Posers and Puzzles
27 Oct 11
Originally posted by talzamirI'll say
A drinking glass has a round base with radius r1, height of h and radius r2 > r1 at the upper rim. On its side on the table the glass goes around in a circle. How many full rotations does the glass do during one such circle?
Theta = 2*pi*(r2-r1)/h
Theta = 2*pi
so number of rotations is
N = (r2-r1)/h
Originally posted by joe shmoI do not think your answer will yield the right trends with respect to parameter variations. For example, consider if (r2-r1) becomes very small (for some constant h). N should get very big because the glass will sweep out a very large circle, right?
I'll say
Theta = 2*pi*(r2-r1)/h
Theta = 2*pi
so number of rotations is
N = (r2-r1)/h
Originally posted by LemonJelloyeah...I suppose so. I did it hastily while I'm slightly under the influence...sorry
I do not think your answer will yield the right trends with respect to parameter variations. For example, consider if (r2-r1) becomes very small (for some constant h). N should get very big because the glass will sweep out a very large circle, right?