The Physics of Foosball

CRACK! A cheer tears from the mouth of the winner as he watches his kinetic energy translate into certain victory. What sport am I talking about? Foosball of course! You may not know it yet, but Colgate has a pervasive underground Foosball culture, as evidence by the multitude of slightly broken Foosball tables populating residence hall basements all over campus. Since my friends and I refurbished our table with a little bit of duct tape and a lot of WD-40, we’ve played almost nonstop throughout the semester. Like any other legitimate sport, Foosball is a test of focus, coordination, raw talent, and lighting fast reflexes. Of course, the need for quick reflexes comes from the incredible speeds the ball can reach during volleys for possession and goal scoring opportunities. In fact, World Cup Table Football (read: Foosball) players can move the ball up to 35mph (15.65 m/s). Although my friends and I don’t shoot quite that fast, it is fascinating to consider the physics behind each goal. Assuming that we play with a 6g ball with a 35mm diameter, and the each player is a uniform rod 7cm long with a mass of 40 grams rotating at its end, exactly how fast would I need to spin my strikers to score a 35mph goal? And what is the total Kinetic Energy of a foosball rolling without slipping at 35mph?

Part A: The Kinetic Energy of a Foosball

In order to determine the rotational velocity of a player hitting a ball and accelerating it to 15.65m/s, we must determine the total Kinetic Energy of the ball in motion.

KE_total= KE_{translational} + KE_rotational

KE_{translational}= ¹/₂ mv²

KE_rotational= ¹/₂  I ω²

ω = ^v/_r

I_sphere= ²/₅ mr²

KE_total= ¹/₂ mv² + ¹/₂  I ω²

KE_total=¹/₂ mv²+ ¹/₂  (²/₅ mr²) ω²

KE_total= ¹/₂ (.006)(15.65)² + ¹/₂ (²/₅ (.006)(.035)²) (15.65/.035)²

KE_total= .74 + .29

KE_total= 1.03 J

Part B: The Physics of a Foosball shot

Kinetic energy is conserved in this elastic collision.

initial KE_player + initial KE_ball= final KE_player + final KE_ball

We assume the ball is at rest before the shot and the player ends up at rest after the collision. Thus,

initial KE_player= final KE_ball

¹/₂  I_player ω² = + ¹/₂ m_Ball v_Ball^’2

where I_rod=¹/₃ ml²

¹/₂   (¹/₃ (.040) (.07 )²) ω² = 1.03

ω=178 rad/s