Rolling Bowling Ball

  • #1
Hi everyone, this problem involves smooth rolling and translational motion:

1. Homework Statement

A bowler throws a bowling ball of radius R= 11 cm along a lane. The ball slides on the lane with initial speed vcom = 8.5 m/s and initial angular speed w0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.21. The kinetic frictional force acting on the ball causes a linear acceleration of the ball while producing a torque that causes and angular acceleration of the ball. When speed vcom has decreased enough and angular speed w has increased enough, the ball stops sliding and then rolls smoothly. d)how long does the ball slide?



Homework Equations


Flinear = ma = -mgμk
torque = rF = Iα
Ki + Ui = Kf + Uf

The Attempt at a Solution


I have figured out the linear acceleration a = -2.1 m/s2, and angular acceleration α = 47 rad/s2 using the fact that Flinear = ma = -mgμk and torque = rF = Iα.

I know that there is smooth rolling if vcom = rω = (.11m)ω

I set up an equation using conservation of energy to solve for ωf, which is whenvsliding should end and smooth rolling should begin.

(1/2)mvi2 + (1/2)I ωi2 = (1/2)mvf2 + (1/2) I ωf2

(1/2)m(8.5m/s)2 = (1/2)m(.11ω)2 + (1/2)(2/5 mr2f2

Canceling out mass and simplifying:

36.125 m2/s2 = .00605ω2 + .00242ω2

ωf = 65.3 rad/s

then, ω = ω0 + αt

65.3 rad/s = 47 rad/s2t

t= 1.4 s

However, the given answer in the book is 1.2s, and I'm not sure what's going on...I've tried to account for rounding errors, but that doesn't appear to be the problem. Am I neglecting friction when I shouldn't be, and if so, how would I calculate energy lost to friction if I don't yet know the distance the ball traveled? Any help would be appreciated, thank you.
 

Answers and Replies

  • #2
BvU
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Are you sure conservation of energy applies here ?
 
  • #3
OldEngr63
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With sliding friction, there is work being done that goes away as heat.
 
  • #4
rcgldr
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Since you've calculated the linear deceleration and angular acceleration, how long does it take before the ball transitions from sliding into rolling motion?
 
  • #5
haruspex
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linear acceleration a = -2.1 m/s2,
Are you taking g as 10 m/s2? Given the precision of the other data, I would use 9.8.
But the most important thing is to follow rcgldr's hint and not assume work is conserved.

By the way, if you only wanted the final velocity and didn't care about the distance or time to that point, a neat trick is to consider conservation of angular momentum. You have to choose the reference axis carefully so that friction can be ignored.
 
  • #6
Thank you! I was able to get the answer using conservation of momentum: rmvi = rmvf + Iω. But what is the deal with rmv, and why wouldn't I use just mv for translational momentum? As in, what is the difference between the two types of momentum in this equation I set up? I know it's correct, but I'm not sure why...
 
  • #7
rcgldr
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You didn't need to use conservation of momentum either. The friction force divided by the mass of the ball is the linear deceleration of the ball, and the friction force times the radius of the ball divided by the angular inertia of the ball is the angular acceleration of the ball. This allows you to calculate the linear speed (which is decreasing) of the ball versus time, and the surface speed of the ball (which is increasing): radius x ω versus time, and at some point in time they are equal and the ball transitions into rolling.
 
Last edited:
  • #8
haruspex
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Thank you! I was able to get the answer using conservation of momentum: rmvi = rmvf + Iω. But what is the deal with rmv, and why wouldn't I use just mv for translational momentum? As in, what is the difference between the two types of momentum in this equation I set up? I know it's correct, but I'm not sure why...
The frictional force contributes a change to the horizontal linear momentum. It can be omitted from the angular momentum equation by taking the axis to be a point in the line of action of the frictional force. The friction doesn't matter then because it has no moment about the axis.
 

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