I had this question when observing something of the same nature with precise measurements.
An object has 2 circular ends (objects) joined together by a cylinder. The length of the cylinder is 5.6cm. One of the ends has a radius of 1.7cm and the other end 2.15cm. Calculate the circumference of the inner and outer circle made by the object when it moves.
I solved it using drawings and construction. They were drawn to scale and I got precise figures.. Though I'm not sure since I didn't calculate any thing....
I also want to ask if there's any formula for calculating this? Especially the angle of deviation of the ends from 90 degrees, the distance from the object's ends to the center of the circle of motion (both the angled (elevated) distance and flattened distance), and the angle from the center to the bigger circular end....
Its easy to get the figures from construction but I just want to know if there's any formula for it...
Imagine extending the axle (cylinder) until it reaches the ground. The entire assembly will pivot around this point (think of a rolling cone). The total length $L$ of this imaginary axle can be found via similar triangles: $${L-5.6\over L}={1.7\over2.15}$$ so $L\approx26.76$. The outer radius on the ground can then be computed using the Pythagorean theorem, so for the outer circumference we have: $$C_{\text{outer}}=2\pi R_{\text{outer}}=2\pi\sqrt{L^2+5.6^2}\approx171.75.$$ For the inner circumference, we can either compute the inner radius in a similar manner or take advantage of the circumference of a circle being directly proportional to its radius and the fact that the two radii are in the same proportion as the radii of the two end discs: $$C_{\text{inner}}={1.7\over2.15}C_{\text{outer}}\approx135.80.$$
The tilt angle of the discs from vertical is the same as the angle $\theta$ the axle makes with the ground, from which $$\sin\theta={2.15\over L}$$ and so $\theta\approx4.61^\circ$.