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

[nahanksh]

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 $$\alpha = a/R$$
Then i got some value of acceleration which turns out to be wrong.

After that, when i used the torque equation $$\tau = I\alpha$$
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($$s to \theta, v to \omega, a to \alpha$$)...

Could someone help me out here..?

 

[Gear300]

When you used $$\alpha = a/R$$, 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).

 

[kerimek]

I would like to point out that rotational kinematics can be a complex topic and it is not uncommon to make mistakes in calculations. In this particular scenario, it is important to note that the bowling ball is sliding down the alley, which means that both translational and rotational motion are occurring simultaneously. This means that the equations used for purely rotational motion may not accurately represent the situation.

In your first attempt, you used the formula \alpha = a/R, which is correct for a point rotating around a fixed axis. However, in this case, the ball is not rotating around a fixed axis, but rather it is sliding and rotating at the same time. This means that the radius of rotation is constantly changing, making this formula not applicable.

In your second attempt, you used the torque equation \tau = I\alpha, which is the correct approach for this scenario. This equation takes into account both the translational motion (through torque) and the rotational motion (through moment of inertia). This is why you got the correct answer using this equation.

It is also important to note that in rotational kinematics, we use different variables and equations compared to translational motion. When dealing with rotational motion, we use angular displacement (\theta), angular velocity (\omega), and angular acceleration (\alpha), while in translational motion we use linear displacement (s), linear velocity (v), and linear acceleration (a). This is because rotational motion involves circular motion, which has its own unique set of equations.

In conclusion, it is crucial to carefully consider the scenario and choose the appropriate equations when dealing with rotational kinematics. It is also important to understand the differences between rotational and translational motion and use the correct variables and equations accordingly.