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Kinematics (Confusing)

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A watch has a second hand 2.0cm long.
    a) compute the speed of the tip of the second hand.
    b) What is the velocity of the tip of the second hand at 0.0s and 15s


    2. Relevant equations

    No Idea...

    3. The attempt at a solution

    I drew a clock...but I don't even know where to start, any tips would be great.
     
  2. jcsd
  3. Feb 13, 2009 #2
    All you need to know is that the clock is circle and circles have a certain number of degrees or radians. From there you can determine the length the hand moves every second. Also s=rθ, ω=∆θ/∆t and v=rω.
     
    Last edited: Feb 13, 2009
  4. Feb 13, 2009 #3

    tiny-tim

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    Hi mattstjean! :smile:

    How long does it take the second hand to go round once?

    How far does the second hand move in that time? :wink:
     
  5. Feb 15, 2009 #4
    Thanks. It seems the simplest questions give me the hardest trouble. I always tend to over-complicate something that ends up being a lot more straight forward.
     
  6. Feb 15, 2009 #5
    Im not sure but the answer to the first one is pi/15 cm/sec

    The second one is (2qrt2)/15. I dont understand which way the velocity goes. Is it the way from the starting point? If it is then it is SouthEast?
     
  7. Feb 15, 2009 #6
    Well, it's not as simple as I thought ... I think I got the first part right. I did:

    r= 2cm (because the second hand is 2cm long and it is ask about from the tip)

    c=[tex]\pi\ [/tex] r 2
    c = 12.5663706 cm

    V = c/t
    V = 12.5663706cm / 60s
    V = 0.209439510 s

    So I said the speed of the tip of the second hand is 0.21 cm/s. Does that sound right?

    As for the second part. I don't really know how I should go about finding the velocity. I'm thinking it might be 0.21cm/s [E] because it's right at the top.

    Any tips for the velocity would be nice. I don't understand what the last guy tried to say.
     
  8. Feb 15, 2009 #7
    I'd rather be steered in the right direction with some hints and tips rather than given the answer. I'll die on my test without knowing how to arrive on the answer. Could you tell me how you get to that answer for the second one? And what does (2qrt2)/15 mean....
     
  9. Feb 15, 2009 #8
    For the second one I'm thinking that the velocity at 0s would be 0.21 cm/s [E] because it's at the very top and at 15s I think it's 0.21cm/s because it's at the very right.

    I was also asked to compute the change in velocity between 0.0 and 15 seconds and the average vector acceleration between 0.0 and 15 seconds.

    For the change in velocity I did:

    r= [tex]\sqrt{}0.21^2+0.21^2[/tex]
    r=0.3cm/s

    tan[tex]\theta[/tex] = 0.21/0.21
    [tex]\theta[/tex] = 45degrees

    so the change in velocity would be 0.3cm/s [45 degrees E of N]

    I don't really know what to do for acceleration...I'm assuming it is 0 because of the constant magnitude of the velocity ...

    Any help would be greatly appreciated,
    Matt
     
    Last edited: Feb 15, 2009
  10. Feb 15, 2009 #9
    i meant (2sqrt2)/15. OK what we know is that the average speed is distance/time. We can easily set the time to 60 seconds. The distance the second hand travels is the circumference because it makes a revolution every 60 seconds. To find the circumference we know radius is 2. And you hopefully know circumference formula. So we get 4pi/60 and that simplifies to pi/15 which is about .21 cm/sec.
     
  11. Feb 16, 2009 #10

    Office_Shredder

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    matt, your velocities look good, except:

    Considering your velocity went from pointing east to pointing south, I would imagine your change in velocity vector (taking the south and subtracting the east) would give you something pointing in the south-west direction.

    To get the average acceleration, just divide the change in velocity by the time that it takes to change
     
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