General motion of a cone on an inclined surface

Question

Suppose that a solid cone is placed horizontally on an inclined surface and is initially at rest. How will the cone move when it starts motion due to its weight?

I know that its motion depends on the incline angle and also on the friction coefficient of the surface (as I observed by doing some experiments), but I can't establish the relation between them. Can anyone help me?

Answer 1

"How will the cone move when it starts motion due to its weight"

to answer this equation you have to obtain the equations of motion.

you have 3 generalized coordinates

• rotate about then z -axis angle $\psi$
• incline plane x position $s_x$
• incline plane y position $s_y$

thus you get :

Kinetic energy

$$T=\frac 1 2\,{{\it \dot{s}_x}}^{2}m+\frac 1 2\,{{\it \dot{s}_y}}^{2}m+\frac 1 2\,{\dot{\psi}}^{2}{{\it J_z} }^{2}-\tau_z\,\psi $$

Where $\tau_z$ is the torque due to the friction forces and $J_z$ is the inertia of the cone z component.

Potential energy

$$U=m\,g \left( \sin \left( \alpha \right) {\it s_y}+{\it s_z} \right) $$

Where $s_z$ is the z component of the COM.

Thus: The EOM's:

$$m\,\ddot{s}_x+F_{Rx}=0$$ $$m\,\ddot{s}_y+F_{Ry}+m\,g\,\sin(\alpha)=0$$ $$J_z^2\,\ddot{\psi}+\tau_z=0$$

with $\tau_z=F_{Ry}\,y_s\quad $ and $F_{Rx}=\mu\,N\,,F_{Ry}=\mu\,N\quad , N=m\,g\,\cos(\alpha)$

According to the EOM's the cone will move "diagonal" on the incline plane and rotate about the cone z- axis

Answer 2

For a solid cone the COM is $\frac{h}{4}$ above its base

Along the incline we can write the following equations.

Forces: $$Mg\sin\theta+f=Ma$$

Torque: $$\frac{fh}{4} = I \alpha$$

In this case $I = \frac{3}{5}m(\frac{r^2}{4}+h^2)$

Rolling: $$\alpha = \frac{a}{h/4}$$

Since we have 3 equations and 3 unknowns $(f, a, \alpha)$ the system can be solved.