How does a particle slide inside a cone?

Question

Small particle is sliding inside a cone without friction. The axis of the cone is vertical. Minimum distance from the tip of the cone Dmin = 30cm, and maximum distance Dmax = 90cm.

What is average by time distance <D> from the tip of the cone to the particle?

Answer 1

I do not understand how a smart person like you would support an incompetent person like John McCain.   I seriously do not get it.    

Answer 2

10 seconds

Answer 3

Ahh, the gravity fairee.  Her intuition about gravity is amazing.  I hear she likes chocolate.  Although, you might tempt her with some Rocky Road.

Answer 4

I already gave two solutions to this question, which are located here:

http://answers.yahoo.com/question/index;...

Let m be the mass of the particle, g the acceleration of gravity,  ÃƒÂŽÃ‚¸ the half-angle of the cone, and let E be the total energy of the particle and J its angular momentum.  Let z be its height above the vertex at any instant in time (related to D by z= D cosθ).

As I showed before, the dependence of z on time is periodic of some period T between two turning points (a minimum height z1 and a maximum height z2) that are the two positive solutions in z of the following cubic equation:

mg (cosθ)^2 z^3 -E z^2 +J^2 (cosθ)^4/[2m(sinθ)^2]=0.

This equation also has a negative solution z3, but that is obviously outside the range of motion of the particle.

Since the motion is periodic, it is not necessary to average over infinite time, one period of motion is enough.  Thus:

&lt;D&gt; =sec θ &lt;z&gt;

where

&lt;z&gt; = [ ∫ z dt]/∫ dt = [ ∫ z dt/dz dz] / ∫ dt/dz dz

= [ ∫ z dz/v(z) ] / ∫ dz/v(z)

and v(z) is dz/dt at the height z. The integrations are between the turning points z1 and z2.  v(z) is given by

v(z)=[2{E-J^2 (cosθ)^4/ [2 m(sinθ)^2 z^2] - mg (cosθ)^2 z}/m]^(1/2),

Noting that v(z)=C [(z-z1)(z-z2)(z-z3)]^(1/2)/z, where C is a constant, and z1,z2,z3 are the three roots mentioned above, it follows that

&lt;z&gt; = [ ∫ z^2 dz/ [(z-z1)(z-z2)(z-z3)]^(1/2) / ∫ z dz/[(z-z1)(z-z2)(z-z3)]^(1/2).

Thus the answer depends on the values of all three of the roots z1, z2, z3, and the angle θ to get &lt;D&gt;.  Since only z1 and z2 are given in the question,  more information should be given  so that z3 can be determined.

I am interested to see what solution the Mullah has.  Even if he has a &quot;shortcut,&quot; the expression he finds for &lt;D&gt;  ( D=secθ &lt;z&gt; ) should be the same, unless one of us has it wrong.

Answer 5

I agree with chris.  

Answer 6

&lt;D&gt; = 65 cm.

Without lost of generality, we can consider a particle of unit mass.

The particle velocity is split in two components: the horizontal and towards the tip of the cone.

From the conservation of the angular momentum it follows that  the horizontal velocity  is equal to M / h.

Here, M is a constant and h is the cone height at a given point. Since  the energy E is conserved, we have

Eq.(1):  k (h&#039; )^2/ 2 + M^2 / (2 h^2) + g h = E,

where k=cos^2 (α/2) and α is the cone angle. Actually, k does not play a role.

Eq.(1) describes one dimensional motion of a particle with mass k in the potential field M^2 / (2 h^2) + g h.

The equation of motion itself is

Eq.(2):  k h&#039;&#039;  = M^2 / h^3 - g.

Multiply Eq.(2) on h:

Eq.(3): k (h h&#039; ) &#039; - k  (h&#039; )^2 = M^2 / h^2 - g h.

Substitute here k (h&#039; )^2 from Eq.(2):

Eq.(4):  k (h h&#039; ) &#039;  + 3 g h = 2 E.

When you average over the time, the first term drops out and

Eq.(5):  &lt;h&gt; = (2/3) E / g.

h&#039; = 0 at points where the height reaches its minimum or maximum. From Eq.(1) it follows

Eq.(6):  E/g =  (h_max^3 - h_min^3)/ (h_max^2 - h_min^2) =

(h_max^2 + 2 h_max h_min + h_min^2)/ (h_max + h_min),

so that

Eq.(7): &lt;h&gt; = (2/3)  (h_max^3 - h_min^3) / (h_max^2 - h_min^2).

The same relation holds for distances from the tip, because D is proportional to h.

Substitute numbers and get the result.

Answer 7

Where did you get this question from? A physics book or a calculus book - I&#039;m curious and I want to know how to do it please.

I bet the special prize is 10 points for best answer.