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?
 A: 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.
A: 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 $B$.
A: 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)
A: 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.)
A: 

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 $\varphi=2\pi~,s(u)=2\pi\,r-\frac{2\pi\,r}{h}\,u$
$$s(0)=2\,\pi\,r~,s(h/2)=\pi\,r~,s(h)=0$$
