How Does Friction Affect the Angular Acceleration of a Sliding Bowling Ball?

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SUMMARY

The discussion focuses on calculating the initial angular acceleration of a bowling ball sliding down an alley with a coefficient of sliding friction of µ = 0.6. The ball, with a diameter of 25 cm and an initial velocity of 11 m/s, experiences an initial deceleration of 5.886 m/s². The correct method to find angular acceleration involves using the torque equation τ = Iα, rather than the incorrect approach of α = a/R, which only applies to tangential acceleration at a point on the ball.

PREREQUISITES
  • Understanding of rotational dynamics and angular acceleration
  • Familiarity with the moment of inertia for a solid sphere, Icm = (2/5)mr²
  • Knowledge of torque equations in physics
  • Basic concepts of translational versus rotational motion
NEXT STEPS
  • Study the relationship between torque and angular acceleration in rotational dynamics
  • Learn about the moment of inertia for different shapes and how it affects rotational motion
  • Explore the concepts of translational and angular kinematics
  • Investigate the effects of friction on rolling versus sliding objects
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators teaching concepts of rotational motion and frictional effects on acceleration.

nahanksh
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Homework Statement


https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/spring09/homework/10/bowling_ball/8.gif
A bowling ball 25 cm in diameter is slid down an alley with which it has a coefficient of sliding friction of µ = 0.6. The ball has an initial velocity of 11 m/s and no rotation. g = 9.81 m/s^2.
Given that the initial deceleration of the ball is 5.886.

What is the initial angular acceleration of the ball?



Homework Equations


For a sphere Icm = (2/5)mr^2.


The Attempt at a Solution



Firstly, i tried to use the formula [tex]\alpha = a/R[/tex]
Then i got some value of acceleration which turns out to be wrong.

After that, when i used the torque equation [tex]\tau = I\alpha[/tex]
I got the different answer and it was correct...

Why did i get the wrong answer at the first attempt?
I am really confused when i could use the transformation formulae([tex]s to \theta, v to \omega, a to \alpha[/tex])...

Could someone help me out here..?
 
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When you used [tex]\alpha = a/R[/tex], I'm assuming the acceleration you used was the deceleration value they gave you. That value (the deceleration value) is the translational deceleration, meaning it is the deceleration of the center of mass of the ball. The equation you used is referring to the tangential acceleration of a point on the ball that is a radial distance R from some reference point (which you probably took as the center of mass).
 
Last edited:

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