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

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

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 $2 y_C$