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A friend gave a challenging math problem to solve for fun, but since I'm a high school calc student, it's too hard for me to figure out.

If a small ball is at the top of a larger stationary sphere (radius = $1$) and it starts to roll down the side, at what point will the smaller ball lose contact with the larger sphere?

I got off to a good start, but then got stuck. Here's my work. If you don't care to read it all, you can skip to the last paragraph:

I first aim to find the velocity. I start with the acceleration of the ball: $a = g\dot{}\sin(\theta)$ where theta is the angle of inclination the ball is rolling at any given instant. The derivative of the circle equation gives the slope of this incline. The circle is modeled by $f(x) = \sqrt{1-x^2}$ and $f'(x)= -x\dot{}(1-x^2)^{-1/2} \ $Therefore: $\ \theta = \tan^{-1}(f'(x))$.

Substituting this back into the original equation, this is where it gets ugly:

$$a = g\dot{}\sin\left(\tan^{-1}(f'(x))\right)$$

I figured I needed to put all of this is terms of time, so I change $x$, horizontal displacement of the ball, to $s(t)$, and $a$ to $a(t)$. Using my current knowledge of kinematics, I can relate displacement to acceleration such that

$s(t) = \int_{}v(t)~dt \ $ and $\ v(t) = \int_{}a(t)~dt$

so if I'm not mistaken

$$s(t) = \int_{}\int_{}a(t)\ dt\ dt$$

So, substituting this all back into my original equation, I get this:

$$a(t) = g\dot{}\sin\left(\tan^{-1}\left(f'(\int_{}\int_{}a(t)\ dt\ dt)\right)\right)$$

Not sure if this is even correct syntax at this point. Anyways, I got everything in terms of $a$! But, I've never formally learned to solve DE's, so I'm not sure what to do next or if this DE is solvable, and if it is, it must be so complex that it's not a practical solution. There must be a simpler way that I'm missing, what do I do? After solving for the velocity, how would I use it to find when the ball loses contact with the larger sphere?

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  • $\begingroup$ Rolling makes things even more difficult, as for a rolling (solid, homogenous) ball, $2/7$ of the total kinetic energy (if I recall correctly) is rotational and not translational. $\endgroup$ – Arthur Jan 30 '17 at 23:21
  • $\begingroup$ @Arthur I didn't know this. I don't know if the problem is intended to take this into account or not, so let's ignore that detail for now :) $\endgroup$ – Ryan Jan 30 '17 at 23:25
  • $\begingroup$ This can be done in the general case for arbitrary moments of inertia (though it genuinely only makes sense for bodies that actually roll unlike an arbitrary parallelepiped shape) and it reduces down to the case of a body sliding in the limit of zero moment of inertia. :) $\endgroup$ – Triatticus Jan 31 '17 at 4:52
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If we ignore the rolling motion of the small ball with mass $m$ (it slides over the spherical surface), we can write these equations:

\begin{cases}mg\cos\theta-R=m{v^2\over r}\hspace{3cm}\text{Newton's law along the radius}\\mgr+0=mgr\cos\theta+{1\over2}mv^2\hspace{1,3cm}\text{Conservation of energy}\end{cases} enter image description here The small mass loses contact with the sphere when the reaction $R$ is zero, so from the first equation:

$$mg\cos\theta=m{v^2\over r}$$ $$mv^2=mgr\cos\theta$$

where $\theta$ is the angle between the radius $r$ of the sphere and the vertical line passing through the center.

Thus, replacing into the second equation: $$mgr=mgr\cos\theta+{1\over2}mgr\cos\theta\longrightarrow 1=\cos\theta+{1\over2}\cos\theta$$ $$\cos\theta={2\over3}$$ $$\theta=\arccos\left({2\over3}\right)$$

Rolling motion

If we take into account the rotation motion of the small ball (we'll consider simple rolling motion, i.e without slip along the surface), we can write equations similar to the previous ones:

\begin{cases}mg\cos\theta-R=m{v_{cm}^2\over r+r'}\hspace{7cm}\text{Newton's law along the radius}\\mg(r+r')+0=mg(r+r')\cos\theta+{1\over2}mv_{cm}^2+{1\over2}I\omega^2\hspace{1,3cm}\text{Conservation of energy}\end{cases}

where $v_{cm}=\omega r'$ is the speed of the center of mass, $r'$ is the radius of the small ball, $\omega$ is the angular velocity of the small ball and $I={2\over5}mr'^2$ is the moment of inertia of the small ball respect to its center.

enter image description here

Solving the system we get:

$$\cos\theta={2m\over 3m+{I\over r'^2}}$$

If $I=0$, as in the first case, we obtain the previous solution.

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  • $\begingroup$ @Ryan I add something for completeness $\endgroup$ – MattG88 Feb 1 '17 at 20:32
  • $\begingroup$ Very nice. I could never do these types of problems. $\endgroup$ – marty cohen Feb 1 '17 at 20:47
  • $\begingroup$ @martycohen Thank you marty!! I think you can do it ;-) $\endgroup$ – MattG88 Feb 2 '17 at 0:21
  • $\begingroup$ @MattG88 you can use the fact that v = ωr to simplify the answer and get that the angle is arccos(10/17). $\endgroup$ – Eliot Mar 8 at 4:05

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