Deriving Angular Momentum in a Particle Inside a Cone Problem

[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 .