Question on Sliding/rolling spherical ball

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SUMMARY

The discussion centers on the dynamics of a bowling ball sliding and transitioning to rolling motion. The ball, with a radius of 11 cm and an initial speed of 7.8 m/s, experiences kinetic friction with a coefficient of 0.13. The linear acceleration calculated is -1.274 m/s², while the angular acceleration is determined to be 28.9545 rad/s², although the latter was initially reported as incorrect. The time taken for the ball to slide is approximately 1.74927 seconds, leading to further calculations regarding distance and final speed upon transitioning to smooth rolling.

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  • Understanding of Newton's Second Law (F = m*a)
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  • Knowledge of moment of inertia for solid spheres (I = (2/5)mR²)
  • Basic concepts of friction and its effects on motion
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  • Learn about the transition from sliding to rolling in rigid body dynamics
  • Explore the implications of angular acceleration in rotational motion
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Homework Statement


A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball (the figure) slides on the lane with initial speed vcom,0 = 7.8 m/s and initial angular speed ω0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.13. The kinetic frictional force http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c11/low_fvec.gifkacting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed vcom has decreased enough and angular speed φ has increased enough, the ball stops sliding and then rolls smoothly. During the sliding, what are the ball's (a) linear acceleration and (b) angular acceleration? (c) How long does the ball slide? (d) How far does the ball slide? (e) What is the linear speed of the ball when smooth rolling begins? Note that the clockwise direction is taken as negative.

Homework Equations


Second Laws: F=m*a and. Torque=I*angular acceleration

I=(2/5)mR^2 ------ this I am not sure. This is probs where i went wrong.

The Attempt at a Solution


For part a, I used Newtons second law
-m*a=friction force
thus a=-frictionconstant * 9.8=-1.274

Part b,
I had this expression, Iα=R*μ*m*g
where i too I to be I=(2/5)mR^2
and i got α=28.9545
But this is WRONG

Part C i got 1.74927... which is correct
Part D is wrong followed by the wrong answer from part B
Part e i got the right answer
 

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I agree with the magnitude of the angular acceleration you got for part b, but note the instruction that clockwise is to be taken as negative. Relate that to the picture.
 
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