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- amilapsn

when does the minute hand and hour hand come one on one after 12 o' clock?

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- amilapsn

when does the minute hand and hour hand come one on one after 12 o' clock?

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- ParthKohli

|dw:1433320516438:dw|It's around 1:05. You can use math to get a more precise answer.

- amilapsn

thnk you

- ParthKohli

Let's say it's 1 hr and \(x\) min. What does it mean when two hands coincide? How do we put it mathematically? It means that we can make an equation. What do we equate? The angles. We equate the angles.
At 1 hour and \(x\) minutes, the angle created by the hour-hand wrt 12 is:\[30^{\circ}\left( 1+\dfrac{x}{60}\right)\]and that created by the minute hand wrt 12 is:\[360^{\circ }\cdot\dfrac{x}{60}\]Equate those two and you get\[x = \frac{60}{11} = 5.4545\cdots\]

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- ParthKohli

Therefore, the time is 1 hour and 5.4545... minutes.
Or around 1:05:27.

- ganeshie8

is it possible to model the situation wid vectors

- dan815

ye

- dan815

we know both angular velocity so

- dan815

u can write position vectors for both and equate them together

- ParthKohli

oh wow

- ganeshie8

t is in hours :
minute hand :
hour hand :
how to solve

- ganeshie8

wolfram is giving up
http://www.wolframalpha.com/input/?i=solve+cos%282pi*60*t%29%3Dcos%282pi*1*t%29%2C+sin%282pi*60*t%29%3Dsin%282pi*1*t%29%2C+0%3Ct%3C2

- dan815

i think its better to write it out as arc length wrt to time

- dan815

and equate the 2 sinusoids and u shud get 12 solutions which is easy to see from the clocks and the periods that are forming

- ParthKohli

\[\rm \omega_{second}= 2\pi / T = 2\pi ~rad / 60 sec = \pi/30 ~rad ~sec^{-1}\]\[\rm \omega_{hour} = 2\pi / T = 2\pi ~rad / (60\times 60)sec = \pi /1800 ~ rad~sec^{-1}\]We equate the positions now? (The position of second hand can be a multiple of that of the hour hand.)

- dan815

|dw:1433321503977:dw|

- ParthKohli

Oh, whoops...

- ParthKohli

\[\rm \omega _{hour} = \frac{2\pi}{60 \times 60 \times 24} ~rad ~ sec^{-1}\]

- dan815

i guess we can just jump straight to arc angles since arclength/r is a constant

- ParthKohli

I mean this isn't any different from what I did earlier. We're equating angles and using fancier terms. :P

- dan815

ya its same thing lol, oky lets try to see vectors then

- dan815

|dw:1433321910738:dw|

- dan815

same simplifcation in polar form

- dan815

its not really ideal to solve in cartesian form so

- dan815

thats why wolfram giving up

- ganeshie8

https://www.desmos.com/calculator/k3awg1pnpm

- dan815

you solving sec and min there eh

- ganeshie8

Oh right :/

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