A bowling ball with a diameter of 21.9cm is rolling down a level alley surface a

Question

A bowling ball with a diameter of 21.9cm is rolling down a level alley surface at 12.1m/s without slipping. Assume the ball is uniform and made of plastic with a density of 850kg/m3 . What is the angular speed of the ball? Calculate the speed (relative to the alley surface) of a point on top of the ball directly above the contact point on the floor. v=? What is the ball's linear kinetic energy? K=? If it now starts to roll up a 30 incline, how far up the incline will it travel before it stops? d=?

Explanation / Answer

Just calculation error

Angular speed = v/r = 12.1/0.219 = 55.25 rad/s

Speed of point at top = v + r = 12.1 + 11.88*0.219 = 24.2 m/s

Linear KE = 1/2mv2 = 0.5*(850*(4/3(0.219)2))*12.12 = 2737.67 J

By conservtion of energy

1/2mv2 + 1/2I2 = mgh (h is the vertical distance)

=> 0.5*12.12  + 0.5*((2/5)*(0.219)2*11.882 = 9.8*h (I of sphere = 2/5mr2)

=> h = 10.46m

dcos60 = h

d = 20.92 m

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