# Finding the final height of a particle moving within an inverted cone

There is a particle moving within this inverted circular cone with an initial tangential velocity of $$v$$ at an initial height of $$h$$, and we see that the cone forms an angle of $$\theta$$ with the horizontal.

The problem is this: to find the final height of the object when it stops climbing up or sliding down the incline. The only forces are gravity and the normal force.

I know how to analyze the motion of an object like this when we are given that it stays at one height. We have a centripetal acceleration of $$a_c = \frac{v^2}{r}$$ towards the center, but this scenario I'm not sure about.

This is the free-body diagram I drew:

And the 2nd law equations I had: \begin{align*} \sum F_x &= N\sin \theta = ma_c \\\\ \sum F_y &= N\cos \theta - mg = ma_{tan} \end{align*}

Now, I thought I had solved the problem using this, but looking back, I'm not sure I would be justified in thinking that the acceleration in the x direction would be $$a_c = \frac{v^2}{r}$$. I'm not sure if the acceleration in this situation is centripetal. If one imagines looking at the cone from above, if the particle is rising, it seems like the path would be some kind of a spiral.

My assumption was, based on the initial velocity, it would either fall, rise, or stay at the same spot, depending on the normal force. If the normal force is large enough it is stronger than gravity so the ball will rise, but as the ball rises, the radius gets larger, so $$a_c$$ gets smaller, and the normal force also gets smaller. I thought the particle would reach basically an equilibrium and stop rising or falling and I wanted to know what height that would be. I got an answer of $$y_f= \frac{v^2}{g}$$ but I don't think this is right. I can go into how I did that but it's just solving the equations that I set up. The problem is I don't think I was justified in setting up those equations.

Can you help me solve this?

Edit: Also, would I be correct in thinking that the tangential velocity will remain constant?

• Try using conservation of energy. Oct 9 '20 at 1:41
• @DavidWhite Okay. I'm not sure if I'm doing this right. So we have $\frac{1}{2} mv_i^2 + mgy_i = \frac{1}{2}mv_f^2+mgy_f$ which implies $y_f = \frac{v_i^2-v_f^2}{2g} + y_i$, but that seems to suggest that the velocity has to change for it to climb, or that it stays at the initial position. I think I've probably set this problem up incorrectly. Oct 9 '20 at 1:56
• Like does it necessarily even reach a maximum/minimum height and stay there? Oct 9 '20 at 2:33
• I'm not sure if the particle will stay at the maximum height. The particle will change its velocity as it reaches its maximum height and the radius of the cone will change at that height. At the maximum height, there will be only horizontal velocity in the cone. This makes the problem somewhat more involved than your equation in your comment. Oct 9 '20 at 15:03
• Assuming the cone is smooth so there is no loss of energy then the particle oscillates between two different heights at which its velocity is horizontal. One of these heights is the initial height, and the other can be found by solving a quadratic equation. Oct 9 '20 at 17:35

The key here is conservation of angular momentum. Assuming the cone is smooth, then $$mv_tr$$ must be constant, where $$v_t$$ is the horizontal speed of the particle. And $$r$$ is proportional to $$h$$. So if the initial tangential speed at height $$h_i$$ is $$v_i$$ then we have

$$\displaystyle v_t(h) = v_i \frac {r_i}{r} = v_i \frac {h_i}{h}$$

Conservation of energy tells us that

$$\displaystyle \frac 1 2 m v^2 + mgh = \frac 1 2 m v_i^2 + mgh_i \\ \displaystyle \Rightarrow v^2-v_i^2 = 2g(h_i-h)$$

In general $$v > v_t$$, but when the particle is at its maximum or minimum height then $$v=v_t$$ and we have

$$\displaystyle v_i^2 \left(\frac {h_i^2-h^2}{h^2} \right) = 2g(h_i-h)$$

We know that $$h=h_i$$ is either a maximum or a minimum height. To find the other extreme point, asssume $$h \ne h_i$$ so that we can remove a factor of $$h_i-h$$ to get

$$\displaystyle v_i^2 \left(h_i+h \right) = 2gh^2$$

This quadratic equation in $$h$$ has two real roots but only one of them is positive - this is the root that you want.