Now, for the physics. With proper bowling form, all of the kinetic energy given to the ball comes from the conversion of gravitational potential energy after the backswing, as the ball should act as though it is in freefall. The arm should not give the ball any kinetic energy. The shoulder should just act as an axis of rotation. The ball I use is 15 pounds (6.8kg) and the average height of my backswing is approximately 3 feet (0.91m). Using this, we can find the net kinetic energy of the ball at the moment before the ball hits the lane.
KEnet = -mg∆h
KEnet = -(6.8kg)(9.8m/s^2)(-0.91m)
KEnet = 60. J
Because the lanes are covered in oil, the ball is not rolling without slipping, so to simplify things, I'm going to treat the ball as a sliding object (the physics of a ball rolling on oiled lanes while also changing direction due to the spin on the ball is too complicated to analyze here). Also, since the oil makes the friction of the ball against the lane negligible, we will assume that velocity is constant. Therefore, we can find the forward velocity of the ball "sliding" down the lane.
KE=(1/2)mv^2
60.J=(1/2)(6.8kg)v^2
v=4.2 m/s
The ball hitting the 6-pin is an elastic collision. Therefore, we can find the velocity of the ball immediately after the collision, because the mass of each pin is 1.5kg. Also, the 6-pin will need the travel at approximately 2.0 m/s after the collision in order to have enough energy to knock the 7-pin over (a bit of an oversimplification).
MBiVBi=MBfVBf + MPVP
(6.8kg)(4.2m/s)=(6.8kg)VBf + (1.5kg)(2.0m/s)
VBf = 3.8m/s
As you can see, bowling involves many different physics concepts, such as energy, kinematics, and rotational motion (which was not covered here). Not all of this information is useful in improving my scores, but it's at least interesting to think about.
Sources:
https://www.thoughtco.com/bowling-pin-rack-420521
https://en.wikipedia.org/wiki/Bowling_pin
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