Jump to content

ahmeeeeeeeeeed

Senior Members
  • Content Count

    39
  • Joined

  • Last visited

Community Reputation

0 Neutral

About ahmeeeeeeeeeed

  • Rank
    Quark
  1. I may think that the limit (r^6 cos^9 (theta) sin (theta) ) / (r^4 cos^6 (theta) + sin^2 (theta)) has a problem of the sin^2 theta downwards , as it may become zero and the limit won't be zero / positive value any more but then it is 0/0 , so what to do next to know whether this (x/x) will turn eventually to be zero in the limit or not ?! And question two , is actually the question in the above comment : why does wolfram alpha say the value of the limit is zero despite the result of m/(1+m^2)^2
  2. Hello I have the limit lim (x^9 * y) / (x^6 + y^2)^2 (x,y)---> (0,0) when I use polar the final result is limit = lim (r^6 cos^9 (theta) sin (theta) ) / (r^4 cos^6 (theta) + sin^2 (theta)) r--->0 and substituting r = 0 , it will give zero * I tried it on wolfram alpha and it gave zero http://www.wolframalpha.com/input/?i=limit+%28x^9+*y%29%2F%28x^6+%2By^2%29^2++as++%28x%2Cy%29+--%3E+%280%2C0%29 but when I use cartezean and try the path y = mx^3 the result turns to be m/(1+m^2)^2 which depends on m so what is it ?!
  3. hello About Einestien's train example I can't get what it really does with the constancy of the speed of light , what would be the difference if , instead of two bolts , we have two balls thrown at the same angle ?! the observer in the train would still see the front ball first Or we are using light because it is what determines simultaneity in our eyes , which means when I see two balls passing by me at different times , I don't have to conclude they weren't thrown simultaneously but when I see Light doing that I will ? I need basically to know what difference does the constancy of the speed of light make here thanks in advance ;
  4. thanks for response , I put photos of my text book just to let you see the question just as I see it , and for my notes I'll explain it as ( sorry , I was answering just as I would answer in the exam. the question is asking for the velocity of the pin (B) at the instant (phi) equals 45 degrees the pin is B , and is attached to the arm AB , and allowed to move just in the circular groove , the circular groove itself moves in the (i cap) direction with a constant velocity = 60 m/s as the X velocity is constant = 60 m/s , the circular groove and also the pin have it so what I'm seeking is the velocity of the pin tangential to the circle ... I treated this problem as a joint kinematices problem , so I devided it into two parts the first part , I'm treating the pin only as a part of arm( AB) , and so I used polar co-ordinates R cap and theta cap where V = dR/dt in the direction of R cap , and R* omega in the direction of theta cap = (.3)* omega since the length of arm AB is constant then dR/dt = zero and the (theta cap) velocity is R * omega We also have the X velocity of the pin so V = (.3 * omega) theta cap + 60 (i cap) ----------------> equation one ----- secondly I treated the pin as part of the circular part we see that both the groove and the pin move in the positive x direction with the same speed , so we can neglect this speed and consider the motion of the pin just circular motion I chose intrinsic co-ordinates with ( tcap) tangential and (n cap) pointing towards the center , I could also use polar again with this . We see that V in the direction tangential in the only velocity compnent here , and it is what I seeking then I resuluted the unit vectors (t cap) and (i cap) in the directions of R cap and theta cap using the geomitry of the figure , inorder to equate like quantities , that resulted in the tangential speed equals (- 61.5) t cap which means opposite in sense to the unit vector (t cap) , which also means that the pin is going up ! .. I can't find this logical as i think under this situation the pin should go down the groove not up the groove and here came my question here . sorry I can't understand this " putting off some of the heavy hitters in maths" Yes , A is stationary , B is the pin , the question as asking for the tangential velcoity of the pin , but of cource getting the \omega of the pin in the half -circular groove would give us the tangential speed .. I'm using polar co-ordinates for arm AB as i said above.. yes the x velocity of the pin is the same value (60 m/s) but I'm seeking the tangential velocity also. that's what I expect , but my way of answering showed that at the instant phi = 45 the tangential velocity well be up not down , showing that phi ( at that instance) is seeking to increase not decrease , which I think is wrong and that's why I asked the question from the beginning ... --------------------------- so what I am seeking now is would you please tell me how I could find the tangential velocity of the pin B at the instant phi = 45 degrees from the givings in the textbook ? I want both its magnitude and direction , and I really want to help me find where is the mistake in my calculations ? thanks in advance well this is not mine if this is what you mean http://www.scienceforums.net/topic/65674-crank-slider-mechanism-velocity-of-piston-angular-velocity-of-link-hp-needed/
  5. Hello , here is a problem http://www.flickr.com/photos/76599498@N07/7070964675/ when I solve it , the pin turns to be going upwards , which I can't find logical , what do you see ,please ? here's my answer http://www.flickr.com/photos/76599498@N07/7070966251/ http://www.flickr.com/photos/76599498@N07/7070966875/
  6. Who said that x is in terms if time ? how ?
  7. no here the threading of nuts at 3:45 , but I was asking about very tiny nuts , but I think no problem in manufacturing a small machine of the same type
  8. But it gave me the Y in terms of X , not the position in terms of time
  9. thanks, I found a vedio on youtube that die forging is the used method
  10. Hello at this problem the car is moving around the circle with constant speed , and it is asking for d^2R/dt^2 for the radar And d^2(theta)/dt^2 for the radar at the instance the speed of car is vertical downwards as shown the speed equals 120 km/hour And the radius of the circle is 450 so I said a® = (d^2R/dt^2) - R (omega^2) but the only acceleration is the central which equals (v^2/ Radius) and in the same direction so we can get d^2R/dt^2 . and when I try to get the (alpha)(d^2 theta / d t^2) a (theta) = R*(alpha) + 2 (omega)* (dR/dt) and a(theta) equals zero and here is my question , what is the value of dR/dt ??? I think it has to be zero , since the velocity is vertical and the unit vector R cap is horizontal so 1- Is Alpha zero ? 2- can dR/dT equal zero at that situation , while d^R/dt^2 has a value ? I think no problem but I wanted to make sure I'm right in this solution
  11. hello, a problem says http://www.flickr.com/photos/76599498@N07/6895404760/ I couldn't know how to start but that's what I did http://www.flickr.com/photos/76599498@N07/6895405486/ http://www.flickr.com/photos/76599498@N07/6895406078/ http://www.flickr.com/photos/76599498@N07/6895403992/ but I don't think this is the way , since I got the acceleration at a special point . I want to ask some questions 1-the acceleration magnitude is variable , due to the variation of the radius od curv. , isn't it ? If so , what is meant by determining the acceleration without specifing a certain moment ? 2- this problem is under the title ( intrinstic co-ordinates) , If I were to use these co-ordinates what is the key to start with ? I can think of the tangential unit vector to be the same as the direction of the derivative of Y with respect to X , but how to employ this ? 3- the radius of curv. at some points turned to be negative , is that ordinary or did I just did some calculation wrong ?
  12. Hello, A Question says Assign a proper manufacturing process for nuts and bolts Can I say die forging then threading or die casting then threading How are they mass produced ? and is there tiny lathes to thread nuts ?
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.