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SOLUTION It is convenient in this problem to treat y as the independent variable and x as the dependent variable So we write the curve as x = y 1/3 Then, at position y, the curve is distance y 1/3 from the axis so the disk generated under rotation will have radius y 1/3 (Fig 817) Thus the disk will have area A(y) = [y 1/3 ]2 Also, since x ranges from 0 to 3 we see that y ranges from 0 to 27 As a result,

25:

FIGURE 817 Kinematic inversion of the mechanism to determine the motion in the second part of the cycle

break; } else if(stringCompare(answer, "Q", true) == 0) { break; } else if(stringCompare(answer, "M", true) == 0) { ConsoleWriteLine("Searching for another link"); } } else { ConsoleWriteLine("No link found"); break; } } while(linkLength > 0);

// Close the response respClose(); } while(uristr != null); } catch(WebException exc) { ConsoleWriteLine("Network Error: " + excMessage + "\nStatus code: " + excStatus); } catch(ProtocolViolationException exc) { ConsoleWriteLine("Protocol Error: " + excMessage); } catch(UriFormatException exc) { ConsoleWriteLine("URI Format Error: " + excMessage); } catch(NotSupportedException exc) { ConsoleWriteLine("Unknown Protocol: " + excMessage); } catch(IOException exc) { ConsoleWriteLine("I/O Error: " + excMessage); } ConsoleWriteLine("Terminating MiniCrawler"); } }

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while the roller retraces its path is governed by the pro le that prescribes the motion of the cam in the rst part of the cycle It is now clear that we can specify the motion of the cam in only half the cycle and that the remaining half is determined automatically, thus limiting control over the complete motion However, by studying the motion in the second half we can design the cam pro le to have better control over its motion Figure 817 shows a kinematic inversion of the basic arrangement of Fig 815, where B0A0A is the con guration during the initial traverse of the roller on the cam B0 A*A* is 0 the con guration when the roller returns to the same point during the reverse traverse on the cam The angle between A0B0 extended beyond B0 and a reference (horizontal) line on the xed link (ie, the cam in this inversion) f, indicates the rotation of the cam in the basic arrangement Figure 817 shows the same angle f* for the reverse traverse The following relationships can be observed from the gure: q * + q = 2p f * - f = 2h (815a) (815b)

Here is a short a sample session that begins crawling at McGraw-Hillcom (Remember, the precise output will vary over time as content changes)

0 27

Linking to http://mcgraw-hillcom Link found: http://stimcgraw-hillcom:9000/cgi-bin/query mss=search&pg=aq Link, More, Quit M Searching for another link Link found: http://investormcgraw-hillcom/phoenixzhtml c=96562&p=irol-irhome Link, More, Quit L Linking to http://investormcgraw-hillcom/phoenixzhtml c=96562&p=irol-irhome Link found: http://wwwmcgraw-hillcom/indexhtml Link, More, Quit L Linking to http://wwwmcgraw-hillcom/indexhtml Link found: http://stimcgraw-hillcom:9000/cgi-bin/query mss=search&pg=aq Link, More, Quit Q Terminating MiniCrawler

Let s take a close look at how MiniCrawler works The URI at which MiniCrawler begins is specified on the command line In Main( ), this URI is stored in the string called uristr A request is created to this URI and then uristr is set to null, which indicates that this URI has already been used Next, the request is sent and the response is obtained The

where h is the angle between the line joining the cam-pivot and the roller center, and the line joining the cam pivot and roller-crank pivot By differentiating Eq 814b with respect to input q, we get df * df dh =2 dq dq dq Eq (815) can be rewritten as df * dq * df dh =2 dq * dq dq dq (817) (816)

Part II:

y 5/3 5/3

.

Denoting df/dq as n and df*/dq* as n*, and using the fact that dq*/dq = -1, Eq (817) becomes n* + n = -2 dh dq

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