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Circular motion - pointers of a clock

  1. Oct 7, 2006 #1
    (a) What's the time between 9h and 10h in which the minute and hour pointers are equal? (b) After midday, what's the first time in which the three pointers are equal?

    For (a), I tried to do:

    [tex]\Theta = \Theta_0 + \omega_{hours} \cdot t[/tex]

    [tex]\omega = \frac{2 \cdot \pi}{T}[/tex]

    [tex]\Theta_0 = 270^o[/tex]

    [tex]\Theta = 270 + \frac{2\cdot \pi \cdot t}{60 \cdot 60 \cdot 12}[/tex]

    [tex]\Theta = \Theta_0 + \omega_{minutes} \cdot t[/tex]

    [tex]\Theta_0 = 0^o[/tex]

    [tex]\Theta = \frac{2 \cdot \pi \cdot t}{60 \cdot 60}[/tex]

    [tex]270 + \frac{2\cdot \pi \cdot t}{60 \cdot 60 \cdot 12} = \frac{2 \cdot \pi \cdot t}{60 \cdot 60}[/tex],

    But the result of this equation is much bigger than 1 hour. So what's wrong?

    I didn't try (b) yet but probably I need some clarification on (a) to give it a shot.
    Last edited: Oct 7, 2006
  2. jcsd
  3. Oct 7, 2006 #2


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    Staff: Mentor

    Between 9h and 10h, the minute hand must pass over the hour hand.

    On a standard 12 hr clock face, 9h represents 3/4 of a revolution and the 10 hr is 10/12 or 5/6 for a rotation. At 9 o'clock, the hr hand would be exactly on the 9 as the minute hand points vertically to the 12 (0 hr). Then as the minute hand sweeps from the 9 toward 10, it must pass the hr hand, which must be beyond 3/4's of the angle between 9h and 10h.

    The minute hand sweeps 360° in 60 minutes or one hour, but the hour hand sweeps only 30° in that same period or 1/12 of the 360°.

    Now if the minute hand sweeps some fraction f of a revolution (less than one hr), it must pass over hr hand.

    So let the minute hand sweep f * 360°, then the hour hand must sweep some fraction f of 30° from its starting point. Then equate the two expressions and solve for f.

    Then the time is just T = f * 60 minutes.

    Does this refer to a seconds hand, in addition to minute and hour hands?
  4. Oct 7, 2006 #3
    Thanks for the answer, but I actually found out where the error was.

    As I considered the angular velocity in rad/s, the initial angle should be in radians, but I used 270 where I should use 3*pi/2. By doing this substitution, I finished up with the same answer given by your method: 49 + 1/11 min.

    About (b), it's the point where the hour, minute and second hands are in the same place (as in midday). By equaling the three equations, I found that t = 0, so the answer should be midnight, right?
    Last edited: Oct 7, 2006
  5. Oct 7, 2006 #4


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    Staff: Mentor

    Certainly at 12:00 all three pointers are at 12. But the question (b) asks for the first time after midday.

    So after midday, the next hr is one o'clock so one has to find some time after 1 o'clock. The second hand sweeps one revolution in 1 minute, the minute hand sweeps one revolution in 1 hr.

    The 1 hr mark is at 30° or 1/12 of a revolution so the minute hand must be about 30° + 30°/12.
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