Ten-ping bowling: Can a ping pong ball knock over a bowling pin?

Question

In this video we see a rather unorthodox bowling technique on display. The gentleman appears to knock over all of the pins using only a ping pong ball. It's a fake, of course: you could never knock over a bowling pin with a ping pong ball. Or could you? How fast would a ping pong ball need to travel in order to knock over a heavy object like a bowling pin? Naturally, you are free to make any simplifying assumptions and order-of-magnitude estimates you like.

Answer 1

I will attempt to answer this question with some basic dynamics and some contact mechanics.

There are two special cases here. a) There is sufficient friction to keep the base of the pin A fixed (imparting a reaction impulse $J_A$ when hit by the ball, or b) The floor is smooth and the pin will translate and rotate at the same time with $J_A=0$. There is also a special configuration where the two cases are equivalent.

What I am varying is the location of the ball hitting $y_B$ (below or above the center of mass C height of $y_C$), and the impact velocity $v$.

Before the impact the velocity state is (positive $x$ is to the right, measured at the center of mass)

• Ball: $\vec{v}_2 = (-v,0,0)$
• Pin: $\vec{v}_1 = (0,0,0)$
• Pin: $\vec{\omega}_1 = (0,0,0)$

The impulses $J_A$ and $J_B$ alter the motion at the center of mass by

• Ball : $\Delta \vec{v}_2 = (\frac{J_B}{m_2},0,0)$
• Pin: $\Delta \vec{v}_1 = (\frac{J_A-J_B}{m_1},0,0)$
• Pin: $\Delta \vec{\omega}_1 = (0,0, \frac{y_C J_A - (y_C-y_B) J_B}{I_C} )$

The final motion of the ball and points A and B are

• $\vec{v}_2^\star =\vec{v}_2 + \Delta \vec{v}_2 = (\frac{J_B}{m_2} - v, 0, 0)$
• $\vec{v}_A^\star =\Delta \vec{v}_1 + \Delta \vec{\omega}_1 \times ( \vec{r}_A - \vec{r}_C ) = (J_A \left( \frac{1}{m_1} + \frac{y_C^2}{I_C} \right) - J_B \left(\frac{1}{m_1} + \frac{y_C (y_C-y_B)}{I_C}\right),0,0)$
• $\vec{v}_B^\star = \Delta \vec{v}_1 + \Delta \vec{\omega}_1 \times ( \vec{r}_B - \vec{r}_C ) =( J_A \left( \frac{1}{m_1} + \frac{y_C ( y_C-y_B)}{I_C}\right) - J_B \left(\frac{1}{m_1} + \frac{ (y_C-y_B)^2 }{I_C}\right),0,0) $

For the case with rough ground, the elastic impact conditions are (along the considering only the x-axis components). The coefficient of restitution is $\epsilon \approx 1$ for pure elastic contact (no stickyness).

• $(\vec{v}_2^\star - \vec{v}_B^\star) = -\epsilon (\vec{v}_2 - \vec{v}_B)$
• $ (\vec{v}_A^\star) =0 $

with solution

• $ J_A = (\epsilon+1) \frac{I_C + m_1 y_C (y_C-y_B)}{I_C + m_1 y_C^2 + m_2 y_B^2} m_2 v $
• $ J_B = (\epsilon+1) \frac{I_C + m_1 y_C^2}{I_C + m_1 y_C^2 + m_2 y_B^2} m_2 v $
• $\Delta v_1 = -(\epsilon+1) \frac{y_B y_C m_2}{I_C+m_1 y_C^2+m_2 y_B^2} v$
• $\Delta \omega_1 = (\epsilon+1) \frac{y_B m_2}{I_C + m_1 y_C^2 + m_2 y_B^2} v$

Now we consider the kinetic energy of the pin, which we have to compare to the energy needed to tip the pin over. This will give us the speed needed to impact.

• $KE = \frac{1}{2} m_1 (\Delta v_1 )^2 + \frac{1}{2} I_C (\Delta \omega_1 )^2 $
• $PE = m_1 g \left( \sqrt{ \left(\frac{b}{2} \right)^2 + y_C^2} - y_C \right)$ where $b$ is the base diameter.

By setting $PE=KE$ the impact speed $v$ is found. More interesting to me is the question of where to impact?

Consider the special case of $J_A=0$ which happens when $y_B = y_C + \frac{I_C}{m_1 y_C}$. This is the instant center of percussion of the pin and it forces a pure rotation by the base without any reactions.

For the case with smooth floor, the elastic impact conditions are

• $(\vec{v}_2^\star - \vec{v}_B^\star) = -\epsilon (\vec{v}_2 - \vec{v}_B)$
• $ J_A =0 $

with solution

• $J_A = 0$
• $J_B = (\epsilon+1) \frac{m_1 I_C}{I_1 (m_1+m_2) + m_1 m_2 (y_C-y_B)^2} m_2 v$
• $\Delta v_1=-(\epsilon+1) \frac{I_C m_2}{I_C (m_1+m_2) + m_1 m_2 (y_C-y-B)^2} v$
• $\Delta \omega_1=-(\epsilon+1) \frac{m_1 m_2 (y_C-y_C)}{I_C (m_1+m_2) + m_1 m_2 (y_C-y-B)^2} v$

and kinetic energy

• $KE = \frac{1}{2} m_1 (\Delta v_1 )^2 + \frac{1}{2} I_C (\Delta \omega_1 )^2$

Consider the special case when $y_B = y_C$ yielding a pure translation with $\Delta \omega_1 =0$.

To optimize the problem we must find the distance $y_B$ which maximizes the kinetic energy. The values that do this are

• Rough floor: $$y_B = \sqrt{ \frac{I_C+m_1 y_C^2}{m_2} }$$
• Smooth floor: $$y_B = y_C + \sqrt{ \left( \frac{m_1}{m_2}-1\right) \frac{I_C}{m_1}}$$

NOTE:

Both of these points fall onto the instance axis of percussion $y_B = y_C + \frac{I_C}{m_1 y_C}$ when $$ m_2 = \frac{1}{\frac{1}{m_1} + \frac{I_C}{m_1^2 y_C^2}}$$ this puts the requirement that $m_2 > \frac{3}{4} m_1$ when the pin is approximately a cylinder with height $2 y_C$

Answer 2

This was a problem that came up in the course of my work, so I am posting it here for my own future reference. I am writing it up publicly since the problem was a little more fun/complicated to solve than I initially expected, so perhaps it will be entertaining for someone else with some spare time on their hands. If anyone can spot an error in my reasoning or find a more elegant solution then I would also be very grateful :)

The objective is to compute the critical velocity $v_c$ of a ping pong ball, which will cause the bowling pin to topple over after the collision. Here are my assumptions. I model the bowling pin as a cuboid block of mass $M$, uniform density, height $h$ and width $w$. A real bowling pin is actually not uniform, but has a lower centre of gravity and is therefore more difficult to push over. This assumption therefore leads to an underestimate of the critical velocity of the ping pong ball. I assume that the ping pong ball of mass $m$ impinges on the very top of the block with velocity $v$ at time $t=0$, and then bounces back with velocity $-v$, transferring $2mv$ momentum to the block. Strictly speaking, this process does not conserve energy, but the correction is small provided that $M \gg m$. Again, this assumption leads to an underestimate of the critical velocity. I also assume that the collision is essentially instantaneous. Finally, I assume that there is sufficient friction between the block and floor so that no slipping occurs.

Given these assumptions, the initial state of the block is specified by the angle between the bottom of the block and the floor, $\theta(0) = 0$, and the initial angular velocity $\dot{\theta}(0) = 2 m v h/I$, with $I = \frac{M}{3}(w^2 + h^2)$ the moment of inertia of the block. The equation of motion for the block is given by Newton's second law $$ \qquad \qquad \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} = \frac{Mg}{2I}(h\sin\theta - w \cos\theta ). \qquad \qquad (1) $$ The physical argument goes as follows. Once the ping pong ball has bounced away, the only force acting on the block is gravity. If the initial angular velocity of the block is very small, then gravity acts to pull the block back down. Conversely, if the initial angular velocity is sufficiently large, then the block will reach a "tipping point", where the block's centre of mass is over the pivot. This point is specified by the critical angle $\tan\theta_c = w/h$. If the block tips further than this, then it will fall under its own gravity. Therefore the critical velocity is the smallest $v$ such that $\theta(t) = \theta_c$ for some $t$.

Now, I'm far too lazy to learn how to solve the nonlinear differential equation $(1)$, so I make a further simplifying assumption, that $w\ll h$. Therefore, if we restrict our analysis to angles $\theta < \theta_c$, the small-angle approximation applies, since $\theta_c\approx w/h$ is itself small. Notice that this assumption again makes the block easier to tip over, and therefore leads again to an underestimate for $v_c$. Within this approximation, we have

$$\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2} \approx \frac{Mg}{2I}(h\theta - w ),$$

which has the solution

$$\theta(t) = \frac{w}{h}[1- \cosh(t/\tau) ] + \frac{2mvh\tau }{I}\sinh(t/\tau),$$

where I have applied the boundary conditions and defined the time constant

$$ \tau = \sqrt{\frac{2I}{Mgh}},$$

which is a characteristic parameter of the block and is therefore independent of $v$.

We find that the block can reach the critical angle $\theta_c \approx w/h$ if there exists a $t$ such that

$$ \frac{2mvh^2\tau}{wI} \tanh(t/\tau) = 1, $$

and clearly this condition can be satisfied for all speeds $v$ greater than the critical velocity

$$v_c = \frac{wI}{2mh^2\tau} \approx \sqrt{\frac{M^2g w^2 }{24m^2 h}}, $$

where the last approximation comes from $I \approx Mh^2/3$, since $w\ll h$. Assuming that the block weighs $M = 2$ kg and has dimensions $h = 30$ cm and $w = 10$ cm, and that the ping pong ball weighs $m = 2.7$ g, we get $v_c \approx 90$ m/s $\approx 300$ kph. The fastest serve of a ping pong ball has been estimated at around 100 kph.