Deriving Angular Momentum in a Particle Inside a Cone Problem

[Tanya Sharma]

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.

Okay I will try it . Logically I am convinced that component of torque around the z axis will be zero ,hence angular momentum would be conserved about the z -axis .

Now I would like to conserve angular momentum of the particle 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 ?

 

[ehild]

The a

Okay I will try it . Logically I am convinced that component of torque around the z axis will be zero ,hence angular momentum would be conserved about the z -axis .

Now I would like to conserve angular momentum of the particle about the apex .

It is not conserved.

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 ?

The angular momentum is a vector quantity. Not all components are conserved.
The magnitude of the angular momentum about a point is the product of r (distance of the particle from the chosen point) multiplied by the magnitude of its velocity and with the sine of the angle between $\vec r$ and $\vec v$.

 

[Tanya Sharma]

Ok .

I would like to go a bit slow and one step at a time.

What should be my next step ,assuming component of angular momentum is conserved around the z -axis ?

 

[ehild]

I think, your first step is to determine the z component of angular momentum vector at an arbitrary point on the cone. It would be a good practice to write out the cross product $\vec r \times \vec v$.

 

[Tanya Sharma]

Should I use cartesian coordinates or sperical coordinates would be easier to work with ?

 

[ehild]

Should I use cartesian coordinates or sperical coordinates would be easier to work with ?

Use Cartesian coordinates expressed with the angles theta and alpha .
It might cause confusion that I labelled the position vector by $\vec r$, and you used l for it, and the problem denoted the radius of the horizontal circle with r. It is better to denote it with something else, ρ, for example.

 

[Tanya Sharma]

Use Cartesian coordinates expressed with the angles theta and alpha .
It might cause confusion that I labelled the position vector by $\vec r$, and you used l for it, and the problem denoted the radius of the horizontal circle with r. It is better to denote it with something else, ρ, for example.

$\vec{\rho} = \rho sin\alpha \cos\theta \hat{i}+ \rho sin\alpha \sin\theta \hat{j} +\rho cos\alpha \hat{k}$

Is it correct ?

 

[Chestermiller]

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

What we have been alluding to in our responses is that:
$$mv_cr=mv_0r_0$$
where v_c is the circumferential component of the particle velocity at a location where the radial position of the particle is r (this r is defined in your figure). This is the conservation of angular momentum equation about the axis of the cone.

Chet

 

[Tanya Sharma]

What we have been alluding to in our responses is that:
$$mv_cr=mv_0r_0$$
where v_c is the circumferential component of the particle velocity at a location where the radial position of the particle is r (this r is defined in your figure). This is the conservation of angular momentum equation about the axis of the cone.

Chet

You mean $mvcos\theta r=mv_0r_0$ ?

 

[Chestermiller]

$\vec{\rho} = \rho sin\alpha \cos\theta \hat{i}+ \rho sin\alpha \sin\theta \hat{j} +\rho cos\alpha \hat{k}$

Is it correct ?

That is the equation for the position vector drawn from the apex of the cone to any location ρ,θ on the cone, where ρ is the distance from the apex and θ is the circumferential angle (longitude).

Chet

You mean $mvcos\theta r=mv_0r_0$ ?

Yes.

 

[Tanya Sharma]

$$\theta = cos^{-1}\left(\frac{v_0r_0}{\sqrt{v^2_0+2gh}(r_0-htan\alpha)}\right)$$

Is it correct ?

 

[ehild]

$\vec{\rho} = \rho sin\alpha \cos\theta \hat{i}+ \rho sin\alpha \sin\theta \hat{j} +\rho cos\alpha \hat{k}$

Is it correct ?

Yes. And what are the components of the velocity if its magnitude is v?
It is useful to practice vectors and their cross product!

 

[Tanya Sharma]

And what are the components of the velocity if its magnitude is v?

I don't think $\alpha$ and $\theta$ would be relevant in finding components of velocity . Is that so ?

 

[Chestermiller]

I don't think $\alpha$ and $\theta$ would be relevant in finding components of velocity . Is that so ?

θ in the equation for ρ is a different θ from θ in your problem. You probably should have used φ in your equation for ρ to represent longitude.

 

[Chestermiller]

$$\theta = cos^{-1}\left(\frac{v_0r_0}{\sqrt{v^2_0+2gh}(r_0-htan\alpha)}\right)$$

Is it correct ?

Yes.

 

[Tanya Sharma]

Yes.

Thanks

 

[Chestermiller]

I don't think $\alpha$ and $\theta$ would be relevant in finding components of velocity . Is that so ?

When I worked this problem in spherical coordinates, the independent variables I used were ρ and φ. In this problem α is constant. Once you have this equation for $\vec{ρ}$, you can differentiate it to get the velocity and acceleration in terms of the time derivatives of ρ and φ. In the end, I worked using the unit vectors and components in spherical coordinates.

Chet

 

[ehild]

In either coordinates, the z component of the angular momentum resulted in L_z=m (ρsin(α))²dΦ/dt. With the notation of the problem, ρsin(α)=r. So L_z=m r*(r dΦ/dt) , but r dΦ/dt=v_c, the component of velocity along the horizontal circle. L=mrv_c, given by Chet in Post #38. It is a good practice to derive it from the vector product .