Marble in a bowl

We consider the conservation of energy with 3 different contributions:

$\begin{aligned} \text{potential energy:} \qquad E_\text{pot} &= m g y \\ \text{kinetic energy:} \qquad E_\text{kin} &= \frac{1}{2} m v^2 \\ \text{rotational energy:} \qquad E_\text{rot} &= \frac{1}{2} I \omega^2 \\ E_\text{total} = E_\text{pot} + E_\text{kin} + E_\text{rot} &= 0 \end{aligned}$

with mass $m$ , velocity $v = r \omega = R \dot \phi$ , angular velocity $\omega$ and moment of inertia $I = \frac{2}{5} m r^2$ for a solid sphere. With $y = -R\sin \phi$ follows

$\begin{aligned} & & E_\text{total} &= - m g R \sin \phi + 0.7 m R^2 \dot \phi^2 = 0 \\ \Rightarrow & & \frac{d \phi}{dt} &= \sqrt{ \frac{g}{0.7 R}} \sqrt{\sin(\phi)} \\ \Rightarrow & & \frac{d \phi}{\sqrt{\sin(\phi)}} &= \sqrt{\frac{g}{0.7 R}} dt\\ \Rightarrow & & \int_{0}^{\pi/2} \frac{d \phi}{\sqrt{\sin(\phi)}} &= \sqrt{ \frac{g}{0.7 R}} T\\ \Rightarrow & & T &\approx \frac{2.622}{\sqrt{\frac{g}{0.7 R}}} = 0.302 \, \text{s} \end{aligned}$

The non-linear differential equation was solved by separation of variables and the specific integral was evaluated numerically (with the help of Wolfram Alpha).

I used the old "back of an envelope" method. In freefall, it would take approximately 0.19 s to fall the 18 cm. Deflection by the bowl would make it take somewhat longer, but not greatly so. 0.3 s is therefore the sane answer. :)

The ball travels along a circle of radius $R = \SI{0.18}{m}$ . It acts much like a pendulum of length $R$ , but static friction is a retarding force. We analyze the forces:

• Force of gravity $mg$ ; the component parallel to the bowl is $mg\sin\theta$ .

• Normal force; has no parallel component.

• Static friction $F_s$ ; directed parallel to the bowl, opposing the motion. It also generates a torque $\tau = F_sr$ on the ball.

For the rotational acceleration we find (with $I = \tfrac25mr^2$ for a solid sphere) $\alpha = \frac{\tau}I = \frac{5F_s}{2mr}.$ For the linear acceleration along the bowl we have $a_{\|} = g\sin\theta - \frac{F_s}m.$ Since the ball rolls without slipping, $a_{\|} = \alpha r$ . Substituting the previous two equations allows to eliminate $F_s$ ; we find $a_{\|} = \frac 5 7 g \sin \theta.$ Relating the linear motion along the bowl to the angle $\theta$ gives $a = R\ddot\theta$ ; thus $\ddot\theta = \frac 5 7 \frac g R \sin \theta.$ Compare this to the equation for a normal pendulum: $\ddot\theta = (g/\ell)\sin\theta$ , it is easy to adjust the well-known equation for the period: $T \approx 2\pi\sqrt{\frac{7R}{5g}}.$ This small-angle approximation is not exactly valid, since the ball's amplitude is $90^\circ$ ; however, the increase in period is only $18\%$ . Finally, the answer to the problem is one-fourth of a full period: $t \approx \frac \pi 2 \sqrt{\frac{7\cdot 0.18}{5\cdot 9.81}} \approx \boxed{\SI{0.3}{s}}.$

I start this by calculating the amount of space that the ball travelled using a relation between the angle that the movement describes and the length of the circumference:

• 2 π -------- 2 π R

• $\frac{π}{2}$ ------- x

• x= $\frac{2*π*R*π}{2*π*2}$ //

• x= $\frac{π*R}{2}$ //
• x= $\frac{π*0,19}{2}$ //
• x=0,3
• Δs = 0,3

Now, I will use conservation of energy, since the ball hasn't lost any energy during his movement, all the potential energy got transformed into kinetic energy:

• Epot = Ekin
• m g h= $\frac{mv²}{2}$
• 10*0,19 = $\frac{v²}{2}$
• v²=3,8
• v=1,95

Now that I have velocity, I can go to the simplest formula of acceleration, and since the initial velocity is zero:

• a= $\frac{Δv}{Δt}$
• a= $\frac{1,94}{Δt}$

That's all we need for now. With these elements, I can use the kinematic equation of Torricelli, notice that we only need the initial and the final state of the object, avoiding the midterm situation:

• v²=vo²+2 a Δs
• 3,8=0+2 $\frac{1,94}{Δt}$ 0,3
• 3,8=(\frac{1,17}{Δt})
• Δt=(\frac{1,17}{3,8})
• Δt=0,30

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The objective of this solution was to avoid the maximum amount of complicated calculum as possible, using the simpliest equations I've found. I hope everything is right, but please let me know if I made some mistakes ( or if it's everything wrong ). Also, I'm really sorry, but I am new here and I don't really know how to work with Latex, so I made the best I could out of the tools that I've found.