# Why do objects with one small circle and one big circle on either side move in a circular path when pushed forward?

When a cylinder, (where both circles on either side are the same diameter), is pushed forward, it will move forward. However in a shape where one circle is a different size to the other, it moves in a circular path. I am sure that the bigger circle moves a greater distance than the smaller circle in a given time, but how does this (or what does) make it curved?

• This question is rather confusing. After reading it several times, I think that you are using the term "cylinder" to refer to a conical frustum. The word "cylinder" refers to a shape in which all of the cross sections are congruent. If the two sides are different sides, then it's not a cylinder. Commented Nov 1, 2020 at 5:07
• Apologies for the confusion. It was unneccesary to refer to a cylinder, but I simply mentioned how it would move in a straight path, whereas an object with one small circle and one big circle on either side, a conical frustum for example, would move in a circular path Commented Nov 1, 2020 at 5:15
• @EmilyWilkins Welcome to Physics SE! I'd suggest you make these changes in the question itself so that people don't have to look at the comments to understand the context. You can edit it and use the points mentioned in the comments Commented Nov 1, 2020 at 10:35

If you rotate a disk of radius $$r$$ with a frequency of $$f$$ rotations per second then the outermost ring has a velocity of $$v=2\pi f r$$ You can show this because in one time period the outermost ring travels a distance $$2\pi r$$ so $$v=\frac{\Delta x}{\Delta t}=\frac{2\pi r}{T}=2\pi f r$$ When you rotate the frustrum around its axis the velocity of its surface is different depending on where you are. If you call the biggest radius $$R$$ and the smallest radius $$r$$ then the velocity on the surface is $$v=2\pi f R$$ and $$2\pi f r$$ respectively. So obviously the side with bigger radius moves faster. When an objects rolls it can't slip (otherwise it wouldn't be rolling) so the distance that the surface travels is also travelled on the ground. You can draw the line of contact between the frustrum and the ground over time. In this picture I drew how this line would look like if I naively used the information from above.

Obviously, this is wrong. The line gets longer implying that the cylinder gets longer. So is there a way which (a) gives the right velocities on the surface (implying the right distance travelled on the floor) and (b) which doesn't stretch the cone? There is! By curving the path both of these constraints are met. These are important constraints because they follow from the geometry of the object and the no-slip condition so you can't just ignore them. This gives the following unique path:

Bonus: Once you assume the frustrum moves in a circle you can solve for the inner radius. If you call this inner radius $$a$$ and the length of the frustrum $$w$$ you get the following picture

After a time $$t$$ the bigger side has traced out an arc with length $$2\pi f R t$$ and the smaller side has traced out $$2\pi f r t$$. Since the big arc is similar to the small arc (the big arc is a scaled version of the small arc) you get $$\frac a{a+w}=\frac{2\pi f r t}{2\pi f R t}=\frac r R$$ Solving for $$a$$ gives $$a=\frac{rw}{R-r}$$ When $$r=0$$ you get a cone that rotates around its tip since $$a=0$$. When $$R=r$$ you get a cylinder and $$a$$ becomes infinite (= a straight path, like Andrew mentioned)

The key thing to note here (which I missed earlier) is that this is a rigid body

Meaning the two circles have a common axis of rotation, Therefore the angular acceleration will be the same for any cross-section of your conical frustum

Let's say the angular acceleration given to the BODY by torque due to friction is $$\alpha$$

With the relation

$$a=r.\alpha$$

we can see that the larger cross-sectional wheel of the conical frustum has a greater magnitude of acceleration. Due to a larger radius. Helping it to sweep the same angle about a point as for all cross-sections of the conical frustum.

Why a circular path?

The "conical frustum " can be imagined to be a cross-section of a cone

We can simplify this to just an axle and the circular base, as you can imagine the apex has zero velocity, and the entire system pivots around it , this applies to each circular cross-section of a cone. Resulting in the entire cone pivoting around the apex. For the "conical frustum," it is just pivoting around some imaginary apex.

• Why are the moments of inertia of the two circles equal? The bigger circle should have a greater moment of inertia since it has a larger radius.
– Toba
Commented Nov 1, 2020 at 9:24
• There is no moment of inertia of the ends if we speak of a rigid body. The treatment of something soft enough escape my current capability. What will happen in a more fluid ensemble? Deformation in addition to a curved path? Commented Nov 1, 2020 at 10:05
• @Toba Thank you for pointing that out. I treated them to be separate entities. Commented Nov 1, 2020 at 10:46
• I don't think the question has been answered. Why does the body move in a circular path, why not a linear one? Why must the imaginary apex have zero velocity? We can imagine a case in which the entire frustrum as well as the imaginary apex moves forward. Why doesn't that happen?
– Toba
Commented Nov 1, 2020 at 11:06
• @JustJohan curve surely. Circle I don't know. It was belong to the part of moment of inertia, which in a rigid body does not pose problem, of course. My comment came in because of the reference on momentum I. It could have been skipped I guess. Actually the question is even a bit surprising as the answer is merely geometric. Commented Nov 1, 2020 at 11:22

Suppose the two ends of the frustrum have radii $$r$$ and $$R$$ ($$R>r$$) and initially touch the ground at points $$A$$ and $$B$$. If the frustrum is rotating about its axis at an angular speed $$\omega$$ radians per second and is not slipping on the ground then in a short time $$\delta t$$ the contact point $$A$$ moves a distance $$\delta d_A = r \omega \delta t$$ and the contact point $$B$$ moves a distance $$\delta d_B = R \omega \delta t$$. If the slant height of the frustrum (the distance from $$A$$ to $$B$$) is $$H$$ then the line of contact with the ground rotates through an angle

$$\displaystyle \delta \theta = \frac {\delta d_B - \delta d_A}{H} = \frac {(R-r) \omega}{H} \delta t$$ radians

so its angular speed is

$$\displaystyle \frac {\delta \theta}{\delta t} = \frac{(R-r) \omega}{H}$$ radians per second

As Justjohn says in their answer, this corresponds with both $$A$$ and $$B$$ tracing a circular path around the imaginary apex of the frustrum which is at a distance $$\frac {rH}{R-r}$$ from $$A$$ and a distance $$\frac {RH}{R-r}$$ from $$B$$.

What is described is a conical frustum with a small base r and a larger base R. As the frustum rotates one full cycle the outer edge of the smaller base travels $$2\pi r$$ and the outer edge of the larger base travels $$2\pi R$$. This causes the frustum to travel in a circle. The circle has the virtual apex of the frustum as its center--since that does not move. (You would need the distance between the bases to completely specify the frustum.)

• Yes, The important point, I think, is that it's really not a physics question at all. Just simple geometry. Commented Nov 2, 2020 at 3:31

A cone is rotate about the blue axes with $$\varphi=\omega\,t$$

the height of cone is h and the base radius is r.

if you look at the projection (Y-Z plane) of the cone , you can obtain the radius $$\rho$$ that perpendicular to the rotation axes

$$\rho=r\,\frac{h-u}{h}~,0\le u\le h$$

with $$~v=\omega\,\rho~$$ and $$\frac {ds}{dt}=v$$ we obtain

$$s(u)=\omega\,t\,\rho=\varphi\,\rho=\varphi\,r\,\frac{h-u}{h}~,\,0\le\varphi\le\,2\pi$$

thus:

with $$\varphi=2\pi~,s(u)=2\pi\,r-\frac{2\pi\,r}{h}\,u$$

$$s(0)=2\,\pi\,r~,s(h/2)=\pi\,r~,s(h)=0$$