# Homework Help: Puzzle On Motion In Circle

1. Jul 10, 2006

### Delzac

Hi all,

The minute hand of a church clock is twice as long as the hour hand. At what time after mid-night does the end of the minute hand move away from the end of the hour hand at the fastest rate?

I totally dun have a clue on how to do this any hints??

By thw way what is linear speed?

Last edited: Jul 10, 2006
2. Jul 10, 2006

### Andrew Mason

How would you express the distance between the ends of the hands in terms of the angle from 12 o'clock (let 12 o'clock be 0)? How would you find the rate of change of that function?

AM

3. Jul 11, 2006

### Delzac

erhh......$$S=R \theta$$ ?? do u need to use calculus for this QNs?(rate of change?) at my level i still do not know how to apply calculus in physic, but i learned it in math alrdy.......

Last edited: Jul 11, 2006
4. Jul 11, 2006

### andrevdh

I interpret this as "when the difference in speed between the two points is the greatest".

Now the speed of both points are different but constant (which you can calculate in terms of R - the length of the hour hand say, but I think it is not necessary to do this in order to answer the question though). It is just their direction that changes. So the question boils down to deciding how these two vectors should be orientated w.r.t. each other so that the difference vector has the greatest length. Then comes the hard part - at what time will the two vectors be at this orientation with respect to each other!

At what orientation of the "speed vectors" do you get the greatest difference between their motion (the "speed vectors" are always perpendicular to the radius)?

Last edited: Jul 12, 2006
5. Jul 12, 2006

### andrevdh

Two runners, A and B, are running in the same direction. Runnner A is running 3 m/s faster than B, or the difference between their motion is the difference vector $$V_{BA}$$ (the velocity of B relative to A). That means runner A sees runner B pulling away from him at a speed of 3 m/s in the indicated direction.

With the hands of the clock we get a similar difference in velocity. Here the direction of motion changes with time. So the two velocity vectors will point in different directions. To get the difference in motion between them you draw the two velocity vectors tails together and the difference in the motion is the vector pointing from the one head to the other.

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6. Jul 14, 2006

### andrevdh

This problem is so quaint that I thought I would post its solution to the benefit of our knowledge seekers.

The biggest relative speed difference between the two hands are when they are at an angle of $$180^0$$ with respect to each other as the diagram shows.

Using the constant angular speed of the two hands in radians per minute

$${\omega}_m = \frac{2\pi}{60} ; {\omega}_h = \frac{2\pi}{12 \times 60}$$

one get the time at which the speed difference is at a maximum

$${\theta}_m - {\theta}_h = \pi$$

$$({\omega}_m - {\omega}_h)t_{max} = \pi$$

which gives the required configuration of the clock hands 32 minutes 44 seconds after midnight. The subsequent times at which similar configurations would be obtained can be found by solving for angles of

$$u \pi$$

where u denotes uneven integer values.

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7. Jul 14, 2006

### Andrew Mason

I am not sure why you say the speed difference would be maximum when the angle between the hands is $\pi$. That gives the maximum separation, which means that the rate of change of separation is 0.

I would think that the maximum separation would occur when a line from the end of the minute hand to the end of the hour hand is perendicular to the hour hand. Proving that mathematically may be a bit difficult to do.

AM

8. Jul 14, 2006

### arunbg

Andrew, I think andrevdh's solution gives the time at which the velocity vectors of the two hands are changing the fastest, which is when they are anti parallel. I think that was what the OP's question was about too, and not about the rate of maximum separation.

You question, as you said, would be tougher to ponder and would certainly involve a lot of geometry :)

Cheers
Arun

9. Jul 14, 2006

### andrevdh

Andrew, if you look at my diagram of the clock hands the velocities of the hands are indicated by the two vectors at their ends. Their relative velocity is given by $$v_{mh}$$, which is the velocity of the minute hand relative to the hour hand. As the little vector addition alongside the clock displays it is a maximum at this configuration of the hands giving us their greatest rate of separation (also look at my post about the runners).

Oh, I now get what you are considering - the maximum rate of change in the length of the line joining the ends of the two hour hands.

Last edited: Jul 14, 2006
10. Jul 15, 2006

### Andrew Mason

The question is open to interpretation. If "move away from" "at the fastest rate" means "greatest difference in velocity", that would occur when the separation was $\pi$ radians as you have said. The problem is that at that point, they are not moving away from each other. An instant later they are approaching each other. Also the length of the hands has nothing to do with the problem.

AM

11. Jul 15, 2006

### andrevdh

The velocity vectors are always perpendicular to the radius vectors for circular motion. So when the hands are at 180 degrees their endpoints are moving in opposite directions with respect to each other as the diagram indicates . This gives us the greatest rate of separation between them. Unfortunately they are at their greatest separation as you remarked, so the rate of directional change is small. Otherwise one would naturally say you should get the greatest rate of change when they pass each other?

I agree with you that the problem is open to other interpretations. That is why I started my analysis with a statement of how I interpret the question (post #4).

What is your analysis of the problem?

Last edited: Jul 15, 2006
12. Jul 15, 2006

### nrqed

Did you use the fact that the minutehand is twice as long as the hour hand? It seems to me that this information suggests that they are are not looking for the maximum relative velocity but max relative speed. One would then write the position vector of the hour hand, the position vector of the minute hand, calculate the distance between them as a function of time and optimize that.

Just my impression.

Patrick

13. Jul 16, 2006

### andrevdh

Putting it that way makes one think of two planets in two different orbits and orbiting with different speeds. The question would then want the position of the planets when the outer planet is moving away from an observer on the inner planet at the highest rate. That is when the distance between them is changing at the highest rate. It then becomes a problem in which one need to rewrite the equations of motion for an observer on the inner planet. This might then have something to do with epicycles - the strange observed motion of planets among the stars.

14. Jul 16, 2006

### Delzac

Thx for the help guys!

15. Jul 16, 2006

### nrqed

Have you solved it?

I have written the expression to maximize and have started to calculate the derivative but it's quite a mess. Sounds straightforward but quite a mess. I did not finish it because I was not sure if that's what you needed or even if you were still interested.
What's your status on this problem?

16. Jul 17, 2006

### arunbg

Well Patrick, since you have begun working on that approach, I will post my comments on the same.

The function relating time and distance between the tips of the two hands of the clock is given by,
$$d^2=5x^2-4x^2cos((\omega_2-\omega_1)t)$$
where x stands for length of the hour hand, d the distance between the hands.
Differentiating twice and setting this equal to zero should give t .

17. Jul 17, 2006

### andrevdh

After the remarks by Andrew and nrged I thought I should investigate how the distance between the two endpoints of the hands changes. Here goes:

The angle between the two hands is given by

$$A = \theta_m - \theta_m = (\omega_m - \omega_h)t$$

giving

$$\dot{A} = \omega_m - \omega_h = \Delta \omega$$

using the cosine rule to determine the distance between the endpoints of the hands one get that

$$\frac{a}{R} = \sqrt{5 -4 \cos(A)}$$

(same as arunbg) time derivative of this comes to

$$\frac{1}{2R} \dot{a}= \frac{\dot{A}\sin(A)}{\sqrt{5 - 4 \cos(A)}}$$

which needs to be maximized.

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Last edited: Jul 17, 2006
18. Jul 19, 2006

### andrevdh

Differentiating the las equation again gets me to (I hope someone is checking this):

$$\frac{1}{2R\dot{A}^2} \ddot{a} = \frac{B\cos(A) - \frac{2\sin^2(A)}{B}}{B^2}$$

where

$$B = \sqrt{5 - 4\cos(A)}$$

multiplying the numerator with B and a little manipulation gives

$$\cos^2(A) - 2.5\cos(A) + 1 = 0$$

solving the quadratic shows that

$$A = 60^o\ !$$

so the greates rate of separation occurs when the angle between the hands are sixty degrees, which occurs 10 minutes 55 seconds after midnight.

Last edited: Jul 19, 2006
19. Jul 19, 2006

### Andrew Mason

A very impressive effort to solve this rather quaint problem!

I'll have a look at it tonight to check it.

AM

20. Jul 20, 2006

### Andrew Mason

This looks right. The second derivative is a bit of a challenge. As I suspected, the maximum rate of separation occurs when the angle of the line between the ends makes a 90 degree angle with the hour hand ($a^2 = 3R^2 = (2R)^2 - R^2$)!

AM