Tuesday, November 29, 2016

Velocity of a Bowling Bal

Over Thanksgiving break a few friends and I went bowling. While watching the bowling ball travel down the lane I realized that I approximate the velocity at which a ball was traveling. Important aspects of this scenario to note are I'm assuming the alley is friction-less and the ball is rolling without slipping. The length of a standard bowling alley lane is around 19 meters and the ball in question has a mass of 5kg.

The Potential energy of the ball equals the mass of the ball times g times the change in height. With the change in height being a little lower than my height when I bring my arm back before releasing the ball. A little less than my height of 1.76m, the approximate height I bring the ball up should be around 1.7m.

PE=mg(h1-h2)=5kg*9.8m/s2 *(1.7m-0m)=83.3 Joules

The total Kinetic Energy of the ball would be the sum of the KE(rot) and KE(trans)

KE(tot)=KE(trans) + KE(rot)                     I(sphere)=(2/5)mr2      w=(v/r)

KE(tot)=(1/2)mv2+(1/2)Iw=(1/2)mv2 + [(1/2)((2/5)mr2](v/r)2)=(7/10)mv^2

Assuming no friction, the entire PE is converted into KE as the ball travels 
KE=PE         (7/10)mv2   = 83.3 J               (7/10)(5 kg)v^2  = 83.3 J    v=4.9m/s

Solving for velocity a value of 4.9 m/s is obtained

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