In theory, this should be easy

23 times ?

 

It depend upon when you start counting the 12 hours.

 

59x2--each click on the clock can have a right hand right angle and left handed one...

 

Yes. If you start counting at 3:00 or 9:00 for 12:00+epsilon hours, with epsilon positive as small as you want, it seems it makes 23 times. If epsilon is negative, that probably makes only 22.

 

Does it have an hour, minute as well as second hand? Or only hours and minutes?

 

OOps--forget previous answer...thinking 24 now...

 

I was thinking second hand before

 

Well, I got the first (3:00) yesterday before going to bed, and I stupidily was happy about myself, so I forgot the other one (9:00)

 

There are others as well.

 

Yes, so difficult the problem

 

I have used a "brute force" method, simply writing down the (approximate) times at which right-angles occur. Highlight the text below for the solution.

Also ... somebody may want to check if I missed any


There are 22 times where right-angles occur.

Note: times are approximate.

12:15
12:50
1:20
1:55
2:25
3:00
3:30
4:05
4:35
5:10
5:40
6:15
6:45
7:20
7:55
8:25
9:00
9:30
10:05
10:40
11:10
11:45

 

There is an analytic way to approach the problem--

The position vector of the big hand is given by (where t=0 is noon)

[tex]\mathbf{x_b}(t) = \sin \omega_b t \mathbf{i} + \cos \omega_b t \mathbf{j}[/tex]

while the position vector of the little hand is given by

[tex]\mathbf{x_l}(t) = \sin \omega_l t \mathbf{i} + \cos \omega_l t \mathbf{j}[/tex]

and you require that they are perpendicular to each other so that

[tex]\mathbf{x_b}(t)\cdot\mathbf{x_l}(t) = 0 \Rightarrow[/tex]

[tex]\cos\omega_b t \cos\omega_l t + \sin \omega_b t \sin \omega_l t = 0 \Rightarrow [/tex]

[tex]\cos (\omega_l - \omega_b)t = 0 \Rightarrow [/tex]

[tex](\omega_l - \omega_b)t_n = \pi n \Rightarrow [/tex]

[tex]t_n = \frac{\pi n}{\omega_l - \omega_b}[/tex]

but

[tex]\omega_b = 2\pi/T_b, \omega_l = 2\pi/T_l[/tex] where the T's are the periods.

So we have

[tex]t_n = \frac{n}{2} \frac{T_b T_l}{T_b - T_l}[/tex]
where
[tex]T_b = 12 h, T_l = 1 h[/tex]

 

Well its fairly easy to work out. angular velocity of each hand is simple.

Angular velocity of minute hand: [itex] \omega_m=\frac{2\pi}{3600} [/itex]
Angular velocity of hour hand: [itex] \omega_h = \frac{2\pi}{43200} [/itex]

Goverened by the rotational dynamics equations:

[tex] \theta = \theta_0 + \omega t[/tex]

We want:

[tex] \theta_m - \theta_h = (2n+1) \frac{\pi}{2} [/tex]

and we get:

[tex] t = \frac{(2n+1) \pi}{2(\omega_m - \omega_h)} [/tex]

First couple of answers are (assuming you start from 12:00):

12:16:21.82
12:49:05.45
..etc

EDIT: I was wondering why my tex had come out as vectors for a second.

 

Oh yeah woops I made a rookie mistake-- should have been (2n+1)/2 \pi as Kurt has it and not \pi n. Duh!

 

wow someone brought maths into it.
i'd say for the first hour
when the minute hand is 15mins past the hour hand
for the last hour
when the minute hand is 15mins befor the hour hand

for the 10 hours in between
at 15mins before and after the hour hand

10*2 + 2*1 =22?

 

Are we counting every second, minute or hour? LOL, J/K.

 

This is pretty easy stuff:


The big hand makes 11 laps relative to the small hand. There are 2 times per lap that it forms a right angle, so the number is 2*11=22.

 

What's with making the text invisible? It doesn't even work, the text is still visible.

 

To prevent spoilers to those who want to solve it themselves is the purpose. You have to use the correct colour however which is e3e3e3 I believe.

Edit: I have seen other forums with some sort of system that blacks out the text of spoilers which you have to mouse over to reveal. I wonder if that could be applied to this forum.

 

The correct color is now e1e1e2 (as of Dec. 2007)

https://www.physicsforums.com/showthread.php?t=100534

Color test for e1e1e2:

This is a test.

 

Comes out as e3e3e3 on photoshop for me. Both should work fine though.