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Change in clock position

  1. Jan 13, 2014 #1
    1. The problem statement, all variables and given/known data
    What are a) average velocity and b) average acceleration of the tip of the 2.4cm long hour hand clock in the interval from noon to 6pm. Use unit vector to express, wth the x axis pointing towards 3 an y axis towards noon.

    2. Relevant equations

    3. The attempt at a solution

    V = deposition/dt

    P2 = -2.4j
    P1= 2.5j
    P2-p1 = -4.8
    -4.8e-3/21600= -2.22e-6j

    b) can I have a leg up?
    I know a = dv/dt but I can't build a mental geometrical set up.
  2. jcsd
  3. Jan 13, 2014 #2
    Check your answer for (a). You seem off by a degree of magnitude.

    For (b), you need initial and final velocities. What are they?
  4. Jan 13, 2014 #3

    A) it should be -4.8e-2/21600= -2.22e-6

    I'll attempt it procedurally:
    Vi = -2.22e-6 sin 90 j
    Vf -2.22e-6 sin (-90) j
  5. Jan 13, 2014 #4
    How did you obtain the initial and final velocities?
  6. Jan 13, 2014 #5
    At 12 noon, the minute hand is at pi/2
    Since this is purely vertical; vi = -2.22e-6 sin pi/2
    At 6pm, the minute hand is at -pi/2
    This is also purely vertical; hence, vf = -2.22e-6 sin(-pi/2)
  7. Jan 13, 2014 #6
    Does the tip of the hour hand move vertically at 12 and 6? And where does the magnitude of velocity come from?
  8. Jan 13, 2014 #7
    It comes from part (a).
    I presume it does since in moving from 12 to 6, the unit vector goes from +J to -J
  9. Jan 13, 2014 #8
    Part (a) is about the average velocity about these two positions. Part (b) requires instantaneous velocities at these two positions.
  10. Jan 13, 2014 #9
    I believe my part(a) is correct. The answer sheet confirms it.
    As for part(b), I do not know how to go about setting up the question.
    I know a = dv/dt but I do not know how should I find vf and vi

    Edit: Do I assume the velocity is the angular velocity or linear velocity?
  11. Jan 13, 2014 #10
    a = dv/dt
    v = ds/ dt
    ds = rΘ/dt=rω
    v = rω/dt
    ∴a = d(rw)/dt
  12. Jan 13, 2014 #11
    Check out again what Voko said in post #8. What is the instantaneous velocity vector at 6 pm (in terms of the unit vector in the x- direction)? What is the instantaneous velocity vector at noon? What is the difference between these two vectors?

  13. Jan 13, 2014 #12
    At 6pm: (0,-2.22e-6j)
    At 12pm (0,2.22e-6j)
  14. Jan 13, 2014 #13
    Part (b) needs linear velocity. But it is related to angular velocity. Again, what are the directions of the instantaneous velocity of the tip of the hour hand at 12 and 6? Image for a second that the hour hand rotates at a visible pace. Over a very short arc at 12 and 6, where is it going?
  15. Jan 13, 2014 #14
    Δv is headed towards the center
    v on the other hand is perpendicular to the hour hand.
  16. Jan 13, 2014 #15
    ##v_i## and ##v_f## are indeed perpendicular to the hour hand. But what are their magnitudes and signs?
  17. Jan 13, 2014 #16
    These are not the instantaneous velocities at those times. The instantaneous velocities of the tip of the hour hand at those times multiply i, not j. The angular velocity is 2π radians divided by 43200 seconds.
  18. Jan 13, 2014 #17
    Why i and not j?
    Why 43200s? If the hand moves from 12 to 6, then effectively, the change in radians is pi; therefore, v = rω
    r = 2.4cm A ω = pi/6(60x60) = pi/21600s
    v = 2.4cm x pi/6(60x60)

    Edit: I think you might be right with instantaneous velocity having an i direction.
    At 12pm, the instantaneous velocity has a +i unit vector; at 6pm, it has a -i unit vector.
    But my contention against 43200s and theta = 2pi stands.
    Last edited: Jan 13, 2014
  19. Jan 13, 2014 #18
    Is instantaneous velocity at 12pm = (-2.22j,0) and at 6pm = (2.22j,0) ?
  20. Jan 13, 2014 #19
    (2.22e-6i,0) and (-2.22e-6i,0) ?
  21. Jan 13, 2014 #20
    2π divided by 43200 is the same thing as π divided by 21600.
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