A clock problem [equation solving]

[mech-eng]

Homework Statement

[/B]
At what time after 4:00 will the minute hand and the hour hand of a clock first be in the same position?

Homework Equations

Angular Velocity_{minute hand}=12w
Angular Velocity_{clock hand}=1w[/B]

The Attempt at a Solution

For attempting a solution first I should understand the question. What does "same position" mean here? They would be in the same position another day i.e after 24 hours later. If I can understand this part I can try an attempt.

Source: Algebra and Trigonometry by Keedy/Bittinger.

Thank you.[/B]

 

[BvU]

Same position means both hands at the same angle wrt the 12 o'clock. E.g. at 12:00

Or a little bit after one o'clock

 

[mech-eng]

Then the answer to the question should be 24 hours later because only after 24 hours later clock hand will be at 04:00 and when it is that position minutes hand will be at 12. Is that correct?

Thank you.

 

[Ray Vickson]

Then the answer to the question should be 24 hours later because only after 24 hours later clock hand will be at 04:00 and when it is that position minutes hand will be at 12. Is that correct?

Thank you.

No, it is not correct. Read the question again. What are the relative positions of the two hands at 1:00? What are their relative positions at 1:30? What do the two answers tell you?

 

[BvU]

Same position means pointing in the exact same direction: big hand right on top of the small hand !

 

[mech-eng]

Same position means both hands at the same angle wrt the 12 o'clock. E.g. at 12:00
Or a little bit after one o'clock

I understand the example but what does "wrt" mean? I have searched it in google but haven't found an explanation for it. I misunderstood same positon thinking it just the same with previous positions which requires 24 hours or one full revolution.

No, it is not correct. Read the question again. What are the relative positions of the two hands at 1:00? What are their relative positions at 1:30? What do the two answers tell you?

Does it tell me their relative positions should be the same i.e zero angle between?

Same position means pointing in the exact same direction: big hand right on top of the small hand !

I.e when they coincide?

1. I should calculate the angle between them at that instance. The angle between every hour is 30 degrees so there is 120 degree between them.
Angular Velocty of minute hand= 360/60=6 degrees/sec.
Angular Velocity of clock hand=360/12*60=1/2 degree/sec.
The Relative Velocity=5.5 degrees/sec.

120 degrees/5.5 degrees/sec=24 seconds. At the time 04:24 the minute hand and clock hand coincide.

Is above correct what are other more practical ways?

Thank you.

 

[BvU]

No it is not correct.
and

so there is 120 degree between them.

is correct at 16:00, but at e.g. 16:24 the hour-hand has moved further !

Your rounding off is disastrous. Don't do it unnecessarily.
And: if you divide degrees by minutes, you get degrees/minute, not degrees/sec.

 

[mech-eng]

No it is not correct.
and
is correct at 16:00, but at e.g. 16:24 the hour-hand has moved further !
And: if you divide degrees by minutes, you get degrees/minute, not degrees/sec.

I understand above but I do not understand the following

Your rounding off is disastrous. Don't do it unnecessarily.

Would you please explain it. I am on a new try.

Thank you.

 

[mech-eng]

A new try: The degree between the minute hand and hour hand is 120 degrees. A clock hand completes a full revolution in 12 hours and a minute hand completes a full revolution in 60 minutes.

Angular velocity of the clock hand=$\frac{360 degrees} {12 \times 60 \times 60}$=1/120 degree/sec.
Angular velocity of the minute hand=$\frac{360 degrees}{60 \times 60}$=1/10 degree/sec.

The formulation should be $120degrees+ (\frac 1 {120})t$=1/10t

If we put t alone one side we get t=1309.09 seconds or 21.81 minutes. This means they will coincide when the time 4:21.

Thank you.

 

[haruspex]

21.81 minutes

Yes.

4:21.

That's not the nearest whole minute.

 

[BvU]

Much better indeed ! Well done. Basically one hand rotates at an angular speed of 1/12 revolutions/hour and the other at 1 rph. So the difference is 11/12 and they will be on top of each other once every 12/11 hour. for the fourth time after 12 that will be at 48/11 hour = 4.3636 hour = 4$^\circ$21.81'=4$^\circ$21'49", rounded off to 4 h 22'

 

[mech-eng]

Yes.

That's not the nearest whole minute.

A bad round-off? Even 21.81 minutes migth not be meaningful. We might be in a different counting system as 60 instead of 10.. So in this part I am confused 21.81 minutes surely meaningless. Would you help at this stage?

Thank you.

 

[haruspex]

21.81 minutes migth not be meaningful.

No, it is quite meaningful. 4.5 hours is meaningful, and is the same as 4:30 and 4h30m and 4°30'.

We might be in a different counting system as 60 instead of 10

Unless specifically stated, the decimal point is always decimal, no confusion. Other systems use clearly different notation, as above and in BvU's post. You may find it ugly, but there is no rule against mixing notations, such as 4°30.5'=4°30'30".

 

[mech-eng]

No, it is quite meaningful. 4.5 hours is meaningful, and is the same as 4:30 and 4h30m and 4°30'.

Unless specifically stated, the decimal point is always decimal, no confusion. Other systems use clearly different notation, as above and in BvU's post. You may find it ugly, but there is no rule against mixing notations, such as 4°30.5'=4°30'30".

I am confused 21.81 minutes. is 21 part minute 81 part second? it is clear that it requres some rounding according to multiples of 60. We do not say 21.81 minutes because 81 is greater than 60.

Thank you.

 

[BvU]

No. 21.81 means 21 + 81/100 minutes. 81/100 $\approx$ 49/60

 

[mech-eng]

Then is it also 21 minutes plus 0.81*60=48.6 rounding to 49 seconds?

And then is the answer 04:21:49 in terms of hour, minute and second?

Thank you.

 

[haruspex]

Then is it also 21 minutes plus 0.81*60=48.6 rounding to 49 seconds?

And then is the answer 04:21:49 in terms of hour, minute and second?

Thank you.

Yes, that is the value to the nearest second.

 

[mech-eng]

So the difference is 11/12 and they will be on top of each other once every 12/11 hour

Would you explain this part of your approach? Yes the difference is 11/12 rph but how do you know they will be on top of each other once every 12/11 hour by just reversing the difference?

 

[BvU]

Would you explain this part of your approach? Yes the difference is 11/12 rph but how do you know they will be on top of each other once every 12/11 hour by just reversing the difference?

one hand rotates at an angular speed of 1/12 revolutions/hour and the other at 1 rph. So the difference is 11/12 and they will be on top of each other once every 12/11 hour.

Big hand is on top of small hand at 12:00.
Big hand does 1 rph, so in 12/11 h it rotates 12/11 = 1/11.
Small hand does 1/12 rph, so in 12/11 h it rotates 1/11 and therefore is on top of small hand again.
Repeat every 12/11 h.

 

[mech-eng]

Big hand is on top of small hand at 12:00.
Big hand does 1 rph, so in 12/11 h it rotates 12/11 = 1/11.
Small hand does 1/12 rph, so in 12/11 h it rotates 1/11 and therefore is on top of small hand again.
Repeat every 12/11 h.

But is this topic modular arithmatic or least common multiple? You saw this very fast this should be belonged to a topic.

Thank you.

 

[BvU]

A simple difference in rotation speeds means the relative angle changes to the tune of 11/12 rph. So one full turn takes 12/11 h.

 

[SammyS]

A bad round-off? Even 21.81 minutes might not be meaningful. We might be in a different counting system as 60 instead of 10.. So in this part I am confused 21.81 minutes surely meaningless. Would you help at this stage?

Thank you.

Well, you could be in base eleven. The answer works out nicely.

In decimal (base ten), you can write the answer as a fraction - in this case a "mixed number"- or as a repeating decimal.

$21\frac9{11}\$minutes or $\displaystyle \ 21.\overline{81}$minutes

after 4:00, of course.