Ball in cone question?

Question

A ball (with moment of inertia I = (2=5)MR2) rolls without slipping on the inside surface of a fixed cone, whose tip points downward. The half-angle at the vertex of the cone is θ. Initial conditions have been set up so that the ball travels around the cone in a horizontal circle of radius L, with the contact points (the points on the ball that touch the cone) tracing out a circle (not necessarily a great circle) on the ball.

What should the radius of the circle of these contact points be, if you want the sphere to travel around the cone as fast as possible?

(You may work in the approximation where R is much less than L. Also, assume that the coefficient of friction between the ball and the cone is arbitrarily large.)

(Although not necessary to solve the problem, you may also assume that the cone is a waffle cone and the ball is made of chocolate ice cream).

Answer 1

Do you need derivatives for this question?

Well, more seriously, I have a few minutes, so I’ll see how far I can get without fully understanding the question, then I’ll return later.  In the following I use brackets to denote vector products, and you will have to imagine arrows above symbols for vectors. I use ‘ to denote complete differentiation with respect to time (t).

The kinetic energy of a rolling sphere is given by

T= ½ m|u|^2 +½ I |ω|^2  (1)

Where u=velocity vector of its center of mass, ω is its angular velocity, m its mass, and I the moment of inertia. With |u|^2 I mean u∙u, |ω|^2 likewise. Both u and ω are time dependent.

Suppose the sphere rolls without slipping and without spin on a surface defined by the equation f(x,y,z)=0 .  Then

ω=[N,u]/a,

where N is the unit normal to the surface and a is the radius of the sphere, given by

N=grad f(x,y,z)/| grad f(x,y,z)|.   Define Q= grad f(x,y,z) for short, then

ω=[Q,u]/(a |Q|).   (2)

Let the vector w(t) give the center of mass of the sphere (measured from the origin of the x,y,z coordinates), let r(t) be the point of contact of the sphere to the surface, and v(t)=r(t)’.  Then

w(t)=r(t)+ a N(t),

and the velocity u(t)=w(t)’  is given by

u(t) = r(t)’+ a N(t)’ = r(t)’+ a F∙v / |Q|- a QQ ∙ F∙v /|Q|^3 ,         (3)

where F is the dyadic of components {{fxx,fxy,fxz},{fyx,fyy,fyz},{fzx,fzy,fz... and

fxy is the second derivative w.r.t. x and y of f, etc.    QQ also represents a dyadic, as does the dot product  QQ ∙ F, so that QQ ∙ F∙v is a vector.  The kinetic energy of the sphere is thus given by Eq.(1) with ω and u given by Eqs.(2) and (3), respectively.  In addition we need the potential energy due to gravity, given by

V=mg ( z + a k∙Q /|Q| ),

  = mg ( z + a fz /|Q| ),    (4)

where k is a unit vector pointing upward (z direction).  Using the constraint f(x,y,z)=0,

the Lagrangean is given by

L=T-V-λf(x,y,z).    (5)

where λ is a Lagrange multiplier. Tomorrow I will derive the equations of motion from eqs.(5), using (2), (3) and  (4), but it will be useful to transform to curvilinear coordinates (cylindrical).

************

Let x=ρ cosφ, y=ρ sin φ.  I  decided that there is no advantage to use the Lagrange multiplier method for this problem, but to substitute  z= cotθ ρ , z'= cotθ ρ' directly, where ρ is the distance of the point of contact of the sphere with the cone to the axis. The Lagrangean comes out to be

L = 1/2 (m + I /a^2)  [(ρ -a cosθ)^2 φ'^2 + cscθ^2 ρ'^2]-mg(cosθ ρ +a sinθ ).

Thus the angular momentum J is a constant, where J is given by

J=(m + I /a^2) (ρ -a cosθ)^2 φ'



The equation for ρ is that for a particle of mass m + I /a^2 with potential energy given by

U=mgsinθ (ρ cosθ+a sinθ^2)+1/2 J^2/[(m + I /a^2)(ρ -a cosθ)^2]

This reduces to that of the previous question in the limit a→0.

Note that when ρ =a cosθ the sphere is at the lowest point, at the "bottom" of the cone.

You can do this problem for an arbitrary surface that is moving in an arbitrary way.  I started doing that yesterday (with explicit time-dependence of f(x,y,z,t), but it turned out to be rather complex, so I decided not to.  Eq.(2), i.e.,

ω=[Q,u]/(a |Q|)

will have to be modified, and F∙v→something else also.

According to the question ρ is a constant, so that ρ'=0 and  dU/dρ=0.  This yields

ρ=a cosθ+ [ J^2 sec θ cscθ/(m g(m + I/a^2)) ]^(1/3)

and for the period P=2π/φ',

P= (2π/J)(m + I/a^2)(ρ-a cosθ )^2 ~ J^(1/3)

Thus P→0  as J→0, which means ρ→a cosθ, which is the rest position at the bottom of the cone - entirely as one would expect!

A comment about the solution I gave to the question asked by Mullah Abdullah and solved by Zo Maar.  The solution I gave is equivalent to that of Zo, but I didn't realize that because I was under the impression that the three roots z1,z2,z3 are unrelated, which is not the case.  The elegant solution Zo gave satisfies the conditions of the question, however, which is to give the solution in terms of the two turning points (extrema), which mine did not.