World’s Longest Bowling Strike

By: Danielle LaPaglia

http://www.youtube.com/watch?v=S_mna_WXnHk

This video shows someone throwing a bowling ball down a hill at a set of ten bowling pins and he gets a strike. I don’t know if it really is a world record or not, but impressive regardless. I will try to find the velocity of the ball when it strikes the pins. The distance the ball traveled is estimated to be 260 ft (approximately 80 m) – 70 m down a hill and another 10 on flat ground before striking the pins. The following estimates and assumptions were made:

Mass of bowling ball = 15 lb = 6.8 kg

Radius of bowling ball = .108 m

Speed of ball when released = 15 mph = 6.7 m/s

The height of the hill = 10 ft = 3.12 m

When the ball is traveling down the hill it is rolling without slipping, so it is assumed that no friction acts on the ball. When the ball reaches the flat ground at the end of the hill, friction is factored in because the ball clearly slows down before it hits the pins. 

Moment of Inertia for bowling ball approx = 2/5 m r² = 2/5 (6.8 kg)(.108 m)² = .032 kg m²

Conservation of Energy

We have to use both translational and rotational kinetic energy here:

∆KE = -∆PE + W_NC

½ m v_f² + ½ I (v_f/r)² – ( ½ m v₀²+ ½ I (v₀/r)²)= -(mgh_f – mgh₀)

½ (6.8 kg) v_f² + ½ (.032 kg m²) (v_f/.108 m)² – ( ½ (6.8 kg) (6.7 m/s)²+ ½ (.032 kg m^s) (6.7 m/s/.108 m)²)= -((6.8 kg)(9.8 m/s²)(3.12 m) – (6.8 kg)(9.8 m/s²) (0 m))

v_f=9.4 m/s at the bottom of the hill

Now, we assume µ between ball and grass = 0.35

∆KE = -∆PE + W_NC

∆KE = -∆PE + W_NC

½ m v_f² + ½ I (v_f/r)² – ( ½ m v₀²+ ½ I (v₀/r)²)= -µmgd

½ (6.8 kg) v_f² + ½ (.032 kg m²) (v_f/.108 m)² – ( ½ (6.8 kg) (9.4 m/s)²+ ½ (.032 kg m²) (9.4 m/s/.108 m)²)=     -0.35(6.8 kg)(9.8 m/s²)(10 m)

v_f= 6.3 m/s

The ball will strike the pins at a velocity of 6.3 m/s. This velocity happens to be very close to the velocity that the ball was released at.