# Particle inside a cone

1. Oct 13, 2015

### Tanya Sharma

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

The only thing I can think of in this problem is energy conservation .

$\frac{1}{2}mv^2_0 = mgh+\frac{1}{2}mv^2$

Not sure how to proceed . I would be grateful if somebody could help me with the problem .

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2. Oct 13, 2015

### Staff: Mentor

Energy conservation is a good start. What does it tell you about the particle at point B?
What else could be conserved?

3. Oct 13, 2015

### Tanya Sharma

It gives us the speed of particle at B . I should have mentioned in the OP that I did managed to get part 2 from energy conservation . I was thinking about part 1 of the question .

I had thought about it , but can't think of one .

4. Oct 13, 2015

### Staff: Mentor

Which conservation laws do you know in general?

The speed is also relevant for part 1.

5. Oct 13, 2015

### Tanya Sharma

Energy conservation , Linear Momentum conservation ,Angular Momentum conservation .

6. Oct 13, 2015

### Staff: Mentor

Well, you used energy conservation, so you have to figure out if one of the other two can be used.

7. Oct 13, 2015

### Tanya Sharma

Linear momentum can not be conserved for sure . So,I am guessing it has to be angular momentum . But the two forces acting on the particle , Normal and the Weight seem to have a torque about the center of the circle ( if I choose center as the reference point ) , in which case angular momentum cannot be conserved .

8. Oct 14, 2015

### TSny

Torque is a vector quantity. Is there any direction in space such that the net torque always has zero component along that direction?

9. Oct 14, 2015

### Tanya Sharma

Hi ,

Thanks for the response :)

If the vector joining the reference point to the point of application of force lies along the force ,the torque would have zero component .

I am not sure if this is what you are asking .

10. Oct 14, 2015

### Nathanael

Conservation laws are shortcuts, but it should always solvable by considering the forces.

There are only two forces acting, gravity and the normal force. In order for the particle to stay in contact with the cone, the normal force must exactly cancel the component of gravity perpendicular to the cone.

So then it's clear that the net force is constant in magnitude and is perpetually directed at the cone's apex.
Taking the vertical component of this net force (which is constant, as you can see if you imagine the net-force vector swinging around) gives you a 1D kinematics problem for the vertical-speed. ("With constant vertical-acceleration, starting from vertical-rest, you fall a vertical-distance h, then what is the final vertical-speed?")

Once the vertical-speed is found, you can relate it to the "down-hill" speed. Since you know the total speed, you can then find the angle θ.

11. Oct 14, 2015

### TSny

If you pick a point on the axis of the cone as origin, can you describe the direction of the net torque when the particle is at some arbitrary point on its trajectory?

12. Oct 14, 2015

### Staff: Mentor

Hi Nathanael,

I don't think that this is correct. There is a component of centripetal acceleration in the normal direction.

Chet

13. Oct 14, 2015

### Staff: Mentor

I've worked this problem using the force balance approach, in spherical coordinates. Even though it was a lot of work, at least I could be comfortable with the results. The key thing that the analysis delivered was that angular momentum around the axis of the cone is conserved. This gave the circumferential velocity component at time t in terms or the circumferential velocity at time zero, the radius at time zero, and the radius at time t. The velocity down the incline can then be obtained from the conservation of energy equation.

Chet

14. Oct 14, 2015

### Staff: Mentor

Exactly. All forces are in radial or vertical direction, never "around the axis".

There is no need to solve anything time-dependent, however. We know the total speed, and we can calculate the (edit) tangential component using angular momentum conservation.

Last edited: Oct 15, 2015
15. Oct 14, 2015

### Staff: Mentor

Looking at it that way, it seems pretty obvious. I guess I could have saved myself a lot of work if I had thought of it.

Thanks.

Chet

16. Oct 14, 2015

### insightful

Did you mean "tangential component"?

17. Oct 14, 2015

### Tanya Sharma

Since the net force on the particle is down the slope towards the apex of the cone ,the net torque about the apex would be zero . Right ?

So, we need to conserve angular momentum about the apex .

Should the angular momentum of the particle at t=0 be $mv_0l$ ,where $l$ is the slant length of the cone . Since $l=\frac{r_o}{sin\alpha}$ , initial angular momentum = $\frac{mv_0r_o}{sin\alpha}$ . Is that correct ?

I would like to admit that I am finding a bit difficult dealing with directions in this problem .

Last edited: Oct 14, 2015
18. Oct 14, 2015

### Nathanael

I'm very sorry for misleading you, but the net force would not always be towards the apex (Chet corrected me).

The important detail about the net force is that none of it is in the horizontal-tangent-direction. Or as mfb put it:
Torque and angular momentum are vectors, and so the equation $\vec \tau = \frac{d}{dt}\vec L$ is really 3 equations, one for each component. There is no point about which the torque completely vanish, but there is a line of points (the cone's symmetry axis) about which one component of the torque will vanish. This corresponds to conservation of that component of the angular momentum.

19. Oct 14, 2015

### TSny

I was thinking about conservation of angular momentum using the vector definitions of torque and angular momentum: $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$ and $\mathbf{L} =m \mathbf{r} \times \mathbf{v}$. There's the basic law $\boldsymbol{\tau}_{net} = d \mathbf{L}/dt$. The latter implies that if a component of the torque vector is zero, then the same component of angular momentum is conserved.

The net force is not necessarily tilted downward at all times. I believe the particle will eventually return to its original height (which requires a net upward force at some times). The important fact is that both the force of gravity and the normal force lie in a vertical plane passing through the axis of the cone (call it the z-axis). So, the net force lies in this plane. If you pick any point on the z-axis as origin for calculating the net torque on the particle, you should be able to deduce whether or not there is a component of angular momentum that is conserved. You can pick the origin at the apex of the cone if you want, but any other point on the z-axis would also be fine.

20. Oct 14, 2015

### Tanya Sharma

I am really sorry , but I am not understanding this problem at all .