} 12 \frac{c m}{s e c} \end{aligned}$
4.	$\begin{aligned} x f e e t \end{aligned}$	$\begin{aligned} 40 r p m \end{aligned}$ ::40 分钟	$\begin{aligned} 8 \frac{f t}{s e c} \end{aligned}$
5.	$\begin{aligned} 12 i n c h e s \end{aligned}$ ::12 英寸	$\begin{aligned} 32 r p m \end{aligned}$ ::32 rrpm 32 rrpm	$\begin{aligned} x \frac{i n}{s e c} \end{aligned}$
6.	$\begin{aligned} 8 c m \end{aligned}$ ::8分钟	$\begin{aligned} x r p m \end{aligned}$	$\begin{aligned} 12 \frac{c m}{m i n} \end{aligned}$
7.	$\begin{aligned} 18 f e e t \end{aligned}$ ::18英尺	$\begin{aligned} 4 r p m \end{aligned}$ ::4 分钟	$\begin{aligned} x \frac{m i}{h r} \end{aligned}$
8.	$\begin{aligned} x f e e t \end{aligned}$	$\begin{aligned} 800 r p m \end{aligned}$ ::800 分钟	$\begin{aligned} 60 \frac{m i}{h r} \end{aligned}$
9.	$\begin{aligned} 15 i n \end{aligned}$ ::15 英寸	$\begin{aligned} x r p m \end{aligned}$	$\begin{aligned} 60 \frac{m i}{h r} \end{aligned}$
10.	$\begin{aligned} 2 i n \end{aligned}$ ::2 英寸 2 英寸	$\begin{aligned} x r p m \end{aligned}$	$\begin{aligned} 13 \frac{i n}{s e c} \end{aligned}$

11. An engine spins a wheel with radius 5 inches at 800 rpm. How fast is this wheel spinning in miles per hour?
::11. 发动机旋转方向盘,半径为5英寸,每小时800分钟。这个方向盘以每小时英里的速度旋转多快?

12. A bike has tires with a radius of 10 inches. How many revolutions must the tire make to ride a mile?
::12. 自行车的轮胎半径为10英寸,轮胎必须进行多少次革命才能开一英里?

13. An engine spins a wheel with radius 6 inches at 600 rpm. How fast is this wheel spinning in inches per second?
::13. 发动机旋转方向盘,半径6英寸,每秒600分钟。

14. Bob has a car with tires that have a 15 inch radius. When he is traveling at a speed of 30 miles per hour, how fast are the wheels spinning in revolutions per minute?
::14. 鲍勃有一辆轮胎车,半径15英寸,当他以每小时30英里的速度行驶时,车轮在每分钟革命中旋转的速度有多快?

15. A circular track has two lanes. The interior lane is 25 feet from the center of the circle and the lane towards the exterior is 30 feet from the center of the circle. If you jog 6 laps, how much further will you jog in the exterior lane as opposed to the interior lane?
::15. 圆形车道有两条道,内陆车道离圆环中心25英尺,向外车道离圆环中心30英尺,如果跑6圈,在外车道上行走比向内车道多远?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::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。

章节大纲

Circular Motion and Dimensional Analysis ::通知动议和内容分析

Examples ::实例

Example 1 ::例1

Example 5 ::例5

Review ::回顾