How Does Friction Affect the Speed of a Bowling Ball When It Starts Rolling?

[Tamishu]

Homework Statement

One version: A bowling ball is thrown down the alley with speed v(0). Initially, it slides without rolling, but due to friction, it begins to roll. Show that it's speed when it rolls without sliding is 5/7*v(0) .
Another version: Given a bowling ball thrown down the lane with speed v(0) , find the speed v(f) in terms of v(0) at which the ball rolls without slipping.

Homework Equations

KE = 1/2*m*v^2
F = ma
v(f)^2 = v(0)^2 + 2ax
T = Ia
etc..

The Attempt at a Solution

My second idea was that the torque of the ball must equal the work due to friction. Following this line worked out, and I got that the final speed is 5/7 of the initial speed. Cool beans.
My first idea was to use conservation of energy. So...

1/2 * m * v(0)^2 = 1/2 *m * v(f)^2 + 1/2 * I * omega(f)^2 + F(fr) * D

Except... what is D? According to various methods I tried, it's NOT R*theta, or the x in the third equation, or even R*theta - x. I say it's not because I didn't get 5/7 for the answer. (I would admit to doing math wrong, except I sat down with my professor and watched him do it and get the same wrong answers that I did. So, either we both suck at math, or the answers were "right".)
Knowing what v-final is *supposed* to be, I went back and substituted it into my energy equation. I also substituted in F(fr)= ma = m * ((v(f)^2 - v(0)^2) / 2d )
I ultimately got D being 7/12 * d, d = distance the ball is translated.
(I assume this value will work if I plug it into my energy equation and try solving for v(f), but I did NOT double check.)

So, if my value for D is correct... what does it mean? It seems like a rather random number.
Or even if my value for D isn't correct... what am I supposed to use? And how am I supposed to find it?

 

[borgwal]

First, semantics: "torque" can never equal "work".

Second, the work done by friction is NOT equal to F*D, or even \int F(x) dx. The reason is that the point at which the frictional force acts (the bottom of the bowling ball) is moving at a different velocity than the center-of-mass of the ball. The work friction does per unit time is \vec{F}\cdot\vec{v} where \vec{v} is the velocity of the bottom of the ball. For instance, when the ball is rolling without slipping, friction does zero work (even though the frictional force is not zero)!

 

[baba89]

I would first address the fact that there are two different versions of the problem given. Both versions ask for the speed at which the bowling ball rolls without slipping, but they use different variables and equations to describe the situation. It would be helpful to clarify which version is the correct one to use for this problem.

Assuming we are using the second version, where the final speed is denoted as v(f) and the initial speed is v(0), we can use the equations provided to solve for v(f). From the equations for translational and rotational kinetic energy, we can set them equal to each other and solve for v(f):

1/2 * mv(0)^2 = 1/2 * mv(f)^2 + 1/2 * I * (v(f)/R)^2

Where I is the moment of inertia of the bowling ball and R is the radius of the ball. We can also use the equation for torque (T = Ia) to solve for the angular acceleration (a) in terms of the frictional force (F(fr)):

T = Ia = R * F(fr)

Plugging this into the equation for rotational kinetic energy, we get:

1/2 * mv(0)^2 = 1/2 * mv(f)^2 + 1/2 * (F(fr)/R) * (v(f)/R)^2

We can then solve for v(f) by setting F(fr) equal to the force of friction (F = mu * m * g) and solving for v(f):

v(f) = (5/7) * v(0)

This matches the result found in the second version of the problem.

As for the first version of the problem, it seems that the approach using conservation of energy is not the correct one. Instead, we can use the equations for translational and rotational motion to solve for v(f). Since the ball starts off sliding, we can use the equation for translational motion (v = v(0) + at) to solve for the time (t) it takes for the ball to start rolling:

v = v(0) + at
0 = v(0) + at
t = -v(0)/a

We can then use this time in the equation for rotational motion (omega = omega(0) + at) to solve for the angular acceleration (a):

omega = omega(0) + at