Brain Teasers ? or 50 Years On........... ?

Don

I think I've got the sums right this time - but sadly the logic is fundamentally flawed ! Essentially I have re-packed the area but whilst this retains the symmetry about the x-axis, it no longer honours the "balance of masses" of the remaining areas.

Maybe time for me to slip back into retirement ???

No - I mustn't admit defeat this early and will give it more thought.

Cheers,

Peter 

Thought of a method but not the math.

Subtracting the area of the small circle from the larger you get the area of the cut disc i.e. 400πcm^2 - 100πcm^2 = 300πcm^2.

Now, to balance the Area of a segment to the right of the COG must be half of this i.e. 150πcm^2.

If you rotate the y axis 90 degrees, you can then consider this as a problem of calculating the area under a curve to determine the then y coordinate of a segment of a circle of radius 20cm where the area will be equal to 150πcm^2.

Or there may be madness in my method

61.4 mm from right hand side (to nearest 0.1mm), on the diameter of the large circle that bisects the small

assuming uniform thickness and density of material, weight is prop to area. CoG is then centre of area = 50% of area each side. Vertical consideration straightforward as symetrical either side of the centre line as drawn.

calc half the area. Imagine vertical line through CoG, that forms a solid chord to its right.

calc height of chord (length from right). I had to look up formula, then tricky because comes out as 424.1= 3.157x - sin x

i didn't know how to solve, but sin is max 1 and min 0, so do calc at both those values gives two close angles, use average to estimate sin and then redo the chord height calc (though probably accurate enough without...) 

Hope I'm allowed a third bite at the cherry ??

Assume that the centre of gravity sits on the centre line and displaced by x to the right of centre.

You essentially have 3 bodies to consider:

1) A small circle diameter 100 mm

2) A large circle diameter 200 mm

3) The shape formed by taking (1) from (2) above.

Now just take moments at the centre line LHS of the large circle and recognise that the moments of shape (1) + moment of shape (3) = Moment of shape (2)

Working in cm (rather than mm) and assuming constant density you end up with:

5 * 25 pi + (10 + x) *75 pi = 10 * 100 Pi

End result is x = 1.6666 cm or 16.6666 mm

I suspect that this can probably be expressed more generally in terms of "R"

Regards,

Peter 

Innocent Bystander posted:

61.4 mm from right hand side (to nearest 0.1mm), on the diameter of the large circle that bisects the small

assuming uniform thickness and density of material, weight is prop to area. CoG is then centre of area = 50% of area each side. Vertical consideration straightforward as symetrical either side of the centre line as drawn.

calc half the area. Imagine vertical line through CoG, that forms a solid chord to its right.

calc height of chord (length from right). I had to look up formula, then tricky because comes out as 424.1= 3.157x - sin x

i didn't know how to solve, but sin is max 1 and min 0, so do calc at both those values gives two close angles, use average to estimate sin and then redo the chord height calc (though probably accurate enough without...) 

Hi Innocent, Nice try. And apologies for initially posting the wrong dimension for the size of the Orange disc, which I hope you didn't find too distracting.

As you can probably see from Peter's last shot above, there is a relatively straightforward solution that avoids chords and sines ! But again, nice try !

Cheers, Don

Here is a revised picture showing the intended dimensions of the two circles.

And here is a picture showing the Moment Arm and the direction of weight associated with each Mass a; A; and (A-a)..............assuming the system is illustrated in a vertical plane.

I have re-drawn (see above) the two circular discs showing the intended dimensions ie Orange Disc = 200mm diameter and Removed disc (White) 100mm diameter.

The Centre of Mass (CoM) of a uniform disc is at its geometric centre.

Based on symmetry, the CoM of both discs is on the X axis and by extension the CoM of the End Product is also on the X axis.

You can find the CoM of the End Product by taking moments about any common point on the X axis. From the point of view of explanation, I think the most convenient point is at the left hand side where the two discs just touch each other. For this reason I have produced a modified diagram showing the Y axis passing through this point. (See above)

For a uniform disc, the Area is directly proportional to its Mass and can thus represent the mass.

Mass of Orange disc  = Pi(100x100)           = Pi(10,000)

Mass of White disc    = Pi(50x50)          = Pi(2,500)

Mass of End Product = Pi(100x100) – Pi(50x50) = Pi(7,500)

The CoM of the Orange disc is 100mm right of the Origin “O”

The CoM of the White disc is 50mm right of the Origin “O”

The CoM of the End Product is “d”mm right of the Origin “O”

These distances are called Lever Arms

Taking Moments about “O” (Moment = Mass x Lever Arm)

Moment of Orange disc  = Pi(10,000)x100 = Pi(1,000,000)……….(1)

Moment of White disc    = Pi(2,500)x50     = Pi(125,000)…………(2)

Moment of End Product = Pi(7,500)x”d”    = Pi(7,500)d…………..(3)

The resultant moment can be found by (1) – (2) = (3)

Note: the “minus” sign is because we are removing material represented by the White disc

Pi(1,000,000) – Pi(125,000) = Pi(7,500)d

Dividing by Pi gives (1,000,000) – (125,000) = (7,500)d

Or          875,000 = 7,500d

Dividing both sides by 7,500 gives d = 875,000/7,500 = 116.6667

Ie the CoM of the End Product is 116.6667mm along the X axis

Or 16.6667mm to the right of the centre of the Orange disc.

Peter's explanation is far more succinct  !!

Oh !

You could take moments about the centre of the Orange disc, as per the original X and Y axes.

The arithmetic is much easier, but the explanation was difficult to describe !

sophiebear0_0 posted:

Hope I'm allowed a third bite at the cherry ??

Assume that the centre of gravity sits on the centre line and displaced by x to the right of centre.

You essentially have 3 bodies to consider:

1) A small circle diameter 100 mm

2) A large circle diameter 200 mm

3) The shape formed by taking (1) from (2) above.

Now just take moments at the centre line LHS of the large circle and recognise that the moments of shape (1) + moment of shape (3) = Moment of shape (2)

Working in cm (rather than mm) and assuming constant density you end up with:

5 * 25 pi + (10 + x) *75 pi = 10 * 100 Pi

End result is x = 1.6666 cm or 16.6666 mm

I suspect that this can probably be expressed more generally in terms of "R"

Regards,

Peter 

 

It's always good when participants set the next teaser......

Hi Peter,

I'm going for R/3 where R = Radius of the cut-out (small circle)

CoM is measured to the right of the centre of the large circle.

Don

Yes I agree (for what's that worth !!)

I followed your suggestion this time and took moments at the centre of the large circle. Makes the sums a wee bit easier. That said, i do agree that taking moments from the point of intersection of the circles makes the solution a little easier to follow.

Nice problem !

Regards,

Peter

sophiebear0_0 posted:

Don

Yes I agree (for what's that worth !!)

I followed your suggestion this time and took moments at the centre of the large circle. Makes the sums a wee bit easier. That said, i do agree that taking moments from the point of intersection of the circles makes the solution a little easier to follow.

Nice problem !

Regards,

Peter

Hi peter,

Solving for "R" as a general case suddenly makes the whole thing look a lot easier. That was a nice suggestion of yours.

Best sharpen up that pencil (this applies to others who are watching) I have another one on standby, in a similar vein, but hopefully a bit more fiendish................well, you have got rid of a few cobwebs, haven't you

From a square sheet of Perspex of uniform density, an isosceles triangle is to be cut from one edge as shown. What is the altitude of the cut such that the remaining Perspex (shown orange), when suspended at any angle at the pivot point “P” (at the apex of the cut) will remain in equilibrium ?

You can assume that the pivot at point P will allow the remaining Perspex to be pivoted at any angle, including “upside-down”.

Now, without making the task any easier, I will post the answer (but not my method) later this evening (unless somebody gets there first), simply so that you will know if your working is on the right track. (My solution involves an ugly-looking quadratic which is surprisingly easy to solve........)

Altitude = a(3 - √3)/2

or 0.634a

Don

I think I have eventually got there ?

The method I have used is similar to the previous teaser with circles:

This time we have 3 shapes:

1) The Square of area a^2

2) The triangle of area (a*p)/2

3) The remainder shape with area = (a^2 - (a*p)/2)

The key to this problem is knowing that the centre of mass of a triangle is located at the intersection of the 3 angle bisectors and occurs a 2/3 along the centre line from the apex. 

By taking moments from the base line mid point you end up with the equality:

a * a^2 = p/3 * (a * p)/2 + p * ( a^2 - (a*p)/2 )

Note these correspond to the moments for shapes 1,2 and 3 respectively.

Solving above ends up with 2 p^2 - 6 pa + 3 a^2 = 0

The quadratic can be solved for p in terms of a which yields your solution above. The other route gives p>a and therefore can be discarded.

Regards,

Peter

Very nicely done Peter. Looks like the old grey matter is still fully functional

Don - I wouldn't go that far about the grey matter !!

It is a very nice problem - I almost feel tempted to check the solution empirically.

I must admit, when you first posted the problem I thought it might be necessary to delve into Sin & Tan of half-angles. Now that would have been really ugly. So thanks for posting the solution early which gave me confidence that is was possibly within my capability ?

I cottoned on to the 2/3 rule for centre of mass of the triangle - and then it should have been plain sailing. But then I messed up with an error in subtracting fractions (primary school error !)

Anyway got there in the end........

Regards,

Peter

All the engines on this aeroplane turn clockwise when seen from behind. The engines turn either propellers or turbo-fans. The aeroplane is performing a turn to the left.

Will the gyroscopic effect of the engines (propeller/turbo-fan) cause the aeroplane to :-

Roll to the right

Roll to the left

Yaw right

Yaw left

Pitch up

Pitch down

Note:- an aeroplane

Rolls around its longitudinal axis

Yaws about its normal axis

Pitches about its lateral axis

Perhaps I should make it clear that the aeroplane is established in a steady, level, left turn.

It is not in the process of rolling into the turn. It is not climbing or descending.

In the diagram, a large gyro-wheel (propeller or compressor/turbine blades)) is spinning clockwise (when viewed from behind) about the horizontal “Z” axis.

The Z axle is (which is generally moving forward, ie towards the letter Z) is being steadily turned to the left. This is the same as if a force (push) is being applied to the rim of the gyro-wheel as shown in the diagram.

The spinning gyro-wheel does NOT behave in the same way as a static wheel when subjected to this push force. But it does behave in a predictable way. The aeroplane behaves in a similar way.

The reaction in a gyro acts in the same direction as the initial force, but 90 degrees in the direction of rotation.

So the gyro rotor tends to pitch up. (but there are other forces acting on an aeroplane during a turn that  also need to be considered)

Hope the diagram helps.

You throw a small ball vertically upwards in real air.

Does it take longer to go up, or to come down ?

That one's easy. It takes longer to come down 'cos it can't come down until it.'s gone up !

sjbabbey posted:

That one's easy. It takes longer to come down 'cos it can't come down until it.'s gone up !

PS I've arranged a "meeting" for you with the Headmaster - Dr Wacker !