Deriving Angular Momentum in a Particle Inside a Cone Problem

[Tanya Sharma]

Homework Statement

Homework Equations

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 .

 

[mfb]

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

 

[Tanya Sharma]

What does it tell you about the particle at point B?

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 .

What else could be conserved?

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

 

[mfb]

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

Which conservation laws do you know in general?

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 .

The speed is also relevant for part 1.

 

[Tanya Sharma]

Which conservation laws do you know in general

Energy conservation , Linear Momentum conservation ,Angular Momentum conservation .

 

[mfb]

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

 

[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 .

 

[TSny]

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

 

[Tanya Sharma]

Hi ,

Thanks for the response :)

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

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 .

 

[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.

 

[TSny]

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

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?

 

[Chestermiller]

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.

Hi Nathanael,

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

Chet

 

[Chestermiller]

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

 

[mfb]

The key thing that the analysis delivered was that angular momentum around the axis of the cone is conserved.

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.

 

[Chestermiller]

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 radial component using angular momentum conservation.

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

 

[insightful]

We know the total speed, and we can calculate the radial component using angular momentum conservation.

Did you mean "tangential component"?

 

[Tanya Sharma]

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?

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 .

 

[Nathanael]

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 ?

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:

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

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.
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[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.

 

[Tanya Sharma]

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.

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

 

[TSny]

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

My approach might not be appropriate if you have not worked with the vector definitions of torque and angular momentum. In a hand-waving way, you can see that neither force tries to push the particle "around" the axis of the cone. This suggests that angular momentum "around the axis" will be conserved.

 

[Tanya Sharma]

In a hand-waving way, you can see that neither force tries to push the particle "around" the axis of the cone. This suggests that angular momentum "around the axis" will be conserved.

Makes sense .

Choosing apex as the reference point , should I express the velocity and position vector of the particle in 3 D coordinate system with apex as origin ? Or do I need to do something different ?

 

[TSny]

If you accept that the z-component of angular momentum, $L_z$, is conserved, then you just want to find an expression for $L_z$. The apex makes a good origin. Consider breaking the velocity vector into a component that is horizontal and a component that points along the line from the apex to the particle.

 

[Tanya Sharma]

Consider breaking the velocity vector into a component that is horizontal and a component that points along the line from the apex to the particle.

horizontal in the sense ?? What is horizontal as per you ?

Would the two components be perpendicular to each other ? Please forgive me if you find these questions silly .

 

[TSny]

I

horizontal in the sense ?? What is horizontal as per you ?

I'm thinking of the z-axis (axis of cone) as perpendicular to the floor. So, the horizontal component of velocity would be the component parallel to the floor. Can you express this component in terms of the speed $v$ and the angle $\theta$ defined in the problem?

Would the two components be perpendicular to each other ? Please forgive me if you find these questions silly .

Try to picture the flat plane that is tangent to the cone at the location of the particle. Both the velocity vector and the line connecting the apex to the particle lie in this plane. Does this help to see whether or not the two components are perpendicular?

Which component does not contribute to the angular momentum?

 

[ehild]

You see that the normal force N has vertical component, but it does not contribute to the torque about the z axis. The horizontal component (in the horizontal plane shown) of N is radial, points to the central axis. Does it have torque about z?
Gravity is vertical, which does not contribute to the z component of the torque, either.

 

[Tanya Sharma]

You see that the normal force N has vertical component

By vertical component , you mean along the z - axis ?

The horizontal component (in the horizontal plane shown) of N is radial, points to the central axis.

By horizontal plane , you mean x-y plane ?

What does $\theta$ represent in your diagram ?

 

[ehild]

By vertical component , you mean along the z - axis ?

Yes.

By horizontal plane , you mean x-y plane ?

Yes.

What does $\theta$ represent in your diagram ?

It is the angle the horizontal projection of $\vec r$ makes with the x-axis (think of the spherical coordinates) . The horizontal projection of N encloses the same angle with the x axis.
Make a paper cone and see (or perhaps there is a funnel in the kitchen :) )

 

[Tanya Sharma]

Okay , so now my first step should be to write out the position vector of a point of the cone with its x,y,z components, taking the angle alpha into account . Right ?

 

[ehild]

Okay , so now my first step should be to write out the position vector of a point of the cone with its x,y,z components, taking the angle alpha into account . Right ?

Both angles, alpha and theta.
Yes, do it, calculate the components of the torque, using the components of the position vector and those of the normal force (and gravity). It will convince you fully.