Particle inside a cone

  • #26
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upload_2015-10-15_9-37-49.png

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.
 
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  • #27
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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 ?
 
  • #28
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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 :) )
 
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  • #29
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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 ?
 
  • #30
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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.
 
  • #31
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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 ?
 
  • #32
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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##.
 
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  • #33
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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 ?
 
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  • #34
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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##.
 
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  • #35
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Should I use cartesian coordinates or sperical coordinates would be easier to work with ?
 
  • #36
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Should I use cartesian coordinates or sperical coordinates would be easier to work with ?
Use Cartesian coordinates expressed with the angles theta and alpha :smile:.
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.
 
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  • #37
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Use Cartesian coordinates expressed with the angles theta and alpha :smile:.
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 ?
 
  • #38
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I am really sorry , but I am not understanding this problem at all :oops: .
What we have been alluding to in our responses is that:
$$mv_cr=mv_0r_0$$
where vc 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
 
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  • #39
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What we have been alluding to in our responses is that:
$$mv_cr=mv_0r_0$$
where vc 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## ?
 
  • #40
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##\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.
 
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  • #41
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$$\theta = cos^{-1}\left(\frac{v_0r_0}{\sqrt{v^2_0+2gh}(r_0-htan\alpha)}\right)$$

Is it correct ?
 
  • #42
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##\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!
 
  • #43
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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 ?
 
  • #44
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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.
 
  • #47
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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
 
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  • #48
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In either coordinates, the z component of the angular momentum resulted in Lz=m (ρsin(α))2dΦ/dt. With the notation of the problem, ρsin(α)=r. So Lz=m r*(r dΦ/dt) , but r dΦ/dt=vc, the component of velocity along the horizontal circle. L=mrvc, given by Chet in Post #38. It is a good practice to derive it from the vector product .
 
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