How Low Can You Get?

I took an empirical approach, sweeping the initial angular velocity over a range. For each initial angular velocity, a time-domain simulation was run until one of the three termination conditions was reached. With a radius of $1$ , I ended up with a minimum height of $h_{min} \approx 0.485$ occurring at initial angular speed $\omega_0 \approx 1.426$ . The angle $\theta$ is taken with respect to the vertical.

Below is a plot of initial angular speed vs the minimum height. If the particle has very little speed initially, it doesn't travel far along the surface, and ends up with a height close to the radius of the sphere. If the particle has very high speed initially, it more or less immediately flies off, resulting in a large final height. Only when the initial speed is at an intermediate value is the final height minimized.

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import math

R = 1.0    # sphere radius
m = 1.0   # particle mass
g = 10.0  # gravity
u = 0.5   # friction coeff

dw0 = 0.001  # increment in initial angular speed
dt = 10.0**(-4.0)  # time step

####################################

w0 = dw0              # w0 is initial angular speed

hmin = 9999999.0
w0_store = 0.0

while w0 <= 10.0:    # evaluate w0 values over a range
                    # for each w0, run the simulation.........
                    # until one of three termination conditions is satisfied

                    # Initialize time domain simulation
    t = 0.0

    S = 0.0         # Circumferential distance covered
    Sd = R*w0
    Sdd = 0.0

    theta = 0.0     # angle with the vertical
    thetad = Sd/R

    # Initialize termination flags to zero

    flag1 = 0   # set high when particle comes to rest
    flag2 = 0   # set high when particle leaves surface
    flag3 = 0   # set high when particle reaches ground without leaving surface 

    while (flag1 == 0) and (flag2 == 0) and (flag3 == 0):

        S = S + Sd*dt         # numerical integration
        Sd = Sd + Sdd*dt

        theta = theta + thetad*dt
        thetad = Sd/R

        Fc = m*(Sd**2.0)/R                # centripetal force
        # Fc = m*g*math.cos(theta) - N

        N = m*g*math.cos(theta) - Fc      # normal force

        F = m*g*math.sin(theta) - u*N      # accelerating force along surface

        Sdd = F/m                        # acceleration along surface

        # Evaluate termination flags

        if Sd <= 0.0:     # come to rest on surface
            flag1 = 1

        if N <= 0.0:      # leave surface
            flag2 = 1

        if theta >= math.pi/2.0:   # reach ground without leaving surface
            flag3 = 1

        t = t + dt

    h = R*math.cos(theta)      # final height upon termination of time-domain simulation

    if h < hmin:               # store hmin and corresponding w0
        hmin = h
        w0_store = w0

    w0 = w0 + dw0

####################################

print dw0
print dt
print ""
print hmin
print w0_store

#>>> 
#0.001
#0.0001

#0.485019792659
#1.426
#>>> 

Let $\theta$ be the angle $\angle TOP$ . While the particle is in contact with the hemisphere and still moving, the particle will experience a normal reaction $mX$ and a frictional force $\tfrac12mX$ as indicated in the diagram (note that the motion is, essentially, two-dimensional). We note that $\theta = 0$ and $\dot{\theta} = \omega$ at time $t = 0$ . For convenience, we introduce the dimensionless parameter $\alpha > 1$ , where $a\omega^2 = \tfrac12(\alpha-1)g$ .

The position vector, velocity and acceleration of the particle are $\mathbf{r} \; = \; a{\sin\theta \choose \cos\theta} \hspace{1cm} \mathbf{\dot{r}} \; = \; a\dot{\theta}{\cos\theta \choose -\sin\theta} \hspace{1cm} \mathbf{\ddot{r}} = a\ddot{\theta}{\cos\theta \choose -\sin\theta} - a\dot{\theta}^2{\sin\theta \choose \cos\theta}$ so we get the equation of motion $ma\ddot{\theta}{\cos\theta \choose -\sin\theta} - ma\dot{\theta}^2{\sin\theta \choose \cos\theta} \; = \; mX{\sin\theta \choose \cos\theta} - \tfrac12mX{\cos\theta \choose -\sin\theta} + mg{0 \choose -1}$ Taking radial and transverse components, we obtain $a\ddot{\theta} \; = \; -\tfrac12X + g\sin\theta \hspace{2cm} -a\dot{\theta}^2 \; = \; X - g\cos\theta$ Eliminating $X$ from these equations, we obtain $\begin{aligned} a(2\ddot{\theta} - a\dot{\theta}^2) & = \; g(2\sin\theta - \cos\theta) \\ \tfrac{d}{d\theta}\left[a\dot{\theta}^2e^{-\theta}\right] & = \; g(2\sin\theta - \cos\theta)e^{-\theta} \\ a\dot{\theta}^2e^{-\theta} & = \; A - \tfrac12g(3\sin\theta + \cos\theta)e^{-\theta} \end{aligned}$ for some constant $A$ . Applying our initial conditions, it follows that $A = \tfrac12g + a\omega^2 = \tfrac12g\alpha$ , so we deduce that $\begin{aligned} a\dot{\theta}^2e^{-\theta} & = \; \tfrac12g\left[\alpha - (3\sin\theta + \cos\theta)e^{-\theta}\right] \\ Xe^{-\theta} & = \; \tfrac12g\left[3(\sin\theta + \cos\theta)e^{-\theta} - \alpha\right] \end{aligned}$ Thus motion continues provided that $\dot{\theta} > 0$ and $X > 0$ , so provided that $f(\theta) < \alpha < g(\theta)$ , where $f(\theta) \; = \; (3\sin\theta + \cos\theta)e^{-\theta} \hspace{2cm} g(\theta) \; = \; 3(\sin\theta + \cos\theta)e^{-\theta}$ Even if it were true that $\dot{\theta} > 0$ for all $0 \le \theta \le \tfrac12\pi$ , the fact that $X = -a\dot{\theta}^2 < 0$ when $\theta = \tfrac12\pi$ means that $X$ will become zero, and so the particle will leave the surface, before it reaches the ground. Thus the third option given in the question never occurs. To make the final height $h$ as small as possible, we want to be able to reach as large a value of $\theta$ as possible.

We note that $f(\theta) < g(\theta)$ for all $0 < \theta < \tfrac12\pi$ . Moreover $f'(\theta) = 2(2\cos \theta - \sin \theta)e^{-\theta}$ , and so $f(\theta)$ reaches a maximum at $\theta = \theta_0 = \tan^{-1}\tfrac12$ . On the other hand $g'(\theta) = -6\sin \theta e^{-\theta}$ , and so $g$ is strictly decreasing on $(0,\tfrac12\pi)$ . Define $\alpha_0 = f(\theta_0) \approx 1.40645358$ . If $1 < \alpha \le \alpha_0$ , then the particle comes to rest on the hemisphere, and the maximum value of $\theta$ is achieved is the solution of $f(\theta) = \alpha$ that is less than or equal to $\theta_0$ . Thus no value of $\theta$ greater than $\theta_0$ is possible in this case. If $\alpha_0 < \alpha < 3$ then the particle does not stop, but eventually leaves the hemisphere when $\theta = g^{-1}(\alpha)$ . If $\alpha \ge 3$ then the particle leaves the hemisphere immediately, and so the largest value of $\theta$ that is achieved is $0$ . Thus the supremum of the values of $\theta$ that can be achieved is $\theta_1 = g^{-1}(\alpha_0) \approx 1.06461595$ . Thus the infimum of the final height above ground that can be achieved is $a\cos \theta_1$ , so that $k = \cos\theta_1 = \boxed{0.48484014}$ .