By: Danielle LaPaglia

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 r

^{2}= 2/5 (6.8 kg)(.108 m)^{2}= .032 kg m^{2}__Conservation of Energy__

∆KE = -∆PE + W

_{NC}
½ m v

_{f}^{2}+ ½ I (v_{f}/r)^{2}– ( ½ m v_{0}^{2}+ ½ I (v_{0}/r)^{2})= -(mgh_{f}– mgh_{0})
½ (6.8
kg) v

_{f}^{2}+ ½ (.032 kg m^{2}) (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/s^{2})(3.12 m) – (6.8 kg)(9.8 m/s^{2}) (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

_{f}^{2}+ ½ I (v_{f}/r)^{2}– ( ½ m v_{0}^{2}+ ½ I (v_{0}/r)^{2})= -µmgd
½ (6.8
kg) v

_{f}^{2}+ ½ (.032 kg m^{2}) (v_{f}/.108 m)^{2}– ( ½ (6.8 kg) (9.4 m/s)^{2}+ ½ (.032 kg m^{2}) (9.4 m/s/.108 m)^{2})= -0.35(6.8 kg)(9.8 m/s^{2})(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.

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