hw7_sol - Dynamics 827 Problems 7.25 and 7.26 A bowling...

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Dynamics 827 Problems 7.25 and 7.26 A bowling ball is thrown onto a lane with a backspin ! 0 and forward velocity v 0 . The mass of the ball is m , its radius is r , its radius of gyration is k G , and the coefficient of kinetic friction between the ball and the lane is k . Assume the mass center G is at the geometric center. Problem 7.25 Find the acceleration of G and the ball’s angular acceleration while the ball is slipping. Problem 7.26 For a 14 lb ball with r D 4:25 in : , k G D 2:6 in , ! 0 D 10 rad = s , and v 0 D 17 mph , determine the time it takes for the ball to start rolling without slip and its speed when it does so. In addition, determine the distance it travels before it starts rolling without slip. Use k D 0:10 . Solution to 7.25 Based on the FBD shown, the Newton-Euler equations are X F x W F D ma Gx ; (1) X F y W N mg D ma Gy ; (2) X M G W F r D I G ˛ b ; (3) where ˛ b is the angular velocity of the ball and where I G D mk 2 G : (4) Since the ball slides (at least at first) over the lane, the force law for this problem is F D k N: (5) The kinematic equations consist only of the following relation: a Gy D 0: (6) Substituting the force law, the kinematic equations, and the expression for the mass moment of inertia into the Newton-Euler equations gives three equations in the three unknowns N , a Gx , and ˛ b whose solution is N D mg a Gx D k g; and ˛ b D gr k k 2 G : Hence, the final answer is a Gx D k g and ˛ b D gr k k 2 G :
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828 Solutions Manual Solution to 7.26 In this problem, the governing equations are identical to those of Problem 7.25. Therefore, we have that the solution of this problem in terms of acceleration of the mass center and angular acceleration of the ball are the same as those of Problem 7.25. To determine when the ball starts slipping, we need to determine the velocity of the point on the ball that is in contact with the ground as a function of time and then determine how long it takes for that velocity to become equal to zero. Observing that the computed accelerations are constant, applying constant acceleration equations, we have that the velocity of the mass center G as a function of time is E v G D .v 0 k gt/ O {: (7) Again using constant acceleration equations, we have that the angular velocity of the ball is E ! b D ! 0 k rg k 2 G t ! O k: (8) Denoting the point on the ball that is in contact with the ground by D and observing that the position of D relative to G is E r D=G D r O | , we have E v D D E v G C E ! b E r D=G D " v 0 C r! 0 gt k 1 C r 2 k 2 G !# O {: (9) The velocity of point D is the velocity with which the ball slips relative to the floor. The time required to achieve rolling without slip is therefore the time it takes for E v D D E 0 , i.e., E v D D E 0 ) v 0 C r! 0 gt r k 1 C r 2 k 2 G !
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