# Rolling a ball into a cone; what should the forces overall be?

Suppose there is a cone, with the apex pointing down, and the top of the cone at height $h$, apex half-angle $\psi$, ball of mass $m$, and the initial velocity into the cone (completely horizontal, tangential to the top radius) $u$ .

My overall aim is to find an expression for angular displacement/velocity/acceleration, given $h, u, m, \psi$ And importantly $t$.

From physics at school, I know that the forces in 2 dimensions can be solved using the equations associated with a banked turn. In this case, the component of weight going down the slope would be $mg \cos\psi - mg\mu\sin\psi$; or $mg(\cos\psi - \mu\sin\psi)$ as the normal force on the slope is $mg\sin\psi$ and Friction $F = \mu*N$.

My question is this: firstly, am I right? Is there also a force in the direction of travel for the ball. Something like $-mg\mu\sin\psi$ ?

EDIT: rethink

I want angular acceleration at $t$ first, so that I can integrate with respect to $t$ to find an expression for angular velocity and displacement.

So, we know $\alpha=$$a\over r$

Therefore I need an expression for $a$, acceleration tangential to the cone's circle, and $r$, the current radius the ball is at on the cone. Both at $t$.

Now I believe that the tangential acceleration is just simply $\mu N \over m$. But what is the overall normal force in this case?

Next, trigonometry gives us the radius, based on the half-angle $\psi$ and the distance from the bottom of the cone $z$: $r = z\times tan(\psi)$, where $z$ can also be equal to height - vertical displacement: $r = (h-Sz)\times tan(\psi)$ (Sz is vertical displacement from top).

So we now need Sz at $t$. To get that I can use force diagrams to get force, then acceleration downwards.

This is what I have so far for the force downwards $mgcos(\psi)(cos(\psi)-\mu sin(\psi))$, so the acceleration is $gcos(\psi)(cos(\psi)-\mu sin(\psi))$. But I am not accounting for the force going up the cone due to the ball's initial velocity $u$.

In Summary:

• What is the overall 3D vector for the force? Where $Fx$ is the force acting in tangent to the circle, (including friction) based on $u,μ,m,ψ$ and $t$. I am really struggling to comprehend the part which causes it to go up the cone, like if you rolled a ball at a ramp with velocity, it would go up the ramp and cause normal forces, and therefore friction. This is a similar problem, but the ball is not going straight up a ramp, its curving into a circle with a variable radius!

I can draw a diagram if anyone is unsure as to the visualization of my problem.

I have also asked this question in physics.stackexchange (https://physics.stackexchange.com/q/141336/62124) : I hadn't thought of asking it there (how stupid, I know) and I think that I'm more likely to find people who know this kind of thing there, but there are still valuable minds in math.stackexhange.

• bmatrix for brackets, pmatrix for parenthesis. Just matrix is more of an array. – Asaf Karagila Oct 8 '14 at 12:24
• The frictional force is proportional to the force of the ball against the cone...which depends not only on the weight of the ball, but on its speed "around the axis". This is the principle used by motorcylists who ride around the inside walls of a cylindrical building. You've ignored that component of force. – John Hughes Oct 8 '14 at 12:27
• @John could you maybe explain further, as an answer? I tried to prove you wrong, but I just went in circles around myself :) I think this is what I needed. the force is dependant on the velocity, but it's the tangential force at time t frame I want (x component of my vector), so that I can find angular acceleration at t, given this and the radius at t. – Sam Walls Oct 8 '14 at 18:32
• @AsafKaragila thanks :) – Sam Walls Oct 8 '14 at 18:34
• Alas, no. I'm not a physicist, and I'd just get tangled up in knots. The essence is this: the ball wants to go straight; to remain in the cone, its velocity vector must change. The required force to make it change is called "centripetal force", and by one of Newton's laws, that force is also exerted by the ball on the cone. The normal portion of that force needs to be accounted for in computing friction. (Anynon-normal part -- is there any??? I don't know --- needs to be accounted for in the ball's "along the cone" motion as well.) – John Hughes Oct 8 '14 at 18:50