If I can hit the pin hard enough it will fall, right?

    Sorry folks this isn't always true.  There are many facts that determine if the pins will fall or not, such as the entry angle of the ball to the force and speed still acting on the ball by the time it contacts the first pin.

   For example, if the ball hits the first pin dead on, you will simply knock down the one and five pin and might knock over the 2,3,8, and 9 pins.  If you had a very slight entry angle such as one that would sweep next the one pin and proceed to follow its entrance angle, say 3º, it would stand a chance at knocking over almost all the pins leaving only the 10 and 7 pin open.  Bad outcome, huh?  Finally, if your shot has an angle of 6º-8º, you stand the best chance of getting a strike.  This is because the ball is forcing a split between the 1 and 3 pin, driving the 1 pin into the two pin and thus all the way down to the 7 pin.  The 3 pin follows suit by knocking down the 6 and 10 pin because the 3 pin is being moved in a complimentary angle to the balls angle of entry.  The remaining three pins are the 5,8 and 9 pins.  The 6 and 8 pin lie in the balls path, and the 9 pin is in the path of pin 6 after it is hit by the ball.  Now go get them turkeys! 

How can the pins spin around and not always fall down?

   The reason this occurs the ball was unable to transfer enough energy to topple the pins center of mass.  The equation to find the center of mass of the pin is h=r/tanØ.  An easy way to find Ø is to place a pin, used with the alley's permission, on a wooden board.  Place a protractor next the board until the board is lined up with the 0º mark on the protractor.  Slowly and steadily tilt one side of the board until the pin tips over.  Plug Ø into the equation to find the height of the center of mass (h).  My results were:  Ø=13º and the radius of the pin bottom is .026 m.

h=radius/tanØ

h=.026 m /tan 13=.11261 m