Optimizing Separation Rate in Rotating Systems

[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?

 

[Andrew Mason]

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?

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

 

[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...

 

[andrevdh]

... at the fastest rate? ...

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)?

 

[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.

 

[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.

 

[Andrew Mason]

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.

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

 

[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

 

[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.

 

[Andrew Mason]

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.

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

 

[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?

 

[nrqed]

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?

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

 

[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.

 

[Delzac]

Thx for the help guys!

 

[nrqed]

Thx for the help guys!

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?

 

[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 .

 

[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.

 

[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.

 

[Andrew Mason]

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

A very impressive effort to solve this rather quaint problem!

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

AM

 

[Andrew Mason]

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.

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

 

[andrevdh]

... the maximum rate of separation occurs when the angle of the line between the ends makes a 90 degree angle with the hour hand ... AM

Did you thought that based just on mathematical/physical insight (those who don't have to learn it the hard way, that is labour through the maths)?

 

[Andrew Mason]

Did you thought that based just on mathematical/physical insight (those who don't have to learn it the hard way, that is labour through the maths)?

Just intuition. Sometimes it works and sometimes it doesn't. I like to use intuition because I am lousy at math.

More intution: I suspect that there may be a way of proving that, regardless of the ratio of length of hands, the fastest rate of separation will occur when the angle of the line between hands is 90 degrees to the slower moving hand.

AM

 

[andrevdh]

My maths aren't that strong either. So I will leave it at that. I had enough.

 

[Andrew Mason]

My maths aren't that strong either. So I will leave it at that. I had enough.

Not quite for me.

There is another way of looking at this. Consider the length as a function of the angle between a and the hour hand which we will call $\beta$ and let the minute hand be n*R (R=length of hour hand). By the cosine rule:

$$n^2R^2 = a^2 + R^2 - 2aRcos\beta$$

which is a quadratic that has solution:

$$a = \frac{-2Rcos\beta \pm R\sqrt{4cos^2\beta +4(n^2-1)}}{2}$$

$$a/R = -cos\beta \pm \sqrt{cos^2\beta +(n^2-1)}$$

Then take the second derivative of a with respect to $\beta$

The second derivative is 0 when $cos\beta = 0$

Then it is just a matter of applying Pythagoras' theorem to determine the length of a and the angles/times.

AM

 

[andrevdh]

I think in planetary configurations this is called quadrature?

You made a little sign error (probably of no consequence to the final conclusion) - the b term should appear as -b in the quadratic solution formula.

It all made sense to me yesterday, but today I am a bit dense. Why would the max rate of separation occur when two planets are in quadrature with the sun? What is so special about this config?

 

[Andrew Mason]

I think in planetary configurations this is called quadrature?

You made a little sign error (probably of no consequence to the final conclusion) - the b term should appear as -b in the quadratic solution formula.

It all made sense to me yesterday, but today I am a bit dense. Why would the max rate of separation occur when two planets are in quadrature with the sun? What is so special about this config?

If the other planet is further from the sun the component of the speed of separation in the direction tangential to the Earth is greatest when the angle is 90 degrees from the Earth because, at this point, the Earth's entire tangential orbital speed is in line with the line between the planets. This is assuming the speed of the planets is constant (ciruclar orbit). It doesn't matter which planet is orbiting faster, since the motion is relative.

AM

 

[andrevdh]

Ahh, so the outer planet need to lag behind the Earth in its orbit (planets orbit anticlockwise viewed from north) and it will be on the observer's meridian (north-south celestial line) at sunset.

 

[Andrew Mason]

Ahh, so the outer planet need to lag behind the Earth in its orbit (planets orbit anticlockwise viewed from north) and it will be on the observer's meridian (north-south celestial line) at sunset.

Not necessarily lag behind - there just has to be a difference in angular speed. If you consider the frame of reference of the other planet (Mars) as 'fixed', the Earth appears to orbit in a circular orbit around the sun at an angular speed of $\omega_{Earth}-\omega_{Mars}$. When that orbital motion of the Earth is directly away from Mars (ie. when a tangent from the Earth's orbit from Earth passes through Mars), the rate of separation is fastest.

AM

 

[andrevdh]

When I first thought about the setup between the two rotating objects I only thought of the inner orbiting object (earth) moving towards the outer orbiting object (mars) like the two hands of the clock. In this config one actually get the maximum rate of approach because as you said the whole of the tangential speed contributes to the approach. That is how I realized that Earth need to be in front of Mars to get the maximum rate of separation (which stated as Mars need to lag behind). Come to think of it the actual angle between the two hands should then be 300 degrees and not 60 to get the maximum rate of separation.