Centripetal motion question: A what time on a clock do the hour and minute hand overlap between 3 and 4? Need work and to understand. h:mm:ss.s format (ex) They overlap between 4 and 5 at 4:21:49.1
Alrighty, first, do you know the angular speeds of each of the hands?
no, but I think I can work it out thats a good place to start
Is there a way to find the velocity without the radius?
Once you know those, you'll also need to think about the starting angles of the hands. I started at 3 o'clock. After that, we can set up an equation: \[\omega_{hour} t + \theta_{hour, starting} = \omega_{minute}t + \theta_{minute,starting}\] and solve for t.
I have the periods and frequencies, but I'm stuck from that.
Yeah, we only need the ANGULAR speed. The minute hand goes all the way around (2 pi radians) once every hour. The hour hand goes all the way around once every 12 hours (or 24, if you're using a 24 hour clock)
so would the minute hand be 2pi/h and hour be 2pi/12h to find the period? (Sorry if im not following well,this is a challenge problem where as everything else we've been doing has nothing to do with angular speed)
Those are the angular speeds. \[\omega_{hour} = \frac{2\pi}{12hr}\] \[\omega_{minute} = \frac{2\pi}{1hr}\]
And no worries. I struggled with angular motion at first, myself. It's not bad once you get the hang of it though, I promise :)
Okay I got those down c: Does theta represent the angle between 3 and 4 on a clock? (just making sure)
No, those are the starting angles of the hands. We'll start at 3 o'clock to make things more simple. At 3 o'clock, at what angle is the minute hand?
90 degrees
or pi/2 in radians
We need to be very careful about defining our angles in this problem. Normally, you'd be absolutely correct. However, the hands are moving clockwise, and we normally measure angles counterclockwise. So, we'll need to use -pi/2. What's the angle for the hour hand at 3 o'clock?
oh I see. pi?
because I assume it wouldnt be 0? im not sure if it flips or just becomes the negative
It's pi/2 ahead of the minute hand. So it'd be 0. The truth is, what we use for these two angles is arbitrary, as long as the hour hand is pi/2 ahead of the minute hand. |dw:1429760020632:dw|
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