A bowling ball is released with speed v o and no rotational kinetic energy. Afte
ID: 1450393 • Letter: A
Question
A bowling ball is released with speed vo and no rotational kinetic energy. After a period of sliding and rotating, the ball enters pure rotational motion. The coefficient of friction between the ball and the ground while sliding is k.
a) Show that the rotational acceleration of the ball during the initial period of sliding is = 5gk / 2r, where g is the acceleration due to gravity and r is the radius of the ball (take it as a solid sphere).
b) Show that when sliding finishes and pure rolling begins, the speed of the center of mass of the ball is vc = 5vo / 7
Remember that the necessary condition for kinetic energy to be conserved is that only conservative forces act during this time period. During sliding the forces acting upon the ball include gravity, the normal force, and a kinetic friction force - consider whether or not these are all conservative forces...
Explanation / Answer
a) Here, Torque is provided by frictional force to rotate the ball .
=> k * M * g * r = (2/5 * M * r2) *
=> k * g = = (2/5 * r) *
=> = (5k * g)/2r
b) Here, time after which sliding finishes and pure rolling begins = (2vo)/(7k g)
=> angular velocity after which sliding finishes and pure rolling begins = * t
= ((5k * g)/2r) * (2vo)/(7k g)
= 5vo/7r
=> speed of the center of mass of the ball , vc = r * 5vo/7r
= 5vo/7
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