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Circular Motion on a Clock Question

  1. Jul 7, 2010 #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 0.0 seocnds? At 15 seconds?
    c) Compute its change in veloicty between 0.0 and 15 seconds.
    d) Compute its average vector acceleration between 0.0 and 15 seconds.

    2. Relevant equations

    a = v^2/r = 4 (pi^2) (r) (f^2)

    3. The attempt at a solution

    a) v = 2.1 E -3 m/s

    b) @t = 0s ; v = 2.1 E -3 m/s [east]
    @t = 15s ; v = 2.1 E -3 m/s [south]

    c) Change in velocity = 2.1 E -3 [45 degrees S of E]

    d) a = 2.2 E -4 m/s^2 [45 degrees S of E]

    Are these the right answers?! Please help :)
     
    Last edited: Jul 7, 2010
  2. jcsd
  3. Jul 7, 2010 #2
    Are you familiar with any vector algebra? Can you think of a way to write a single equation that will give the vector (including direction) as a function of time?
    Does that sound like something you may have covered in class? It would require the use of sine and cosine functions.

    If not then yes your answers look fine for parts A and B.
    However, C is not that simple. Remember, change in anything is (final - initial).
    Can you subtract the two vectors and find the resultant?
     
  4. Jul 7, 2010 #3
    Uuuh, for the first part, I'm not quite sure...I'm not that good at physics, so we might have covered it but I'm not..recalling? D: How can I write B in a single equation that wil give the vector as a function of time?

    As for C, If the magnitude of the velocity stays the same, wouldn't (final - initial) equal zero?

    Thankyou, I appreciate your help! :)
     
  5. Jul 8, 2010 #4
    For C, no they wouldn't be zero. Draw the two vectors, remember you have one going East and the other going Sourth. Draw them both coming from the same origin, so a line from (0,0) to (-2.1E-3 , 0) is your east vector, and another from (0,0) to (0 , -2.1E-3) is your south vector.
    You subtract vectors by drawing a line from the tip of the original to the tip of the final vector. This line is the new vector, and is clearly not zero.

    As for B the equation would be:
    [tex]\vec{v} = 0.0021 [cos(- \frac{\pi}{30} t)\hat{i} + sin(- \frac{\pi}{30} t)\hat{j}][/tex]
    does that look familiar?

    From the way the problem sounds and the fact that you were using "east" and "south" instead of radians, I don't think you are there yet, so don't worry if you don't understand the above equation.

    So you had A and B right, for C just draw the vectors and it should be clear.
    Then for D, the direction of the vector should be the same as the direction of the vector in C.
    To find the magnitude just do the x and y components separately.
     
  6. Jul 8, 2010 #5
    Oh, okay, I think I understand. And no, I definitely have never seen that equation before :P
    So I tried C and D again and my new answers are:

    c) Change in velocity = 3.0 E -3 m/s [45 degrees S of W]
    d) Average acceleration = 3.1 x 10 E -4 m/s^2 [45 degrees S of W]

    Is this correct now? D:
     
  7. Jul 8, 2010 #6
    C looks right, but I'm not sure about your part D still.

    How did you go about finding that answer?
    For example, in the y direction you go from 0m/s to -2.1m/s so the acceleration in the y direction would be: (vf - vi) / t
    (-2.1m/s - 0m/s) / 15s = 1.4 E -4 m/s^2

    do the same for the x axis, and then those are your components of the acceleration, and you already know the direction (you had the direction right the magnitude is off), so all that is left is to find the magnitude. The method should be very similar to part C at this point.
     
  8. Jul 8, 2010 #7
    Oooohhh,
    Okay so would the average acceleration be 2.0 E -4 m/s^2 ?
     
  9. Jul 8, 2010 #8
    It would :D
     
  10. Jul 8, 2010 #9
    Yeessss : D
    Thank you so much ! I really appreciate it :)
     
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