How Does Friction Affect the Angular Momentum of a Bowling Ball?

[Kaln0s]

Homework Statement

This has 4 parts, I'm confident about 2.

1. The remaining problems involve a bowling ball with a uniformly distibuted mass of 6 kg and a radius of 0.20 m. The ball is initially thrown at a speed of 10 m/s, with a backspin of 6 rad/s (this means it has an angular velocity in the opposite direction of what it would have if it were rolling without slipping). The coefficient of static friction between the ball and bowling alley is 0.15, and the coefficient of kinetic friction is 0.10. What is the moment of inertia of the ball about its center, in kg m^2?

For this.. uniform distribution, MI = 2/5 MR² = 2/5 (6)(.2)² = 0.096 kg m².

2. For the bowling ball in problem 5, what is the magnitude of the frictional force (in N) acting on it as it is initially released?

0.1*6*10 = 6 N

It's the next 2 that I'm confused on..

Homework Equations

t = rFsin0

The Attempt at a Solution

3. What is the magnitude of the torque on the bowling ball, about its center, produced by the frictional force? Your answer should be in Nm.

For this I want to use T = rFSin(theta), but I don't know the angle and am confused.

4. So, as the ball slides down the alley, the torque due to friction slows down its backspin and makes it start spinning the other way until it is rolling without slipping. At the same time, the friction slows down the ball's translational motion as well. What time elapses between the ball's initial release and when it begins rolling without slipping? Your answer should be in seconds.

I have no idea how to do this part. :tongue:

 

[Delphi51]

For #3, the angle between the force of friction (horizontal) and the moment arm which is a radius to the point of contact (vertical) is 90 degrees.

If it was a linear problem you would use F = ma to find the deceleration and then
V = Vi + at to find the time. For rotational motion, use the rotational analogs to these formulas.

 

[nlightner]

Hello,

I would like to clarify a few things in your content before providing a response. Firstly, the moment of inertia of a solid sphere with uniform mass distribution is given by (2/5)MR², not (2/5)MR^3 as mentioned in your attempt at a solution. Additionally, in problem 5, the magnitude of the frictional force should be calculated using the coefficient of kinetic friction, not static friction, as the ball is initially in motion.

Now, for problem 3, the angle theta in the torque equation can be determined using the given backspin angular velocity of 6 rad/s. Since the ball is initially thrown with a backspin, the frictional force will act in the opposite direction of the backspin, making the angle between the force and the displacement of the ball 180 degrees (or pi radians). Thus, the torque can be calculated as T = rFsin(pi) = rF(-1) = -rF, where r is the radius of the ball and F is the frictional force. The magnitude of this torque will be equal to the frictional force (6 N) multiplied by the radius of the ball (0.2 m), giving a final answer of -1.2 Nm.

For problem 4, we can use the equation of motion for rotational motion, theta = theta0 + w0t + (1/2)alpha*t^2, where theta0 is the initial angle, w0 is the initial angular velocity, alpha is the angular acceleration, and t is the time elapsed. We know that theta0 and w0 are both zero, and we can calculate the angular acceleration using the torque equation from problem 3 (alpha = T/I). The moment of inertia of the ball about its center is again (2/5)MR², giving an angular acceleration of alpha = -6 Nm/(0.096 kgm²) = -62.5 rad/s². Plugging these values into the equation of motion, we get -62.5t^2 = 0, giving t = 0 seconds. This means that the ball will start rolling without slipping immediately after it is released.

I hope this helps clarify your confusion and provides a thorough response to the given content. Please let me know if you have any further questions.