A bowling ball is launched down a bowling lane at a resturant. When released, th

Question

A bowling ball is launched down a bowling lane at a resturant. When released, the ball is sliding but not rolling. The frictional force on the ball by the floor exerts a torque about its center of mass, and the ball starts to roll faster and faster, even as the same frictional force causes a decelleration of the center of mass. Eventually when the rolling catches up with the sliding, the ball stops sliding and continues to roll the rest of the way down the lane. If ball is released at a velocity of 8 m/s, how fast will it be moving when it gets to the end of the lane 60 feet away? How long does it take to travel 60 feet to the head pin? Take the coefficient of sliding friction between the ball and the lane to be 0.20.

Explanation / Answer

apply the kinematic equation V^2 = u^2 + 2aS

a = u g

a = 0.2 * 9.8

a = 1.96 m/s^2

1 feet = 0.3048 m

so

V^2 = 8^2 + (2 * 1.96 * 60* 0.3048)

V = 11.86 m/s

-------------------

S = 0.5 gt^2

t^2 = 2 * 60 *0.3048/1.96

t = 4.32 secs

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