Question

At a typical bowling alley the distance from the line where the ball is released (foul line) to the first pin is 60ft. Assume it takes 5.0s for the ball to reach the pins after you release it, if it rolls without slipping and has a constant translational speed. Also assume the ball weights 12lb and has a diameter of 8.5 inches.

A- Calculate the rotation rate of the ball in rev/s

B- What is the total kinetic energy in pound-feet? Ignore the finger holes and treat the bowling ball as a uniform sphere

Answer 1

A) The rotation rate of the ball in rev/secs is ;  10.79 rev/secs

B) The total kinetic energy of in pound-feet ignoring the finger holes is : 37.6157 pound-feet

Given data :

mass of ball = 12 Ib

diameter of ball = 8.5 inches

Radius = 4.25 inches = 0.10795 m

Time = 5 secs

distance travelled = 60 ft  = 18.288 m

A) Determine the rotation rate of the ball

First step : calculate the velocity of ball

V = distance / time

  = 18.288 / 5 = 3.66 m/s

Next step :

angular velocity ( w ) = V / r  

                                  = 3.66 / ( 0.10795 ) = 33.90 rad/sec

convert to rev/sec = 33.90 / π

                               = 10.79 rev/secs

B) Determine the total kinetic energy

given that ball rolls without slipping

Vcon = Rw

Total K.E = 1/2 IW² + 1/2 mv²

               = 7/10 MR²W²

Total K.E = 7/10 *  5.44 * ( 0.10795 )² * 33.90²

               = 50.99 ≈ 51 J

convert to pound-feet = 37.6157 pound-feet

Hence we can conclude that The rotation rate of the ball in rev/secs is ;  10.79 rev/secs  and The total kinetic energy of in pound-feet ignoring the finger holes is : 37.6157 pound-feet.

Learn more about translational speed : https://brainly.com/question/9078768