This is a generalization of the question Particle sliding on a sphere when we also have friction given by $F_f = \mu N$.

See the following figure:

enter image description here

Before doing anything, we can imagine what friction will do. It is clear it will reduce the velocity and thus, delaying the "jumping" point from the sphere to a large angle $\theta_J$ (with respect to the vertical). I have tried two approaches:

1) Numerically via Newton's equations

Newton's equations:

$$r: \quad mg \cos(\theta) -N = m \dot{\theta}^2R $$ $$t: \quad mg \sin(\theta) - \mu N = m \ddot{\theta}R, $$ where $R$ is the radius of the sphere.

Substituting one into the other and rearranging I get:

$$\ddot{\theta} = \mu \dot{\theta}^2 + \frac{g}{R} \left( \sin{\theta} - \mu \cos {\theta} \right),$$ I believe this ODE equation does not have an analytical solution (at least that is what WolframAlpha says) so I have integrate it numerically considering $\theta(0) = 0$ and $\dot{\theta}(0) = 0$ as the boundary conditions, as well as requiring the friction force to be smaller than the weight projection in the tangential direction (otherwise the particle won't move).

Then I have computed numerically the moment where the normal $N$ is zero for the first time, that is, for the "jumping" angle $\theta_J$ and velocity $\dot{\theta}_J$ : $$N(\theta_J, \dot{\theta}_J) = mg \cos \theta_J - mR\dot{\theta_J}^ 2 = 0.$$ From this condition I have been able to extract the "jumping" angle and velocity in function of $\mu$, $\theta_J (\mu)$ and $\dot{\theta_J}(\mu)$. Suprisingly, after all this "complex" computations I got what seems linear relationships for $\theta_J (\mu)$ and $\dot{\theta_J}(\mu)$ (for $m=1$, $R=1$, $g = 9.8$, $\theta_0 = 0.57º$ and $\dot{\theta}_0 = 1.5 \text{ rad/s}$ ): enter image description here enter image description here For the first plot, the linear fit is given by: $$ \theta_J = a\mu + b $$ for $a=21.07 \pm 0.3$ and $b=42.0$. The interceipt $b$ is the value expected for $\theta_J (\mu = 0)$, which is given by (from the problem without friction): $${\theta_J}(\mu = 0) = \cos^{-1} \left(\frac{2}{3} + \frac{{\dot{\theta}_0}^2R}{3g}\right) = 42.0.$$

This linear behaviour stands out as the ODE is highly nonlinear! So I suspect there might be a simpler way to compute the "jumping" points which avoids computing the full trajectory and makes use of the "jumping" condition given by $N(\theta_J, \dot{\theta}_J) = 0$ to get (perhaps) an analytical linear solution for $\theta_J (\mu)$ and $\dot{\theta_J}(\mu)$?

2) Via Energy conservation Following the non-friction case, the easier way to compute it is considering energy conservation to find a relationship between the "jumping" velocity and angle $\dot{\theta_J} (\theta_J)$. I have tried it but I get stuck. I started considering the loss of mechanical energy is the work of the friction force: $$\Delta E_m = W_{F_f}$$ Now, the left hand side is: $$\Delta E_m = mgh_J - \frac{1}{2} m v_J^2 = mgR(1- \cos \theta_J) - \frac{1}{2} m R^2 \dot{\theta_J}^2, $$ where $v_J$ is the linear jumping velocity and $h_J$ the height from the top of the sphere where the ball "jumps".

And the right hand side: $$ W_{F_f} = \int_0^{\theta_J} F_f R d\theta = R m \mu \left( \int_0^{\theta_J} g \cos \theta d\theta - \int_0^{\theta_J} R \dot{\theta}^2 d\theta \right) = R m \mu \left( g \sin \theta_J - R \int_0^{\theta_J} \dot{\theta}^2 d\theta \right), $$ where at the second equality I have substituted the expression for the normal $N(\theta, \dot{\theta})$. The question is how to approach this last integral... I substituted $\dot{\theta}^2$ from the 2nd order ODE from the previous approach and managed to integrate both terms and obtained the same expression as in the left hand side, basically finding that $$ 1 = 1,$$ which is not very useful.

QUESTION: Any ideas or approaches to solve this problem with an apparent simple solution? Is it true that $\theta_J (\mu)$ and $\dot{\theta_J}(\mu)$ follow linear relationships or is it just an artifact of the range studied and farther out the functions have different behaviours?

Comment: I have to say too I am using as initial conditions $\theta_0 = 0.57º$ and $\dot{\theta}_0 = 1.5 \text{ rad/s}$ so the particle does not get stuck before moving. That is because it will only move if, initially (which is the worse case), $W_t > F_f$, for $W_t$ the tangential weight. This implies there is a condition for a maximum $\mu$ given $\theta_0$ and $\dot{\theta}_0$: $$ \mu \leq \mu_\text{max} \equiv \frac{g \sin \theta_0}{g \cos \theta_0 - \dot{\theta}_0^ 2 R}.$$ For the initial conditions taken, $\mu_\text{max} = 0.0129$.

  • $\begingroup$ The residuals don't look randomly distributed about that line - the points are first consistently below, then consistently above, then consistently below, the line. It seems far more likely to me that the line is merely a good approximation for what's going on. $\endgroup$ Oct 2, 2019 at 14:15
  • $\begingroup$ @probably_someone see my edit. I think the first plots might have some small fluctuations which might be biased because of the method of computation. $\endgroup$
    – Puco4
    Oct 2, 2019 at 14:59
  • $\begingroup$ Check your simulation, I think that sum think is wrong, because if the friction coefficient get larger the angle must be smaller? I also did the simulation the result is “quasi linear “ $\endgroup$
    – Eli
    Oct 2, 2019 at 15:20
  • $\begingroup$ The log scale actively undermines what you're trying to show. If you're trying to make the point that something looks linear, you should show that the differences between the data and the line are randomly distributed about zero, and that these differences are consistent with zero to within the uncertainties of the simulation. The log scale hides these differences for higher values of $\mu$, and it also distorts them: adding 0.01 to a point will move it a different distance than subtracting 0.01. In any case, see the answer I posted. $\endgroup$ Oct 2, 2019 at 15:20
  • 1
    $\begingroup$ @Eli I think it is the opposite, a higher friction means the particle will stay more time in the sphere and leave at a larger angle (the angle is with respect to the top of the sphere) $\endgroup$
    – Puco4
    Oct 2, 2019 at 15:22

1 Answer 1


This is my simulation result , so I don't see the linearity . the simulation is done with Maple program where I stop the simulation if the normal force N is less then zero.

enter image description here

Edit: I did the simulation with this equations

Position Vector is :

$$R= r\,\left[ \begin {array}{c} \cos \left( \varphi \right) \\\sin \left( \varphi \right) \end {array} \right] $$

so I get this equations of motions (Euler Lagrange)

$${\frac {d^{2}}{d{\tau}^{2}}}\varphi \left( \tau \right) - \left( -{ \frac {\mu\,\sin \left( \varphi \left( \tau \right) \right) }{r}}-{ \frac {\cos \left( \varphi \left( \tau \right) \right) }{r}} \right) g-\mu\, \left( {\frac {d}{d\tau}}\varphi \left( \tau \right) \right) ^{2} =0$$

and the normal force:

$$N=m \left( \left( {\frac {d}{d\tau}}\varphi \left( \tau \right) \right) ^{2}r-g\sin \left( \varphi \left( \tau \right) \right) \right) $$

Data :




for each $\mu$ the simulation run and stop when $N\le 0.001$

Least-square fit give you


  • $\begingroup$ yeah, maybe I kept with a too low $\mu$ regime... Just for clarification, could you add a linear regression in your plot to visualize? Then I will accept as correct the answer. But I will try first too to get to higher $\mu$ to see if I find the same, did you use the same values for the initial conditions? $\endgroup$
    – Puco4
    Oct 2, 2019 at 16:14
  • $\begingroup$ I really don't understand how can you get to $\mu= 0.3$ with the same initial conditions I wrote, because for the particle to start moving has to fulfill $F_f < W_t$, where $W_t$ is the tangential weight and this gives a maximum value for $\mu$: $\mu_{\text{max}} < \frac{g \sin \theta_0}{g\cos \theta_0 - \dot{\theta}_0^2 R}$, which for my initial conditions is $0.0129$ $\endgroup$
    – Puco4
    Oct 2, 2019 at 17:00
  • $\begingroup$ @Puco4 I add more information for you $\endgroup$
    – Eli
    Oct 2, 2019 at 17:42
  • $\begingroup$ thank you for you answer, but I still have a couple of doubts: 1) i believe you compute the Normal with a negative value, as the normal should be equal to the weight in the radial coordinate - the centripetal acceleracion, not the other way around. 2) I continue without understanding how you can manage to take this initial conditions and compute until this $\mu$, which violates equation from Comment1 (edit 1). You are in a regime where the friction force dominates the weight!! $\endgroup$
    – Puco4
    Oct 2, 2019 at 19:26
  • $\begingroup$ To question 1 my friction force is $\mu\,N$ so if N is negative it is ok. I don’t have yet answer for your question 2 $\endgroup$
    – Eli
    Oct 2, 2019 at 19:34

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