Hollow Circle at the edge!

Classical Mechanics Level 3

Consider a ring which mass is standard distributed .

The ring is place at one corner of the table as the picture shown above.

What is the minimum ratio of area $\frac{A}{B}$ when the ring is just going to fall from the corner?

If the answer can be represented by $\frac{a\pi -b}{c\pi+d}$ which $a,b,c,d$ is positive integer and the fraction is the simplest form. Then find $a+b+c+d$ .

The answer is 6.

1 solution

Kelvin Hong
Aug 27, 2017

What determine the fall of the ring? the table doesn't matter at all, but the two point which the ring meet the table at the edge do.

And it is not to hard to proof that, at this condition, the ring must touch the table corner as the third point which them meet at the edge of table.

If we want ratio be minimum, we needed to minimum the area of A and maximum the area of B.

When the area A reach minimum, distance of the center of ring to the two edges of the table will be the same.

So , let the radius be $R$

$A= \frac{1}{2}\pi R^2 -R^2$

$B= \frac{1}{2}\pi R^2 + R^2$

Do the ratio, we will get

$Ratio = \frac{\pi -2}{\pi +2}$

$a+b+c+d =1+2+1+2=\boxed{6}$

Hollow Circle at the edge!

The answer is 6.

1 solution

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